If I know the semi-major axis, the density of the planet, and orbital period.. how can I find the radius of the satellite? I can't think of any formula to use.

  • 2
    $\begingroup$ Just to be clear, are you trying to figure out a physical property of the (natural?) satellite, or one of its orbital properties? $\endgroup$
    – Alphecca
    Dec 4 '18 at 1:57
  • 1
    $\begingroup$ You can't, from that info. You can calculate the planets radius, if you assume that the given density is its mean density. $\endgroup$
    – PM 2Ring
    Dec 4 '18 at 7:31
  • $\begingroup$ ahhh a lovely typo and wasted time. haha. Thanks for the help. I will use Keplers third law to solve this. $\endgroup$
    – loneWolf
    Dec 4 '18 at 17:07
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    $\begingroup$ @loneWolf can you double check this answer and see if it might be the radius of the planet, not the artificial satellite, that you are looking for? $\endgroup$
    – uhoh
    Dec 5 '18 at 0:43
  • $\begingroup$ These are independent datas. You might have some very uncertain estimations based on the distrubtion of the satellite sizes in that (or similar) systems, but that it is all. $\endgroup$
    – peterh
    Dec 6 '18 at 11:43

How to find the radius of a satellite not knowing the mass:

If I know the

  • semi-major axis,
  • density of the planet, and
  • orbital period..

how can I find the radius of the satellite? I can't think of any formula to use.

I have a hunch that you've misunderstood the question, and that it asks for the radius of the planet not the artificial satellite, and that's certainly doable. Otherwise, why would the planet's density be a given?

The equation that links semi-major axis, mass of the planet, and orbital period is:

$$T=2 \pi \sqrt{\frac{a^3}{GM}} $$

where $G$ is the gravitational constant 6.674E-11 m^3 s^-2 kg^-1.

Moving things around we get

$$\frac{4 \pi^2 a^3}{T^2G} = M $$

Density is mass per volume, or:

$$\rho = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4 \pi R^3}$$

Moving that around we get

$$R = \left(\frac{3M}{4 \rho \pi}\right)^{1/3}$$

Put in the earlier result for $M$, we get

$$R = a \left(\frac{3 \pi}{\rho T^2G}\right)^{1/3}$$

For Earth's average density, Wikipedia says 5514 kg/^3, for its radius, try 6,371,000. meters. For the ISS at 400 km (400,000 meters) above the Earth, use 92.5 minutes (5550 sec). It should all work out nicely!

The reason this works with average density, even though Earth's density is higher in the core than the surface is the application of Newton's shell theorem.


As the comments suggest, in general you cannot. So long as the satellite's mass is a small fraction of the planet's mass (or a planet vs. the Sun's mass), the difference between a circular orbit and the elliptical orbit about the system center of mass is too small to allow accurate calculation.

But if you do know the mass of the planet (or sun), then knowing the shape of the elliptical orbit will tell you the mass of the satellite, and since you state you know the density -- I'm assuming that was a typo in your original question, finding the radius (of a perfect sphere) is trivial.


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