# Kepler's Third Law confusion

• B
Abu
I am confused about the units used in Kepler's 3rd law. Is T supposed to be in years or days? Is R supposed to be in kilometers or meters? Is there ever an instance where one combination of units is preferable over another (for example, if you want to use the answers from Kepler's third law to find another variable in a different equation, such as mass). After searching online, it was suggested to use Astronomical Units, but I've never been taught with those.
I know it is a simple question, just can't find a clear and concise answer.

Thanks.

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Bandersnatch
The clear and concise answer is: It doesn't matter what units you use, be it AU and years, kilometres and seconds, yards and fortnights - as long as you're consistent.

Since Kepler's 3rd has the form:
##T^2 \propto a^3##
It's a statement about proportionality. The factor of proportionality is ##\frac{4\pi^2}{GM}##. So if you were to use the expression as:
##T^2 = \frac{4\pi^2}{GM} a^3##
then you'd have to make sure units in G and M are consistent with those you use for T and a. E.g. if you use AU for distance, you need to make sure you rescale G to be in units of ##N AU^2/kg^2##.

But, since it's a statement about proportionality, it tells you that the expression
##\frac{T^2}{a^3} = const## for all planets (of a given star).

This means, that you'd normally use it to compare two planets around the same star. So, let's say you use it for two planets 1 and 2:
##\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}##
...and as long as you remembered to use the same units for both planets, all the units work out regardless of what they are.

For example. I've just came up with fanciful units for distance [glork] and time [splork].
Planet 1 orbits at a distance of 10 glorks, with a period of 1 splork.
Planet 2 orbits with a period of 20 splorks.
You want to know something about the orbital distance of the second planet, so you write:
##\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}##
##a_2 = \sqrt[3]{\frac{T_2^2 a_1^3}{T_1^2}}##
And you read the right hand side as 'that many glorks'.
Or, equivalently, you write:
##\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}##
##\frac{a_2}{a_1} = \sqrt[3]{\frac{T_2^2}{T_1^2}}##
And you read that as ##a_2## is 'that many times' larger/smaller than ##a_1##. Where 'that many times' is dimensionless.

vanhees71, Abu and Ibix
Abu
The clear and concise answer is: It doesn't matter what units you use, be it AU and years, kilometres and seconds, yards and fortnights - as long as you're consistent.

Since Kepler's 3rd has the form:
##T^2 \propto a^3##
It's a statement about proportionality. The factor of proportionality is ##\frac{4\pi^2}{GM}##. So if you were to use the expression as:
##T^2 = \frac{4\pi^2}{GM} a^3##
then you'd have to make sure units in G and M are consistent with those you use for T and a. E.g. if you use AU for distance, you need to make sure you rescale G to be in units of ##N AU^2/kg^2##.

But, since it's a statement about proportionality, it tells you that the expression
##\frac{T^2}{a^3} = const## for all planets (of a given star).

This means, that you'd normally use it to compare two planets around the same star. So, let's say you use it for two planets 1 and 2:
##\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}##
...and as long as you remembered to use the same units for both planets, all the units work out regardless of what they are.

For example. I've just came up with fanciful units for distance [glork] and time [splork].
Planet 1 orbits at a distance of 10 glorks, with a period of 1 splork.
Planet 2 orbits with a period of 20 splorks.
You want to know something about the orbital distance of the second planet, so you write:
##\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}##
##a_2 = \sqrt[3]{\frac{T_2^2 a_1^3}{T_1^2}}##
And you read the right hand side as 'that many glorks'.
Or, equivalently, you write:
##\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}##
##\frac{a_2}{a_1} = \sqrt[3]{\frac{T_2^2}{T_1^2}}##
And you read that as ##a_2## is 'that many times' larger/smaller than ##a_1##. Where 'that many times' is dimensionless.
Thank you for your great explanation, that really clears things up now. I really appreciate it