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so I attemted to solve the following: planet X of mass $8.8 \times 10^{27}$ kg has a moon of an unknown mass which orbits every 17 days from a distance of approximately one million km ( center to center). What is the approximate mass of moon?

Given: $M = 8.8 \times 10^{27}$kg

m = ?

f (frequency)= 17

$d= r= 10^6$km= $10^9 m$

mechanical system chosen the moon. from newtons 2nd law (sigma) $F = m\times a$ $F(g) = m\times a(c)$ here I assume that the only force applying on the moon is gravity of the planet X, an it has a circular acceleration so I got: $G\times M\times m/r^2 = m\times 4π^2 \times d \times f^2$

However, the $ m$, mass of moon cancels out, and I'm supposed to find it. so?

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Here are some similar situations to support the conclusion that the mass of the Moon will cancel.

  • Imagine you are asked to make the same calculation for a satellite orbiting the Earth. All geosynchronous satellites are at the same altitude, irrespective of the mass of the satellite, which can be considered an artificial "moon".
  • Kepler's law does not depend on the mass of the planets, which can be rightly considered as "moons" of the Sun.
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    $\begingroup$ so, the mass of the moon can be anything? $\endgroup$ Apr 17 '17 at 23:09
  • $\begingroup$ If you check the possible duplicate you will realize that the mass of the moon cancels out of your calculation. $\endgroup$ Apr 17 '17 at 23:10
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When the moon is much smaller than the planet, then the mass of moon doesn't matter. But if it's large, then the period will be reduced.

The Kepler equation can solve for the sum of the masses of the two objects.

$M_1 + M_2 = 4 \pi^2 \frac{a^3}{G T^2}$

For large moons and with very precise values for the other components, you can tease out the approximate mass of the moon. Of course for most planet/moon situations, $M_1 + M_2 \approx M_1$, and the mass of the smaller object is lost in the noise.

Your question is even odder because the numbers don't work even for a zero-mass moon. If the mass and distance are correct, the period is much too long. It's not consistent.

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  • $\begingroup$ @SilentWisper Putting your numbers into BowlOfRed's equation gives an approximate combined mass of $3 \times 10^(26) \,\rm kg$. $\endgroup$
    – Farcher
    Apr 18 '17 at 6:08

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