This page answers questions about gravity. The questions are:
Questions about black holes, which are surrounded by extremely strong gravity, are answered on the Black Holes Page.
There turn out to be 4 fundamental forces in the Universe, from which all other forces are derived. These fundamental forces are the strong nuclear force, the weak nuclear force, the electromagnetic force, and the force of gravity.
If the four fundamental forces did not exist then we would not exist, either, so they are rather important to us. However, we don't really know yet where our fundamental forces come from or why they are exactly the way they are, so perhaps if our fundamental forces didn't exist then there might exist other fundamental forces, and then the Universe would probably look very different from what it looks like today.
The strong nuclear force keeps quarks together in hadrons (such as protons and neutrons), so without it there could be no atoms, and therefore no carbon and oxygen and hydrogen and nitrogen and all of the other elements that we and the world around us are made of.
I don't know a simple equation of force as a function of distance, comparable to Newton's Law of Gravity. http://en.wikipedia.org/wiki/Strong_nuclear_force says that the strong nuclear force stays constant beyond a certain small distance (about as big as a hadron, 10^{−15} m), and is then about 10,000 N (roughly the weight on Earth of 1000 kg or 2000 lb). That that force does not decrease with distance means that quarks cannot get very far from each other. That explains why nobody has ever seen a loose quark ― they are always bundled into hadrons.
The weak nuclear force acts upon certain leptons and quarks. (Some examples of leptons are electrons and neutrinos.) Such exchanges can make a quark's type change, so that for example a neutron can change into a proton. I don't know an equation to describe this force. This force acts on distances up to about 10^{−18} m, roughly 1/1000th of the size of a proton. This short range is related to the fact that the virtual particles that spread the force are relatively heavy.
The electromagnetic force ties electrons to the cores of atoms and is the basis for all of chemistry. Without the electromagnetic force there would be no molecules (sets of atoms tied together in a particular order) and no chemistry, and therefore no living things (which are all made of molecules that are themselves made up of carbon atoms and oxygen atoms and hydrogen atoms and other types of atoms).
The force of gravity is the only fundamental force that in practice can be felt at great distances. It keeps planets together and keeps stars together and makes stars shine, and prevents us from floating away into space.
Gravity is an attracting force that exists between all things with mass or energy. Gravity is considered to be one of the four fundamental forces of nature, next to the electromagnetic force, the strong nuclear force, and the weak nuclear force. The latter two forces only play a role in the nuclei of atoms, where they are important for nuclear processes and radioactivity.
The electromagnetic force makes negatively electrically charged particles and positively electrically charged particles attract each other, and makes similarly charged particles repel each other. The electomagnetic force binds negatively charged electrons to positively charged atomic nuclei and plays a major part in any chemical reaction, and also (of course) in electricity and in magnetic effects.
If there were no strong nuclear force, then there probably would not be any elements other than hydrogen, and therefore no stars or planets or people, because without the strong nuclear force there could not be two or more protons close together in the nucleus of an atom, because those all carry positive electrical charge and hence have a very strong electric repulsion to one another. The strong nuclear force must be very much stronger than the electromagnetic force, at the small scale of the atomic nucleus. The nuclear forces diminish much faster with increasing distance than the electromagnetic forces do, so outside atomic nuclei the electromagnetic force wins from the atomic forces.
The force of gravity diminishes with distance equally fast as the electromagnetic force: both diminish with the square of the distance. However, at small scales gravity is much less strong than electromagnetic forces. You can lift a small iron object against gravity from the ground with even a small magnet (with magnetic forces): that small magnet attracts the iron object more strongly than the whole planet does that the object was lying on! That gravity is yet the most important force on astronomical scales is because gravity, unlike electromagnetism, knows no negative "charge", and because the universe as a whole is electrically neutral. Gravity always attracts, but electromagnetic forces can attract as well as repel. With about as much positive as negative electrical charge at large scales, the electromagnetic forces of attraction and repulsion even out on large scales, and so play no large role there.
The history of ideas about and knowledge of gravity is interesting, but unfortunately too long to tell here. As far as we know, Isaac Newton from England was the first person to begin to understand gravity (in a scientific way), and that was in the 17th century. His laws of gravity form the foundation of the study of motion in the universe even today.
At the beginning of the 20th century, Albert Einstein invented the General Theory of Relativity, from which the laws of gravity of Newton follow through simplification. Only in situations where matter travels at speeds close to that of light do the answers of the theories of Einstein and Newton start to differ substantially. In more ordinary situations, those differences are very small, but yet sometimes measurable. In all tests that have been done so far it has been found that Einstein's theories are correct (to within the accuracy of the test), and that Newton's theories are therefore too simple.
The best description we currently have of how gravity works is a formula from the General Theory of Relativity of Albert Einstein. That formula is small but needs a lot of explanation with difficult mathematics (for example, tensor calculus), so it is only suitable for people who are familiar with that difficult mathematics.
Fortunately, we can usually make do with Isaac Newton's Law of Gravity, which uses only simple mathematics. Big differences between the results of Newton's formula and of Einstein's formula show up only if the gravity is enormously stronger than on Earth, or if you measure very, very accurately.
The Law of Gravity of Isaac Newton is:
\begin{equation} F = \frac{G M m}{r^2} \end{equation}
In this formula, \(F\) is the force of gravity between a mass \(M\) and another mass \(m\) which are each concentrated in a point. \(r\) is the distance between the two point masses, and \(G\) is the universal gravitational constant. If you measure \(M\) and \(m\) in kilograms, \(r\) in meters, en \(F\) in Newtons (a unit of force, symbol N), then \(G\) has the value 6.672 × 10^{−11} m³/(kg s²). For example, the force of gravity between two point masses of 10 kg each at a distance of 1 m is equal to 6.672 × 10^{−11}×10×10/(1×1) = 6.672 × 10^{−9} N.
Newton's Law of Gravity holds also for some masses that are not concentrated into a single point. For things that are point symmetric (so that you can rotate them arbitrarily and they still look the same) you can just take the distance between the centers of the spheres, and then the Law works as well as for point masses. For arbitrary things, the Law isn't totally accurate but still reasonably accurate if you take the distance between the centers of mass for \(r\), and especially if the distance between the things is much greater than the size of the things. For very accurate answers, you have to take the exact shape and distribution of mass of the things into account.
The force of gravity acting on a person with mass \(m\) at sea level on Earth (with \(M\) = 5.976 × 10^{24} kg and \(r\) = 6,378,000 m) is equal to \(F = 9.80×m\) newton. The 9.80 is the acceleration of gravity at the surface of the Earth. This is usually written as \(g\) in formulas. If you divide a force in newtons by 10, then you find how heavy that force would feel if it were pressing down on you. For example, a force of 10 N feels like a weight of 1 kg.
The barycenter or center of gravity of a set of objects is a weighted mean of the position of those objects, where the weight of each position in the mean is equal to the mass of the object. If \(x_i\) is the x coordinate of object \(i\), and \(m_i\) the mass of object \(i\), then the x coordinate \(x_\text{c}\) of the center of gravity of the set of objects is equal to
\begin{equation} x_\text{c} = \frac{\sum_i x_i m_i}{\sum_i m_i} \end{equation}
with similar formulas for the other coordinates.
The "objects" of which you determine their common center of gravity can themselves be sets of (smaller) objects. In that case you should take for the position of such a subset the center of gravity of that subset, and for the mass of the subset the total mass of the members of the subset.
Every set has a center of gravity, by definition. A center of gravity is not itself an object, but is an idea that is useful to understand certain aspects of the motion of celestial objects (or other particles).
The neat thing about a center of gravity is that the location and motion of such a center of gravity cannot be changed by forces between the members of the set with which the center of gravity is associated. This is because of the Third Law of Newton, which says that if A exerts a force upon B, then B exerts a force of equal magnitude but opposite direction upon A, so the effects on the center of gravity cancel. For example, if some planet breaks apart due to some enormous internal explosion (which is very unlikely to happen), then the center of gravity of the collection of bits and pieces after the explosion is equal to the center of gravity of before the explosion (i.e., you cannot tell from the motion of the center of gravity that an explosion has happened).
There is no obligation for all members of a set to orbit around their common center of gravity, because even sets that don't orbit at all have a center of gravity. A center of gravity does not itself exert any forces. The center of gravity does not exist independently from the set of objects with which it is associated. It is just an idea, but a useful one.
It is gravity that keeps satellites and space stations in an orbit around the Earth, at least if that orbit does not intersect the Earth, because then they would crash, and if that orbit isn't too low above the Earth, because then they'd be slowed down by friction with the outermost layers of the atmosphere of the Earth, and then they'd fall down after all. Gravity also keeps the Moon in its orbit around the Earth, and the Earth and the other planets and comets and asteroids in an orbit around the Sun, and the Sun in an orbit around the center of the Milky Way Galaxy.
Johan Kepler (1571 - 1630) used accurate observations done by Tycho Brahe to deduce three laws that govern the motion of planets around the Sun:
Nowadays we can make these laws a little more general, and then they hold not just for planets but for all celestial bodies without propulsion of their own in undisturbed orbits around a much heavier celestial body: Orbits can not only be ellipses, but any kind of conical section, so they can also be circles, parabolas, hyperbolas, and straight lines.
The constant of proportionality in the Harmonic Law is related to the mass of the star that the planets orbit around. We can write the Third Law of Kepler like this:
\begin{equation} M P^2 = a^3 \end{equation}
where \(M\) is the mass of the star compared to that of the Sun (so the Sun has \(M = 1\)), \(P\) is the orbital period measured in years, and \(a\) is the length of the semimajor axis of the orbit (which is roughly equal to the average distance from the star), compared to the orbit of the Earth (i.e., measured in Astronomical Units).
You can use the Third (harmonic) Law of Kepler to quickly deduce the size of a planet's orbit from its orbital period. Suppose, for example, that a new asteroid is discovered with an orbital period of 5 years. How large is its orbit? According to Kepler's Third Law, we then have \(1 × 5^2 = a^3\) or \(a^3 = 25\) or \(a = 2.92\) AU.
And if around a star with three times the mass of the Earth a planet is discovered at the same distance as that of the Earth from the Sun (i.e., 1 AU), then for that planet we have \(3 × P^2 = 1^3\) or \(P^2 = 1/3\) or \(P = 0.577\) year.
From the Laws of Kepler and from more accurate observations you can deduce a method to predict the positions of planets. The Sky Positions Page describes such a method.
This law means that your speed in a circular orbit around the Sun is lower if your orbit is further from the Sun. This may sound strange: you have to accelerate to go higher, but in a higher orbit you go slower.
The wrong assumption here is that if you accelerate you'll end up in a higher circular orbit. If you accelerate briefly from a circular orbit (at fixed distance) around the Sun, then you end up in a different orbit, but that orbit must of course also pass through your previous position in the circular orbit at the old distance, so in that new orbit you're not always at the same distance from the Sun, so that orbit cannot be a circle. If you briefly accelerate in the direction in which you were moving already in your circular orbit, then the point at which you start accelerating becomes the perihelion of your new orbit.
You then end up in an elliptical orbit or a parabolic orbit or a hyperbolic orbit (depending on how strongly you accelerate), and in that orbit you'll speed depends on your distance from the Sun. Near the perihelion of an elliptical orbit, your speed will be higher than the speed in a circular orbit at the same distance, and near the aphelion of an elliptical orbit, your speed will be lower than the speed in a circular orbit at the same distance. If \(v_q\) is the speed at perihelion, \(v_Q\) the speed at aphelion, and \(v_k\) the speed in a circular orbit midway between the perihelion and aphelion of the elliptical orbit, then
\begin{equation} v_q × v_Q = v_k × v_k \end{equation}
One could say that the typical speed in your new orbit is equal to the speed in a circular orbit at the typical distance of your new orbit (but then you'll have to explain what you mean by "typical", as I did above).
That your speed decreases when you get further away from the Sun is because it takes an effort to move away from the Sun against the gravity of the Sun. This working-against-gravity is also one of the reasons why a ball slows down the higher it gets on the side of a hill, and speeds up when it rolls down again.
A planet that is four times further from the Sun has an orbital speed that is half as great, but the gravity is sixteen times weaker there. Why is the orbital speed still relatively high there?
Gravity makes the planet fall towards the Sun, but the orbital speed makes the planet move sideways away from the Sun, and in a circular orbit those two effects and in exact balance, such that the distance to the Sun remains the same. How great must the orbital speed be when the distance to the Sun gets four times (or some other factor) as large?
For the answer to this question we look at a very small piece of the circular orbit.
The deviation that you get in a circular orbit relative to the straight line that touches the circle (nearby the meeting point of the straight line and the circle) is inversely proportional to the radius of the circle. If the circle gets four times as large, then it deviates (at the same small distance from the meeting point) only one fourth as much from the straight line.
If the orbital speed is only half as much, then the planet needs twice as much time to move across the same small distance from the meeting point. Then gravity has twice as much time (for the same travelled distance) to pull the planet away from the straight line. Height differences due to gravity are proportional to the gravity itself and to the square of the time (according to the high school formula \(h = \frac{1}{2}gt^2\)), so at the same distance from the Sun, in twice as much time, gravity will pull the planet four times as far from the straight line, but at four times the distance from the Sun gravity is sixteen times weaker, so with half the orbital speed the gravity will be able to pull the planet away from the straight line by only one fourth as much (times four because of the halved orbital speed, but divided by sixteen because of the weaker gravity), which fits exactly for a circle that is four times bigger, as we saw before.
That orbital speed is not inversely proportional to distance, but to the square root of the distance, is related to the square of time in the formula for height differences due to gravity.
In a system of two celestial bodies that orbit around their common center of gravity in circular orbits, there are five points in which the gravity of the two bodies and the centrifugal force in the circular orbit of the point are exactly balanced. Those points are the Lagrange points, named after the mathematician who first calculated them.
If a thing with negligible mass is exactly in such a Lagrange point, then it can remain there for a long time without needing an engine.
The five Lagrange points are traditionally indicated as L_{1} through L_{5}. The first three Lagrange points are on the line through the center of the two bodies: L_{1} lies between the two bodies, L_{2} lies on the far side of the ligher body, and L_{3} on the far side of the heavier body. L_{4} and L_{5} lie on the points that form equilateral triangles with the center of the two bodies: they are equally far from both bodies. L_{4} precedes the lighter body, and L_{5} follows the lighter body.
What happens if you move a little bit away from a Lagrange point? If the point is stable, then you'll drift around the point but stay in its neighborhood, rather like a marble that rolls around a bowl without friction. If the point is unstable, then you'll eventually drift ever farther away from the point, like a ball that rolls off the top of a hill. Only the last two Lagrange points can be stable, and that only if the mass ratio of the two celestial bodies is 25 or more.
There are asteroids (the so-called Trojans and Greeks) in the L_{4} and L_{5} points of the Sun-Jupiter system. The L_{1} and L_{2} of the Sun-Earth system are sometimes used to park satellites in that observe the Sun continuously (in L_{1}) or study the Earth's magnetosphere (in L_{2}).
Calculate Lagrange points yourself.
The gravity of the Earth continues forever, but gets weaker as you go higher above the ground. The rule is as follows: every time that you get twice as far away from the center of the Earth, the gravity is only a quarter as strong. The radius of the Earth is about 6400 km so if you are 6400 km above the ground (and so twice 6400 km from the center of the Earth), then the gravity is only a quarter as strong as it was at the surface. The gravity of other things also continues forever and declines with the square of the distance.
Mass appears in two fundamental equations in physics. Newton's second Law of Motion states that "the change of motion is proportional to the motive force impressed", which boils down to the equation
\begin{equation} F = m_\text{i} a \label{eq:second} \end{equation}
where \(F\) is the applied force, \(a\) the imparted acceleration, and \(m_\text{i}\) the proportionality factor which is fixed for a specific body and which is called the inertial mass.
Newton's Universal Law of Gravity expresses the force of gravity between two bodies in terms of characteristics of the two bodies and the distance between their centers of gravity:
\begin{equation} F_\text{g} = \frac{G M_\text{g} m_\text{g}}{r^2} \label{eq:gravity} \end{equation}
where \(F_\text{g}\) is the force of gravity, \(M_\text{g}\) is the gravitational mass of one of the bodies, \(m_\text{g}\) is the gravitational mass of the other body, and \(r\) is the distance between their barycenters.
If you equate the \(F_\text{g}\) from equation \eqref{eq:gravity} with the \(F\) of equation \eqref{eq:second}, then you can calculate the acceleration that the force of gravity produces. It is
\begin{equation} g = a_\text{g} = \frac{F_\text{g}}{m_\text{i}} = \frac{G M_\text{g}}{r^2} \frac{m_\text{g}}{m_\text{i}} \end{equation}
The equivalence principle states that the gravitational mass of a body is identical to the inertial mass of the body, \(m_\text{g} ≡ m_\text{i}\), so then
\begin{equation} g = \frac{G M_\text{g}}{r^2} \label{eq:gravacc} \end{equation}
which is independent of the mass \(m_\text{i} = m_\text{g} = m\) of the body being accelerated. Because the equivalence principle holds, the acceleration due to gravity that a body experiences is independent of the mass of the body. For example, the Moon experiences a certain acceleration because of the gravity of the Earth. If you replace the Moon by a car, then that car will experience the exact same acceleration at that same location.
However, the equivalence principle only compares accelerations experienced by different bodies in the same situation at the same location. For example, it does not say anything about the acceleration of the other body. If you replace the Moon by a car, then the acceleration of the car is the same as the acceleration of the Moon was, but the acceleration of the Earth gets very much smaller, because it is now attracted by a much less massive body.
We compare two systems: In system A, the Earth and the Moon are at their average distance but are held still. System B is as system A but has a second Earth instead of the Moon. Now release the bodies in both systems to fall towards each other under the influence of their mutual gravity.
The two Earths in system B will collide sooner than the Earth and the Moon in system A will (assuming that all of them fall freely from a starting position with zero relative speed). The Moon from system A starts out at the same acceleration as the Earths from system B (because of the equivalence principle), but the Earth in system A starts out at a much lower acceleration because the mass of the Moon that attracts it is so much lower (because equation \eqref{eq:gravacc} still contains the mass of the other body). The sum of the accelerations of the two bodies in system B is therefore greater than the sum of the accelerations in system A, so the bodies in system B get closer more quickly than those in system A.
Gravity goes with mass. Everything that has mass also has gravity. The gravity is proportional to the mass, so if the mass increases by a certain factor then the gravity also increases by the same factor.
So, gravity does not come just from planets. Gravity belongs with everything that has mass, so it also belongs with you. However, the gravity that you generate is very much weaker than the gravity that a planet generates, and is also very much weaker than the electromagnetic forces that determine the characteristics of all things around us (except for radioactivity), so usually you do not notice that things on Earth also generate gravity.
Albert Einstein has taught us that space and time are interwoven, so that how long something is or how long something lasts is not the same for everyone even if everyone has perfect measuring devices. Those kinds of effects are only important for things that go almost as fast as light. For example, some very fast cosmic rays can get down to fairly low in the atmosphere because for them time goes slower, because of their very great speed (or, according to them, that the atmosphere is thinner) and it takes longer (according to their clock) for them to fall apart than it would for similar but slower cosmic rays.
Space-time (the combination of space and time) is curved by nearby mass, just like a heavy ball makes a dent if you put it on a stretched rubber sheet. We notice curvature of space-time as gravity. If you roll a marble across the rubber, then the marble will deviate from a straight course because of the dent, just as if the marble is attracted by the ball that makes the dent, but really the marble deviates only because the narrow strip of rubber that it rolls along is not level, whether that is caused by the nearby ball or by something else. In the same way, it appears as if the Earth's gravity pulls at the Moon and so keeps it in an orbit around the Earth, but really the Moon is kept from following a straight path because space-time on her path is not flat, which is due to the presence of the Earth.
So, nothing special happens in a planet for it to generate gravity. You can notice the gravity of a planet only because it has such a very large mass.
In explanations of the General Theory of Relativity, space-time is often compared to a stretched sheet of rubber, or something similar. If you place a heavy ball on the sheet of rubber then it makes a depression in the sheet. If you then place a marble on the rubber, too, then that marble can roll into the depression toward the ball. Why does this help to explain the General Theory of Relativity? Because a sheet of rubber is two-dimensional but space-time is four-dimensional. Moverover, isn't it gravity that makes the marble go toward the ball after all?
It is impossible to explain all characteristics of gravity and space-time using two-dimensional examples, because space-time is four-dimensional. The two-dimensional examples can clarify at most some characteristics, and other characteristics of those examples may not look like gravity and space-time at all.
The purpose of the example with the sheet of rubber is to show that the motion of an object can depend on the circumstances very close to that object, in such a way that it looks like there is a force acting between that object and other objects that are further away, when there isn't such a force at all.
A heavy ball on the sheet of rubber will make a depression, and a small marble placed elsewhere on the rubber can then roll down the depression toward the ball. If you look only at the motion of the marble and don't know anything about why the marble moves, then you might think that the marble is attracted by the ball, like the force of gravity, because the motion of the marble (ever faster as it approaches the heavy ball) looks like what you'd expect if the marble were attracted by the ball through gravity.
But there is no noticeable force of gravity between the marble and the ball. The motion of the marble is not because it is attracted by the heavy ball, but because the piece of rubber that the marble moves across is not horizontal. It does not matter why the rubber isn't horizontal. If we remove the heavy ball, and give only the small part of the sheet of rubber where the marble is at a particular time just the right amount of slope, then the marble will move as if the heavy ball were still there.
The marble moves because of the properties of "space" (the sheet of rubber) at the location where the marble is, without noticeable forces working directly between the marble and other objects in that "space".
The General Theory of Relativity says that gravity works in a similar fashion: gravity does not act directly between two objects, but instead the mass of the objects bends space-time in a large area around each object, and the objects react to how bent space-time is at their location, regardless of the cause of that bending.
Why do we have to make things complicated with space-time bending, if things work just as well with gravity acting directly between objects? That is because those two different models don't give exactly the same results in all cases. Where the two models give different results, then it turns out (so far) that the results from the General Theory of Relativity are better than those from the classical theory of gravity by Newton. If that weren't the case, then we wouldn't need the idea of bending of space-time, and didn't need the example with the sheet of rubber, either.
If there had been no gravity at all, then there would be no stars or planets or galaxies or black holes or people, but only clouds of gas and dust, fairly evenly spread across the Universe and much colder than it ever gets at the South Pole.
If you could cancel the gravity between the Sun and the Earth, then the Earth would fly away in a straight line in the direction in which it happened to be going when the gravity quit, and at the speed relative to the Sun that it happened to have when the gravity stopped, because the first Law of Newton says that something upon which no forces act moves at a constant speed along a straight line. The orbital speed of the Earth is about 30 km/s or about 1 AU per 58 days, so the Earth would then move away from the Sun at about that speed.
However, as far as we know it is impossible to turn off gravity, because it is a characteristic of mass. If there is mass somewhere, then there is gravity, too.
The only circumstance in which you don't feel any weight is when you are in free fall. In space, you can keep this up for as long as you want, but near the surface of the Earth you can do this only for a short while, for example in an airplane that flies up and down along the proper parabolic trajectory. Search the internet for "vomit comet" for examples. To get into and out of that trajectory again you need to make turns that make you feel more weight than you have on the ground. Astronauts train for weightlessness inside such an airplane.
Astronauts train for space walks by going under water. If you wear a proper diving suit, then you are not weightless, but you do float.
You can only feel gravity if something is working against it from the outside. If you stand on the ground, then the gravity of the Earth pulls you down, but the ground pushes back up equally strongly against your feet, so you stay where you are but it feels as if your head and the rest of your body is pushed down to your feet, and that feeling is your weight.
An astronaut in space is weightless, but still has the same amount of mass (the same number of kilograms or pounds) as on Earth, so mass and weight are not the same. Your mass is how much material is in your whole body. Your weight is with how much force the ground (or whatever else you stand on) presses you up. If you stand on Earth or the Moon or another celestial body, then your weight is proportional to your mass but also to the strength of gravity (the gravitational acceleration). If gravity is weaker, then you're pulled down less strongly and the ground pushes you back up less strongly (so you can remain standing up), so then you weigh less. You weigh less on the Moon than on Earth because gravity is weaker on the Moon than it is on Earth.
There are two fundamentally different kinds of weighing instruments. There is a kind that compares weights or masses, and there is a kind that measures forces. There are no common names to distinguish these two kinds. I'll call something of the first kind a mass balance and something of the second kind a spring balance.
With a mass balance, you have to add or remove or shift counterweights until the thing of which you want to measure the mass is exactly balanced by the counterweights of which you already know the masses. How far you have to shift the counterweights or how much counterweight you have to use tells you what the unknown mass is. This measurement is independent of gravity (except that it doesn't work if there is no gravity at all), so it measures the mass and not the weight.
If you stand on a spring balance then you squeeze a spring inside the balance because of the force of gravity that pulls you down. If you have more mass, then you squeeze the spring further and then the balance shows a greater value. If you weigh yourself at a place with less strong gravity, then you squeeze the spring less and then the balance shows a lesser value. Measurements with a spring balance depend on the strength of the local gravity, so they measure weight and not mass.
Most spring balances measure in a unit of mass (such as kilograms or pounds), though they should properly measure in a unit of force (such as the newton). You should interpret kilograms or pounds indicated by a spring balance as "kilograms or pounds in the standard gravity of the Earth".
What happens if you take a mass balance and a spring balance to the Moon and weigh yourself there? The mass balance will indicate the same mass as on Earth, because the counterweights have decreased in weight by the same proportion as your weight has, so they still keep you in balance. The spring balance will indicate about six times less on the Moon than on Earth, because the gravity is about six times weaker on the Moon than on Earth.
Even on Earth the Earth's gravity isn't equally strong everywhere, so your weight on Earth depends a bit on where you are. At the poles, gravity is about half a percent (one part in 200) stronger than at the equator, so you weigh about half a percent more at the poles than at the equator. For someone with 50 kg of mass this makes a difference corresponding to about a quarter of a kilogram.
The weight of a thing depends on its position on Earth, mostly because the Earth rotates around its axis. If the Earth didn't rotate around its axis, then the weight would vary much less from one place to another. Then the weight would vary only because of very small differences in gravity related to mountains or heavy rock layers nearby (or instead the lack of them).
When you're in a merry-go-round, then you feel a centrifugal force pushing you away from the center. This force gets stronger when the merry-go-round goes around faster or when you're further from its center (closer to the outside edge). In the same way, a centrifugal force acts on anyone standing on the rotating Earth. Just like in a merry-go-round, the force gets stronger when you are further from the axis of rotation, so it is strongest at the equator and absent on the poles. The difference is modest: only about half a percent.
The next table shows how strong the force of gravity is at the surface of various celestial bodies compared to that on Earth, and how much you'd weigh there if you weighed 50 pounds or kilograms on Earth, and how high you could jump there if you can jump 1 meter = 100 centimeters high there. All planets, the Sun, and all moons on which you'd weigh more than on Pluto are included in the table. The acceleration of gravity at the surface of the Earth is on average 9.8 m/s^{2}.
Table 1: Gravity on the Sun, Planets, and Moons
Name | g | weight | jump |
---|---|---|---|
Sun | 27.964 | 1398 | 3.5 cm = 1.4 in |
Mercury | 0.376 | 19 | 265 cm = 8 ft 9 in |
Venus | 0.905 | 45 | 110 cm = 3 ft 8 in |
Earth | 1.000 | 50 | 100 cm = 3 ft 3 in |
Mars | 0.379 | 19 | 264 cm = 8 ft 8 in |
Jupiter | 2.530 | 127 | 40 cm = 1 ft 4 in |
Saturn | 1.064 | 53 | 94 cm = 3 ft 1 in |
Uranus | 0.905 | 45 | 110 cm = 3 ft 8 in |
Neptune | 1.137 | 57 | 88 cm = 2 ft 11 in |
Pluto | 0.067 | 3 | 14.9 m = 48 ft |
Io | 0.185 | 9 | 5.4 m = 18 ft |
Moon | 0.164 | 8 | 6.1 m = 20 ft |
Ganymede | 0.146 | 7 | 6.8 m = 22 ft |
Titan | 0.139 | 7 | 7.2 m = 24 ft |
Europa | 0.133 | 7 | 7.5 m = 25 ft |
Callisto | 0.128 | 6 | 7.8 m = 26 ft |
Triton | 0.080 | 4 | 12.5 m = 41 ft |
Something is weightless when the gravity that acts on it is not hindered, and if that thing has no propulsion of its own. Something that floats free in space is weightless. That could be an astronaut, an artificial satellite, or a space ship, but also the Moon, or a planet such as the Earth, or the Sun.
If gravity has free reign, then you are weightless, but gravity then does have noticeable influence on your path through space: The gravity then changes your direction or your speed or both. The direction of an astronaut in an orbit around the Earth changes all the time, because the astronaut is in a curve all the time.
People don't fall from the Earth because of the force of gravity between those people and the Earth. All things have mass and attract each other with a force of gravity that gets greater when the masses get greater and that gets smaller when the centers of the masses get further away from each other. The force of gravity of some things always points at the center of mass of that thing. People don't fall from the bottom of the Earth because the gravity of the Earth pulls towards the center of the Earth, no matter where you are. If you live at the other side of the world, then you stand with your feet on the ground just like here. People at the other side of the world think that we stand at the bottom of the Earth, and they may wonder why we don't fall off.
If you float in space around the Earth, then the gravity from the Earth still pulls at you, but then you don't feel it because there isn't anything like the ground to work against the gravity. Instead, the gravity changes your orbit through space. Without the gravity, you'd float in a straight line at a constant speed. Because of the force of gravity of the Earth, you go down faster and faster (if you fall), or you go in a curved orbit around the Earth.
If you jump up from the floor, then for a few moments the ground cannot push you up against the force of gravity, so then for a few moments you don't feel gravity, and seem weightless.
No matter which part of the surface of Earth you are on, the Earth's gravity always pulls towards the center of the Earth. That's why people at the "bottom" of the Earth can stand up straight with their feet on the ground (or on the deck of a ship) just as well as people at the "top" of the Earth.
There is a force of gravity between any two things that have mass. You have mass (measured in kilograms or pounds), so there is a force of gravity between you and any car, chair, planet, other person, or any other thing. Of things outside the Earth, such as planets and the Sun, you don't feel the gravity because there isn't anything to work against that gravity. Of things on Earth, such as cars or other people, you don't notice the gravity because it is so much smaller than the gravity of the Earth, because the mass of the Earth is very much greater.
The force of gravity between you and a friend (or another person) at 2 meters distance is about as strong as the weight of a grain of sand with a size of about 0.03 mm. And every time that the distance gets twice as large, the force of gravity gets four times smaller. So, you don't notice the force of gravity of other people or things. If you drop something from 100 m high, then the force of gravity of the Earth lets that take only about 4.5 seconds. If you and a friend float in empty space at 100 m from each other, with the very same speed in the beginning, then it will take 125 days before your own forces of gravity make the two of you bump into each other.
Calculations show that in August of 2003 Mars came closer to the Earth than it had done for about 60,000 years (though not so much closer than during recent perigees that we needed to worry about it). To do these kinds of calculations you need to include not just the gravity of the Sun and the positions of the Earth and Mars, but also the gravity (and hence the positions) of the other big planets, especially of Jupiter, which can perturb the orbits of the other planets. The theory of planetary perturbations was developed only in the late 1700s, so people couldn't have done the calculation before then (with sufficient accuracy).
However, to include all perturbations down to the level that is important for the calculation of the closest approaches of Mars requires an enormous amount of calculations, especially if you want to do it for all of the approximately 30,000 times that Mars has come close to Earth over the 60,000 years mentioned. I cannot imagine that anyone would want to do all of those calculations by hand, so such calculations were practically beyond our reach before the advent of computers around the middle of the 20th century, and even then you'd probably have had to wait a couple of decades before there was sufficient computer power available that there was some to spare for this problem, which is after all not very important in the grand scheme of things.
Comets go faster when they are closer to the Sun because of the same reason that a ball or other thing goes faster when it is lower (closer to the Earth): because of gravity. The gravity of the Sun pulls at the comet and lets it go faster. The closer the comet is to the Sun, the faster it goes. This holds also for planets or spacecraft or other things: the closer they get to the Sun, the faster they go. And it holds also for other attractors than the Sun: the closer a spacecraft or satellite is to the Earth, the faster it goes (if the gravity of the Earth is stronger there than the gravity of the Sun).
If the comet goes straight at the Sun, then it will vaporize because of the great heat before it can crash into the Sun.
If the comet does not go straight at the Sun but aims a little to the side, then it will fly around the Sun in a big curve. That is because in a curve you always feel a centrifugal force that pushes you away from the center of the curve. (The center of the curve is the middle of the circle of which the curve is a small part.) You can also feel such centrifugal force if you are in a car that goes through a curve: then you are pushed into the side of the car that is furthest from the center of the curve. If the car goes faster or if the curve is sharper (so the circle is smaller), then the centrifugal force is stronger.
If the comet gets closer to the Sun, then its speed increases and its curve gets sharper, so then its centrifugal force (pushing away from the Sun) also gets stronger. Eventually, the centrifugal force wins from gravity, and then the comet moves away from the Sun again.
You can also explain the behavior of the comet by looking at its energy. A thing such as a comet that is attracted by gravity has two kinds of energy: energy of speed (kinetic energy), and energy of location (potential energy). The energy of speed is greater if the comet goes faster, and the energy of place is greater if the comet is further away from the Sun (opposite to the direction in which gravity pulls). The gravity tries to pull the comet to the place where the energy of place is least (at the surface of the Sun).
According to the Law of Conservation of Energy, energy cannot just disappear or appear, and you have to be able to tell where the energy goes or where it came from. If the comet gets closer to the Sun, then it has less energy of place, so then it must have more energy of speed, for the total energy to remain the same. If the comet goes away from the Sun again, then it exchanges energy of speed for energy of place, so then it goes slower again.
If no energy is transformed into heat (by friction), and if the comet has no propulsion of its own, then the speed of the comet depends only on its distance from the Sun. For example, if the comet on its way to the Sun goes 40 kilometers per second when it passes the orbit of the Earth, then it will also go 40 kilometers per second when it passes the orbit of the Earth again on its way back out. And for this you don't even need to know where exactly the comet went: that's the fun part of such calculations of energy.
If two different solar systems get as close to one another as their diameters, then there are significant tidal forces from each star on the other star's system, and then it is very likely that some of the members of the systems get ejected from their system and get lost in space, or perhaps get redirected into the center of their solar system, or maybe get caught by the other solar system. The outermost members of each solar system are affected first.
It is thought that the Oort Cloud of comets at the outskirts of the Solar System reaches to about 50,000 AU, which is about 0.8 lightyears, so if another Sun-like star gets within about 2 lightyears of the Sun, then it is expected to perturb the closest part of the Oort Cloud and scatter Oort-Cloud comets into different orbits, including orbits into the heart of the Solar System.
If two solar systems pass through one another (i.e., the two stars pass each other within the orbits of their planets), then it is very likely that most if not all of the planets escape into space. It is very unlikely that the two stars will end up orbiting around each other, because there isn't anything obvious to slow the second star down enough for it to remain in the neighborhood of the first one.
The gravity of the Moon can move huge amounts of water, causing the tides. Yet the it doesn't seem to make any difference to our weight whether the Moon is pulling at us from high overhead or from below the Earth.
The force of gravity between the Earth and something or someone on Earth is about 300,000 times greater than the force between the Moon and that same thing or person (because the Moon is much further away than the Earth is and also far less massive than the Earth is), so changes to your weight because of the gravity of the Moon are at most about one part in 300,000, which is not measurable.
The main reason why the tides make huge amounts of water move is because water is fluid, so it moves very easily, and because the oceans provide enormous distances across which the water can move. Freedom of movement is important: after all, tides don't have measurable effects in lakes.
Not much is necessary to make water move. If a large shallow dinner plate filled with a thin layer of water is tilted over only a very small angle, then all of the water moves to the lowest part and gathers there, until the surface of the water is once more horizontal. If there is enough time for the surface of the water to become horizontal again, then the depth of the water at the lowest end has typically become much greater than it was when the plate was horizontal.
If you tilt a full bucket instead of a dinner plate, then likewise water moves to the lowest part until the surface is once more horizontal. The distance by which the water is "higher" at the low end of the bucket than when the bucket was horizontal (i.e., the increase of the depth of water there) increases with the tilt angle but also with the diameter of the bucket: if you tilt a glass by the same angle, then the depth of the water at the receiving end increases far less than it did in the much wider bucket, and if you tilt a swimming pool instead, then the "tide" is greater than in the bucket.
The gravitational forces from the Moon have the same effect as tilting the surface of the water over only a very small angle, of at most something like 1/300,000th of a degree. Because water in the oceans is free to move around the whole planet (the "bucket" is very large indeed), this very small angle still adds up to tides of a few feet. In a lake it corresponds to "tides" that are too small to measure.
If classical gravity is the only force that matters, then two celestial bodies (such as a star and a planet, or two stars, or a planet and a moon) can form a bound system forever, which means that the two bodies do not get further apart than some fixed distance. Such a system is called a two-body system.
If you add a third body to such a system, then its gravity affects the orbits of the other two bodies, and then the system is no longer necessarily stable. This means that there may be a time in the future when at least one of the bodies is ejected from the system. How long you have to wait for this to happen depends on the details of the orbits and masses of the bodies.
In the Universe it happens often that the motion of a particular celestial body is determined mostly by only one other source of gravity. In such a case the body moves almost as it would if those two bodies were the only once in the Universe.
For example, the motion of the planets, asteroids, and comets in our Solar System is determined mostly by the Sun, and the motion of the moons in the Solar System is determined mostly by the planet or asteroid around which they orbit.
Even in the cases where the motion is determined mostly by one source of gravity, all other sources of gravity do have some influence, but that influence is then much smaller. The simplest case that can be investigated is the three-body problem, but that is still quite complicated, unless you impose restrictions that make it simpler.
Favorite restrictions are to assume that the three bodies move in the same plane (so that not three but only two coordinates of location need to be followed), and that the mass of one of the three bodies can be ignored compared to that of the other two, so that the motion of the two massive bodies does not notice the light body at all. In that case the two massive bodies move as in the two-body problem, which has been solved.
If you also assume that the two massive bodies follow circular orbits, then there are five locations where you can put the light body and have it, too, follow a circular orbit in which it keeps the same relative orientation with respect to the massive bodies. Those five locations are the Lagrange Points. There are other effects that you can find in the three-body problem but not in the two-body problem, such as orbit-orbit resonances and the escape of one of the bodies from the system.
If a body (for example, a planet) orbits around two or more other bodies (for example, stars) that are much closer together than they are to the planet, then the combined gravity of the stars looks like the gravity of a single star with the combined mass at about the place were the original stars were, so then the whole system looks like a two-body system, with one planet and effectively one star. Such a multi-body system can be reasonably stable, which means that the planet can remain in its orbit for a long time.
It is very unlikely (perhaps even impossible) for a planet to have a regular orbit (i.e., one that repeats) if it passes through a region of space where the dominant gravitational influence comes from more than one star (effectively). If there are two dominant stars (effectively), then that dangerous region of space is a surface, roughly spherical in shape and centered on the least massive of the stars. I'll refer to that surface as the tidal boundary.
So, for a planet to have a regular orbit in a system with multiple stars, the planet's orbit must either be sufficiently close to one of the stars (so it stays on one side of the tidal boundary), or else so wide that it comfortably goes around all stars (so that the stars effectively act as one star).
If another star is on course for our Solar System, then the associated tidal boundary gets closer to the Sun when the star does, because the distance of the tidal boundary from the Sun is a fixed fraction of the distance between the stars. If the star gets close enough to the Sun, then the tidal boundary sweeps across the orbits of the planets, so then the orbits of the planets are no longer stable and the planets are likely to be ejected from the Solar System.
The radius \(r_\text{tide}\) of the tidal boundary is roughly equal to
\begin{equation} r_\text{tide} = 0.4 \left( \frac{M_\text{small}}{M_\text{large}} \right)^{1/3} r \label{eq:getijde} \end{equation}
where \(M_\text{small}\) is the mass of the least massive star, \(M_\text{large}\) the mass of the most massive star, and \(r\) is the distance between the two stars. (The same arguments and formula hold if the two bodies are of kinds other than stars.) The factor 0.4 is an estimate of mine. If the two stars have equal masses, then the point midway between the two stars (at \(0.5 r\)) is obviously part of the tidal surface, so the factor must be smaller than 0.5. On the other hand, the factor must be large enough that existing regular orbits in the Solar System are allowed by it.
For a system with the Sun, a planet, and a moon of the planet, we can rearrange the formula to the following
\begin{equation} k_\text{moon} = \frac{r_\text{moon}}{r_\text{planet}} \left( \frac{M_\text{sun}}{M_\text{planet}} \right)^{1/3} \end{equation}
where \(r_\text{moon}\) is the orbital distance of the moon from the planet, \(r_\text{planet}\) the orbital distance of the planet from the Sun, and \(k_\text{moon}\) a number that indicates what fraction the size of the moon's orbit is of the radius of the tidal surface, i.e., the relative influence of tidal forces of the Sun on the moon's orbit around the planet. The orbit of the furthest moon of Jupiter, Sinope, yields \(k_\text{moon}\) = 0.31. I don't know of any other moon in our Solar System that has a greater \(k_\text{moon}\). For the orbit of our Moon, \(k_\text{moon}\) = 0.18. Clearly, the factor from the first equation must be greater than the greatest \(k_\text{moon}\), so it must greater than 0.31. The 0.4 I picked seems a reasonable compromise.
In the natural sciences, a resonant phenomenon is a phenomenon where a certain combination of effects reinforces itself in such a way that the total effect is much greater than you would have guessed based on the small original effects. Usually the reinforcement happens only if the effects occur with one of the natural frequencies of the resonant phenomenon.
The nice sounds that come from a well-tuned musical instrument are the result of resonances. You strike or string or blow the instrument in a way that generates a wide range of frequencies, but we get to hear only the sound with the natural frequencies of the instrument which are resonantly amplified.
In astronomy, resonance is also used as the name for a case where small effects that can usually be ignored now do have a measurable impact because they happen exactly in step with something else. There are two kinds of resonances of which there are many examples, namely orbit-orbit resonances and spin-orbit resonances. They are explained in more detail below.
If an object is caught in a resonance and if one of the important periods of the resonance changes sufficiently slowly because of other causes, then the mutual gravity can maintain the resonance and then the other important periods change as well. A change of periods means that energy has been exchanged, so resonances are important ways to exchange energy.
An orbit-orbit resonance occurs if two celestial objects periodically have the same positions compared to one another, so that their mutual gravity keeps it that way (in a stable resonance), or instead makes them quickly move away from that situation (in an unstable resonance), even though that mutual gravity is far smaller than the gravity between the Sun and any one of the celestial objects. In the case of orbit-orbit resonance between two celestial objects, a certain multiple of the orbital period (year) of the one object is exactly equal to a certain multiple of the orbital period of the other object.
For example, Pluto is caught in a 3:2 orbit-orbit resonance with Neptune: Every two orbits of Pluto take the same amount of time as every three orbits of Neptune. Things in Lagrange points are caught in 1:1 orbit-orbit resonances. For example, the Trojan and Greek groups of asteroids are caught in a 1:1 orbit-orbit resonance with Jupiter, and the Hilda group of asteroids in a 3:2 orbit-orbit resonance with Jupiter. Unstable orbit-orbit resonances (2:1, 3:1, 5:2, and so on) cause the so-called Kirkwood gaps in the distribution of astroids over lengths of the semimajor axis of their orbit.
One can also have orbit-orbit resonances between more than two celestial objects, but those can be more complicated and then the orbital periods need not be exact multiples of one another. For example, the moons Io, Europa, and Ganymede of Jupiter are all caught in a single orbit-orbit resonance.
A spin-orbit resonance occurs when a celestial body periodically has the same orientation after a whole number of orbital periods (years) of the object. In that case, a certain multiple of the rotation period (day) of the object is equal to a certain multiple of the orbital period (year) of the object. I do not know if you can have unstable spin-orbit resonances just like you can have unstable orbit-orbit resonances.
All the large moons in our Solar System are in 1:1 spin-orbit resonances: Those moons always show the same side to the planet, just like our Moon does relative to the Earth. It appears that the planet Mercury is caught in a 3:2 spin-orbit resonance: Three times the rotation period (day) of Mercury is equal to two times the orbital period (year) of Mercury.
Because of the tidal forces from the Moon, most seas and oceans on Earth experience two cycles of tides (two high tides and two low tides) a day. This means that there are two high tides on opposite sides of the Earth, which try to keep themselves aligned with the Moon. There are two tides and not just one because of subtle interaction between the force of gravity between the Moon and both tidal bulges (which is less on the far bulge than on the near bulge) and the centrifugal force from the orbit around the Earth (which is greater on the far bulge than on the near bulge).
The tidal forces act not just on water, but on all material in the Earth, so the Earth itself is squeezed a bit in one direction and stretched in another direction such that it seems that the Earth has two low but wide bulges that try to keep themselves aligned with the Moon, but we don't notice the land bulges because the rocks of the Earth can't move as freely as the water of the oceans can.
Likewise, because of the tidal forces from the Earth, the Moon is not perfectly spherical, but has a low but wide bulge at the near side and another low but wide bulge at the far side.
Now suppose that the Moon rotates (compared to the stars) in just a tad more or less time than it needs to orbit. Then the Moon hasn't quite completed one rotation (compared to the stars) or has instead completed just a tad more than one rotation when it has completed one orbit, so then the Moon will appear to have rotated just a tad as seen from Earth, so then the tidal bulges of the Moon are a tad off-center compared to the line through the center of the Earth and the Moon. The bulges try to adjust themselves, under the influence of the tidal forces, but they always lag behind a bit because it takes time to get going and because there is friction that hinders instant movement. The net result is that the bulges are never quite aligned properly, and this produces a net torque that acts to slow down the rotation if the rotation period is shorter than the orbital period, or to speed up the rotation if the rotation period is longer than the orbital period.
So, in any situation where two celestial bodies orbit around each other, their rotation rates will adjust themselves (slowly) until they reach a situation where the rotation rates are equal to their orbital period (i.e., a 1:1 spin-orbit resonance), except if they get caught in a higher spin-orbit resonance. The latter may happen, but (apparently) only if the orbit deviates sufficiently from a circle. An example of this is Mercury, which is in a 3:2 spin-orbit resonance with the Sun.
The adjustment goes very much faster for smaller objects and for objects that are closer together. In the Earth-Moon system, the Moon moved into the 1:1 spin-orbit resonance a long time ago, but the Earth is still very far from the 1:1 spin-orbit resonance. The Earth's rotation is slowing down, but at such a slow rate that it is very unlikely to reach a 1:1 spin-orbit resonance before the Sun turns into a red giant in about 4 thousand million years and makes life on Earth very unpleasant.
If you drop a block while you're standing on the Moon, then the block will fall to the Moon and not to the Earth, because the Moon has gravity of its own that pulls objects towards the center of the Moon.
If there is no atmosphere, then the speed at which a block falls is not affected by the mass of the block, and then blocks of, for example, 10 kg and 50 kg (or 10 lb and 50 lb) fall equally fast (if they are released in the same way at the same time). If there is an atmosphere, then falling objects are slowed down because of the force of friction from the atmosphere, and then the size, shape, and density of the object does affect the speed of falling. In that case, a block of 50 kg will fall faster than a block of 10 kg with the same size and shape in the same orientation.
If you could drill a tunnel straight through the Earth to the other side and if you jumped into one end of that tunnel, then you would not come shooting out the other end of the tunnel, because
You cannot get higher than before (unless you work at it or have some sort of engine), just like each next bounce of a bouncing ball never gets as high as the previous bounce. This is an application of the Law of Conservation of Energy.
The air in the tunnel would slow you down (especially if the tunnel is narrow) so you'd lose lots of speed and wouldn't have enough left to reach the other side.
The Earth rotates around its axis, so, while you're going down the tunnel, the Earth would rotate the tunnel to another orientation. You would not go along with that rotation, because you're not connected to the Earth while you're falling down the tunnel, which means that one wall of the tunnel would move into your path, so you'd start hitting the wall, which would slow you down, so you'd not have enough speed left to reach the other side.
If the Earth did not rotate, and if there was no air in the tunnel, then you could keep falling from one side to the other side of the Earth and then back again, always reaching just the same height on both sides, kind of like a ball at the end of a swinging pendulum (except that a pendulum cannot keep going forever, because of resistance from the air).
According to my calculations, you'd take 19 minutes to reach the center of the Earth, which you would pass at a maximum speed of 9.9 kilometers per second. After another 19 minutes you'd reach the other side of the tunnel, and after twice more 19 minutes (after 76 minutes in total) you'd be back at your starting point.
If the Earth did not rotate but there was air in the tunnel, then you could not attain a high speed because of resistance of the air, and then you'd keep falling from one side to the other side of the Earth but getting less high up the tunnel every time, until finally you'd end up motionless at the center of the Earth, similar to how an empty swing that you made swing ends up motionless again after moving to and fro over ever smaller distances.
If you jump out of an aeroplane, then after a while you'll reach an equilibrium speed, at which there is an equilibrium between the force of gravity that tries to make you go faster and the drag forces from the air that try to make you go slower. That equilibrium speed (also called "terminal velocity") goes down if the gravity gets weaker and also if the density of the air increases.
If we assume that the air in the tunnel had everywhere the same density as at the surface, and that the tunnel was so wide that you would not get extra drag because of the closeness of the walls, and that you tried your best to go as fast as possible (by aiming your body straight down, with your arms tightly along your body) then you'd accelerate to about 320 kilometers per hour within about a minute, and would maintain about that speed for several hours. After about 8¾ hours you'd reach your maximum speed, of about 340 kilometers per hour, and then your speed would gradually decrease, because the gravity gets weaker the closer you get to the center of the Earth. Only after 27⅔ hours would you finally pass the centr of the Earth, at a speed of only about 4 kilometers per hour. You'd then continue to swing around the center, passing the center about once every 26 minutes, but you'd keep closer to the center after every swing. After the first passage you'd get about 420 meters beyond the center, but the second time that greatest distance would be only about 250 meters, and it would continue to decrease.
If the tunnel were narrow, then you'd get extra drag forces, so then you would not reach that 340 kilometers per hour. In addition, the air in the tunnel would not really have the same density everywhere, but would get denser the lower you went (just like the air at sea level is denser than it is high in the mountains), and that would limit your speed even more.
Of course, one cannot really dig a tunnel all the way through the Earth, because the Earth is fluid below the relatively thin crust, so such a tunnel would fill up again immediately.
In the center of the Earth, there are equal amounts of mass in all directions at any given distance, so gravity pulls equally hard in all directions, so an object at the center of the Earth would feel zero net gravity. Such an object would be under enormous pressure, because it does feel the weight of all mass that lies on top of it.
Outside of the Earth, gravity behaves as if all mass of the Earth were concentrated in the very center point, but inside the Earth gravity does not behave like that.
Gravity comes from mass, not from a particular point. Only in the precise center is the gravity equal to zero.
This holds also in the center of a black hole, if the law of gravity as we know it is still valid there (which we don't know).
If the Earth were hollow inside, and all of its mass were in the surface shell, and the mass distribution was spherically symmetric, then you'd be weightless everywhere in the hollow space inside the Earth.
The green line in http://www.splung.com/kinematics/images/gravitation/variation%20of%20g.png shows the variation of the strength of gravity with distance from the center of the Earth. Gravity is zero in the very center, then rises fairly linearly to about 109% of the surface gravity at about 55% of the distance to the surface, then drops back to about 100% and stays there until the surface. Beyond the surface the gravity decreases with the square of the distance from the center.
The red line in the same graph shows the variation of gravity if the Earth were equally dense everywhere. Then the strength of gravity would increase linearly from 0 in the middle to 100% at the surface. That in the "real" Earth the variation of gravity is different is because the Earth is not equally dense everywhere, but is a lot denser near the middle than near the surface.
How strong gravity would be inside a black hole depends on how the mass inside a black hole is distributed, and up to what point the laws of gravity that we know remain valid inside a black hole.
The edge ("event horizon") of a black hole is not a physical thing, but is an invisible boundary. If you pass beyond that boundary from the outside then you cannot go back anymore, but conditions do not suddenly change when you pass through the boundary (unlike when you pass the boundary of the Earth from the outside ― that hurts).
As long as you don't pass any objects on your way to the center of the black hole, the gravity will only increase ― at least up to the point where our laws of gravity break down.
If the mass inside a black hole is squeezed together in the middle of the black hole, to enormously greater densities than we've ever seen, then gravity near that mass can be enormously stronger than we have ever seen. Perhaps the laws of gravity that we know don't describe gravity well anymore in those circumstances. Because we have never been able to investigate gravity as strong as that, we don't know what adjustments might be needed to our laws of gravity. Maybe then the stength of gravity won't increase anymore when you get closer to the mass.
Nobody knows how the matter inside a black hole is distributed. From measurements of the gravity outside the black hole you cannot tell what the exact distribution is of the matter inside the black hole. One can create many models that all have different distributions of mass but yet give the exact same gravity outside the black hole. One of those models is the model in which all of the mass is concentrated in a single point, which is then called the singularity, because according to the classical law of gravity the gravity must then be infinitely great there.
In the same way, measurements of the gravity outside of the Earth are insufficient to say precisely how the mass inside the Earth is distributed. We can still say something useful about that distribution because we may assume that the matter inside the Earth is of the same types that we've studied in the laboratory, and we can calculate how that kind of material must be distributed to generate the observed gravity field.
We can't do that very well for a black hole, because we don't know what the characteristics are of the matter inside a black hole.
It seems reasonable to me to assume that a black hole is mostly empty, but we just don't know if the matter is really all concentrated in just a single point without dimensions, or if perhaps it is a small ball, or if it is a strange cloud in which "our" laws of nature aren't valid anymore.
The important point is that you cannot tell from the gravity field outside of an object what the exact distribution of matter inside that object is. At 100,000 km from the center, the gravity of Earth looks just like the grvaity of a black hole with the same mass as the Earth. You cannot easily tell from measurements of the gravity field at large distances whether the gravity is caused by an Earth or by a black hole.
Suppose that you set a ladder of hunderds or even thousands of kilometers or miles long straight up on the ground and then step off the top of that ladder. Where would you end up?
That ladder would go along with the rotation of the Earth, so you'd end up in an orbit around the center of the Earth, and would not fall straight down.
If your ladder is shorter than 23,460 km (14,580 mi) then your orbit would intersect the surface of the Earth, which means that you'll crash into the ground, but just not at the bottom of the ladder. If your ladder is longer than 23,460 km then it depends on the length of the ladder and on the latitude where it is placed whether you'll end up in an orbit around (as opposed to through) the Earth (assuming that the ladder is exactly vertical). The minimal height above the surface, in kilometers, is equal to
\begin{equation} h_1 = \frac{29840}{\sqrt{\cos(φ)}} - 6378 \end{equation}
for latitude \(φ\).
For example, at 50 degrees north or south latitude, the minimum height is \(h_1 = (29840/\sqrt{\cos(50°)}) - 6378 = 30840\) km so at that latitude the ladder would have to be at least 30,840 km (19,170 mi) long to prevent us from crashing to the Earth.
If the ladder is taller than 46,580 km, then you might escape from the Earth's gravitational pull altogether, again depending on the height and location of the ladder. The minimum height above the surface, in kilometers, is
\begin{equation} h_2 = \frac{53230}{\cos(φ)^{2/3}} - 6378 \end{equation}
At 50 degrees north latitude, this would be 48,380 km, so if we set up a ladder there that is taller than that, then we would end up in an orbit around the Sun.
These calculations aren't very useful in practice, because we cannot make ladders that tall that are strong enough, so such a ladder would collapse under its own weight. In addition, the rotation of the Earth would cause large lateral forces on the ladder, which it would probably not be able to withstand, either.
If you remove a large chunk of the Earth, then the force of gravity from that chunk disappears from the Earth, too. So, if the shape of the rest of the Earth remains the same, then the effect of removing a chunk from the Earth is the same as the effect of adding "anti-gravity" (that does not attract but instead repulses) to the part of the Earth where the chunk is, to balance out the positive gravity from the chunk. "Anti-gravity" added to the Earth would push the oceans away from the location of that anti-gravity, so removing the chunk would likewise make the oceans move away from the newly created hole. If you were standing on the surface near the edge of the hole, then it would feel as if the surface slopes up towards the edge, as if the edge were a mountain.
Of course, if a big chunk did disappear from the Earth, then the shape of the rest of the Earth could not stay the same. Most of the Earth below the surface is fluid and this fluid would quickly flow to fill up the giant hole left behind when the chunk disappeared. This would be bad for our health, just like whatever caused the chunk to be ripped from the Earth in the first place.
It is difficult to define "correct" and "incorrect" when discussing models of real things. We can never know all details of whatever thing we investigate, so we can never prove that any model that we construct of our subject is equivalent to its subject in every detail. The best that we can do is to determine how accurate the predictions can be that come out of our model, when we compare those predictions with the outcomes for the real subject. The more accurate the predictions are, the more "correct" the model is.
The model in which the planets follow elliptical orbits with the Sun at one of the foci of the ellipses and that equal areas of the ellipse are covered in equal times (Kepler's Laws) allows calculation of the positions of the planets with good (but not perfect) accuracy. The good accuracy of the results shows that the model is useful. The errors are due to gravity from the other planets and due to relativistic effects, and are relatively small.
Kepler's Laws in Cartesian coordinates give pretty accurate predictions. If you want to convince everybody that your model of planetary orbits is better than the one based on Kepler's Laws in Cartesian coordinates, then you'll have to show that your model gives more accurate predictions than the standard one.
If you move to a particular different coordinate system then the orbits of the planets look different. However, the choice of coordinate system is a matter of convenience which doesn't say anything about the "true" shape of the planetary orbits. The shape is connected to the coordinate system. It is meaningless to say what the shape of the orbit is if you don't also say what the coordinate system is relative to which the shape is described. (If you don't mention a coordinate system, then everyone will assume that you mean the usual Cartesian coordinate system.)
If you move to a different coordinate system, then not just the orbits of the planets look different, but also the laws of nature and the formulas from which to predict the positions of the planets will look different, because those laws and formulas must still yield predictions that accurately describe the real world
The choice of coordinate system does make a difference for how complicated and long the formulas are. This does not make other coordinate systems less "correct" than one in which the formulas are simple, but does make it less convenient to work in for this particular problem.
So, to solve some mathematical problem or to calculate some astronomical predictions, you are welcome to use any coordinate system that you want. If I prefer some other coordinate system, then I can take your results and translate them into my preferred coordinate system. The results will mean the same thing for the real world.
Moving to a different coordinate system does not prove that Kepler's Laws are wrong. Kepler's Laws can be moved to that different coordinate system, too, and they'll be just as accurate there as they are in the standard Cartesian coordinate system. Kepler's Laws turn out to be pretty accurate.
I think it goes a bit too far to say that Newton's Law of Gravity is "in conflict" with Einstein's theories of relativity ― as if Newton is always wrong and Einstein is always right. Einstein is better than Newton, but Newton is quite useful if you take into account its limitations, and Einstein is incomplete, too, and will someday be replaced by something better.
There are at least two things wrong with Newton's Law of Gravity, if you compare it with Einstein's Theories of Relativity:
The Law of Gravity of Newton assumes that the influence of (a change of) gravity can be felt everywhere in the Universe without any delay ― it travels at infinite speed. According to Newton, the force of gravity that we feel from a distant object points in the direction where that object is right now, and has a strength that depends on the mass that that object has right now.
According to the Theories of Relativity of Einstein no physical motion goes faster than at the speed of light ― not even the influence of the force of gravity. The influence of gravity travels in the form of gravity waves at the speed of light. The gravity that we feel now from the distant object depends on the position and mass of that object at the moment that the gravity waves that reach us now left that object, not on the position and mass that that object has right now.
This works the same as for sound (the sound of an airplane appears to come from the location where the airplane was when it made that sound, not from the location where the airplane is right now, and the sound has a loudness that depends on the condition of the engine at the time when the sound was generated, not on the condition of the engine at the time that you hear the sound) and for light (we see the Sun as it was 8 minutes ago, because the light took 8 minutes to get here) and for waves on water.
Even in a situation without motion and change (so speed does not matter anymore) the General Theory of Relativity by Einstein yields different results than the Law of Gravity by Newton. For common situations the difference is very small, but for exceptional situations (where gravity is very strong) the difference becomes very large.
There is much evidence that the Theory of Relativity by Einstein always yields more accurate results than the Law of Gravity by Newton ― that the answers from Einstein fit the observations better than the answers from Newton do. However, for common situations the differences are so small that the answers from Newton are perfectly fine to use, and Newton's formulas are a lot simpler to use than those of Einstein.
A major defect in Einstein's theories (and also that of Newton) is that it does not include any quantum effects. Quantum theories describe the other three fundamental forces of nature (also) at very small length scales, but we do not (yet) have a good quantum theory of gravity. It seems very likely that Einstein's theory of gravity doesn't work anymore (i.e., does not give accurate results anymore) at those very small length scales, but that is very difficult to test because at those length scales gravity is so much weaker than the other fundamental forces of nature.
At this moment, the theories of Einstein are the best we have, and good enough for all practical cases that we have been able to test so far, which is the same situation that Newton's equations were in a few centuries ago.
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Last updated: 2016−02−07