I have a homework question asking me to calculate the mass of a planet given the semimajor axis and orbital period of its moon. I think I'm meant to assume the moon's mass is negligible because otherwise that's impossible as far as I'm aware. I attempted to use Kepler's 3rd Law, $$M=\frac{4\pi^2a^3}{GT^2}$$ But I come out with an absurdly large mass, several orders of magnitude too large. Which should be no surprise given $G$ is a very small number and $a$ is a very large number. I have a semimajor axis of $3.8\times10^8$ meters and a period of $1.512$ days. Apparently I can't just plug these in to calculate the planets mass. How do I figure this out?

Additional detail: My class is working on velocity and acceleration in polar coordinates with vectors. I see none of that being necessary here, it seems to me that it should be solvable using Kepler's Laws although I may be wrong about that.

  • $\begingroup$ Your semi major axis is very small for your orbital period. $3.8\times 10^8$ is barely more than one light-second, which is about the Earth-Sun distance, but the orbital period of the Moon is about 28 days, so you need quite a bit of mass ($\sim 350$ Earth masses?) to make the numbers work. $\endgroup$ – ZeroTheHero Mar 18 '17 at 2:29
  • $\begingroup$ ... and Keeler's law should work. $\endgroup$ – ZeroTheHero Mar 18 '17 at 2:33
  • $\begingroup$ @ZeroTheHero: I believe the Earth-Sun distance is about 8 light-minutes, I guess it's the Earth-Moon distance that is about 1 light-second, but then, it seems, the mass of the planet is much smaller than that of the Earth. $\endgroup$ – akhmeteli Mar 18 '17 at 2:36
  • $\begingroup$ right but my point is: if the Earth-Moon system yields a period of 28 days for the Moon at about the same distance from Earth as your system, the planet in your example must be much more massive than Earth to reduce the period by ~19. $\endgroup$ – ZeroTheHero Mar 18 '17 at 2:39
  • $\begingroup$ I should be getting a mass about the size of Jupiter. Instead I get a mass of 6340 suns $\endgroup$ – griffin175 Mar 18 '17 at 2:54

Since the distance Earth-Moon is about the same as in your example, you can write $$ M_p T^2_s\approx M_{Earth} T^2_{Moon}\quad \Rightarrow\quad \frac{M_p}{M_{Earth}}\approx \frac{T^2_{Moon}}{T^2_s}=19^2\sim 350 $$ Note the mass of Jupiter is ~320 times the mass of Earth, so you have a Jupiter-sized planet.


Write $M_s=x M_{Earth}$, i.e. use the mass of the Earth as a convenient unit of mass (rather than kg). From the data we know that $T_s\approx (1/19) T_{Moon}$ and use $T_{Moon}$ as a convenient unit of time (rather than days). Hence we find $$ x~\sim (19)^2\sim350, $$ meaning your planet is about $350$ Earth masses.

A more precise calculation would be based on $$ \frac{M_pT_s^2}{a_s^3}=\frac{M_E T_M^2}{a_M^3} \quad \Rightarrow \quad \frac{M_p}{M_E}=\frac{a_s^3T_M^2}{a_M^3 T_s^2}\, . $$

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  • $\begingroup$ This answer uses the Earth's mass as well as the period of the moon (Earth's moon). This is information outside of the parameters of the problem. I need to calculate the mass given only the moon's (of this specific system) orbital period and semimajor axis. $\endgroup$ – griffin175 Mar 18 '17 at 3:12
  • $\begingroup$ Your answer is off by about 31.5 Earth masses because you used a system that approximates this system. This moon has negligible mass and a slightly different radius. My point is, refer to the original question, "given a satellite's orbital period and semimajor axis". Imagine I have no access to information outside this question and go from there. I know the solution, I don't know how to get there. $\endgroup$ – griffin175 Mar 18 '17 at 3:22
  • $\begingroup$ @griffin175 please see my edit. Why can I not choose my units of mass and time as above? Well... I'm sorry I cannot help you more: I'm out of explanations. $\endgroup$ – ZeroTheHero Mar 18 '17 at 3:22
  • $\begingroup$ @griffin175 ... which I can't understand :( You can choose the units as you wish. You must be making a conversion mistake somewhere... If you sort it out please post as I would like to know. Best!! $\endgroup$ – ZeroTheHero Mar 18 '17 at 3:27
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    $\begingroup$ I figured it out. T just needed to be converted from days to seconds. $\endgroup$ – griffin175 Mar 18 '17 at 4:46

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