Its rotation period is 45 hours. Its satellite is at the distance of 15000 km. Given this data, and assuming the satellite's orbital period the same as the main body's rotation period, 45 hours, what is the mass of the main body?
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1 Answer
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We can use Kepler's Third Law, the Law of Periods, to solve this problem. The law can be stated mathematically as:
$T^2 = \frac{4\pi^2}{G(M+m)} a^3$
Where
$T$ is the orbital period,
$G$ is the gravitational constant,
$M$ is the mass of the primary,
$m$ is the mass of the secondary, and
$a$ is the semimajor axis.
Solving for $(M + m)$ with the given values returns a system mass of approximately $8 \times 10^{22}$ $\mathrm{kg}$.
Other data are necessary to isolate the mass of the satellite from the primary.
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1$\begingroup$ Given the mass of the sattelite is negligible, this gives mass about 6 times that of Pluto... Apparently, it is NOT tidally locked... $\endgroup$– AnixxFeb 15, 2017 at 2:35