Question

An artificial satellite is in a circular orbit d = 720.0 km above the surface of a planet of radius r = 3.95 x 103 km. The period of revolution of the satellite around the planet is T = 2.65 hours. What is the average density of the planet? density = 3.39 x1016 kg/m3

Answer 1

First, find the mass of the planet

Consider the artificial satellite

Since the satellite is in a circular orbit, the gravitational force experienced by the satellite is equal to the centripetal force.

$F_{g}=F_{c}$

$\frac{GMm}{(r+d)^{2}}=\frac{mv^{2}}{(r+d)}$

$\frac{GM}{(r+d)}=v^{2}$

$\frac{GM}{(r+d)}=(\frac{2\pi (r+d)}{T})^{2}$

$\frac{GM}{(r+d)}=\frac{4\pi^{2} (r+d)^{2}}{T^{2}}$

$M=\frac{4\pi^{2} (r+d)^{3}}{T^{2}*G}$

$M=\frac{4\pi^{2} *(3.95*10^{6}m+7.2*10^{5}m)^{3}}{(2.65*60*60s)^{2}*6.674*10^{-11}}$

$M=\frac{4\pi^{2} *(4.67*10^{6})^{3}}{(2.65*60*60)^{2}*6.674*10^{-11}}$

${\color{Red} M=6.619534*10^{23}kg}$

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Volume of the planet, $V=\frac{4}{3}\pi r^{3}$

$V=\frac{4}{3}\pi *(3.95*10^{6}m)^{3}$

${\color{Red} V=2.581546*10^{20}m^{3}}$

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Density = Mass/Volume

Density of the planet, $\rho =\frac{M}{V}=\frac{6.619534*10^{23}kg}{2.581546*10^{20}m^{3}}$

ANSWER: ${\color{Red} \rho =2564.174kg/m^{3}}$

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