Previously unknown celestial bodies are regularly discovered in our Solar System, such as comets, asteroids, satellites of planets, and also objects in the Kuiper Belt beyond the orbit of Neptune, such as the object known as 2002 LN₆₀ that was in the news in early October 2002. Except for comets, those celestial bodies are almost by definition small and dim, because all bright celestial bodies were easier to spot. (Comets are special in this regard because they are usually far from the Sun and very dim, but can also become very bright when they spend a short period of time near the Sun.) How can we determine the size of such a dim celestial body?
There are two basic methods for determining the size of a dim celestial body, depending on the measurements that are available. In any case you need to know the distance of the body, and you can calculate that if you know the orbit of the body, which can be derived (with some difficulty) from a series of observations of the position of the body between the stars over a period of at least a few weeks (but the longer the better).
If the object looks just like a star through your telescope, then you can only measure an upper bound on the apparent size, from which follows (with the distance) an upper bound on the diameter. I refer to that diameter as the "blur diameter".
From measurements of the brightness (magnitude) of the object and the distances of the object from the Sun and the observer, you can calculate a lower bound on the diameter of the object. I refer to this lower bound as the "white-ball diameter", because it is the diameter of a well-reflecting white ball that looks equally bright as the object.
If you have measurements of the object in infrared light of various wavelengths, then you can derive an estimate for the temperature of the object, from which (with the distance from the Sun) follows the albedo of the object. From the white-ball diameter and the albedo follows the true (geometrical) diameter.
If \( w \) is the apparent diameter of the body, measured in arcseconds, and \( D \) the distance of the body from the observer, in AU, then the diameter \( d \) of the body, in kilometers, is equal to
\begin{equation} d = 725wD \end{equation}
For example: when Jupiter is closest to the Earth, then its apparent diameter (as seen from Earth) is about 46 arcseconds, and its distance about 4.2 AU. It then follows from the equation that the diameter of Jupiter is about 140,000 km. For Pluto, the corresponding values are 0.16 arcseconds, 29 AU, and 3400 km, and for 2002 LN₆₀ 0.044 arcseconds, 42 AU, and about 1350 km.
To be able to distinguish details of apparent diameter \( w \) with a telescope, the diameter \( d_\text{tel} \) of the telescope (measured in centimeters) must be at least about
\begin{equation} d_\text{tel} = \frac{10}{w} \end{equation}
but if the diameter of the telescope is close to this lower bound, then the telescope must be of very high quality (and thus very expensive) to be able to really distinguish the smallest possible details. Another problem is that the unquiet atmosphere of the Earth makes it difficult to distinguish apparent sizes of less than about 0.5 arcseconds, even with a very large telescope. Telescopes larger than about 20 cm have this problem.
For example: to be able to distinguish Jupiter, with an apparent size of 46 arcseconds, from a star, your telescope must have a diameter of at least 10/46 cm. To be able to distinguish Pluto from a star, your telescope must be at least 63 cm in diameter, but Pluto appears so small that the atmosphere of the Earth makes it very difficult to determine the exact apparent size of Pluto from the ground. To be able to distinguish 2002 LN₆₀ from a star, you need a telescope of at least about 230 cm diameter. The Hubble Space Telescope (HST), with which the apparent diameter of 2002 LN₆₀ was measured, has a diameter of 240 cm and is not troubled by the Earth's atmosphere, so this measurement was at the very edge of what the HST can do.
If \( D \) is the distance from the observer, \( D_\text{sun} \) the distance from the Sun, \( d \) the diameter of the body, and \( A \) the albedo (a number between 0 and 1), then the following formulas provide a reasonable estimate for the smallest magnitude \( m \) (the greatest brightness) of the body:
\begin{equation} m = m_0 + 5\log(D×D_\text{sun}) \end{equation}
\begin{equation} m_0 = 16,4 - 5\log(d_\text{white}) \end{equation}
\begin{equation} d_\text{white} = d\sqrt{A} \end{equation}
In these formulas, \( \log \) is the logarithm to base 10, \( \sqrt \) the square root, \( m_0 \) the absolute magnitude, and \( d_\text{white} \) the "white-ball diameter", the diameter of a white (100% diffusely reflecting) ball that would appear as bright in the same location as the body.
For example: the magnitude of 2002 LN₆₀ was reported as 18.5, and its distance as about 42 AU. From this follows an absolute magnitude of about 2.3 and a white-ball diameter of about 670 km.
The albedo \( A \) can be estimated from the temperature \( T \) (measured in kelvin) that follows from the infrared part of the spectrum of the body, according to
\begin{equation} A = \left( \frac{T\sqrt{D_\text{sun}}}{279} \right)^4 \end{equation}
With the albedo \( A \) and the white-ball diameter \( d_white \), the true (geometrical) diameter \( d \) equal to
\begin{equation} d = \frac{d_\text{white}}{\sqrt{A}} \end{equation}
For 2002 LN₆₀ we can go the other way. From a diameter of 1350 km and a white-ball diameter of 670 km follows an albedo of about 0.25, and a temperature of about 30 K.
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Last updated: 2020-07-18