Question

An artificial satellite is in a circular orbit above the surface of a planet of radius the period of revolution of the satellite around the planet is . What is the average density of the planet?

Answer 1

Answer:

$12548.5 kgm^{-3}$

Explanation:

Complete statement of the question is

An artificial satellite is in a circular orbit d = 310 km above the surface of a planet of radius r = 2050 km. The period of revolution of the satellite around the planet is T = 1.15 hours. What is the average density of the planet?

$r$ = radius of the planet = 2050 km = 2.05 x 10⁶ m

$d$ = distance of the satellite above the surface of planet = 310 km = 0.310 x 10⁶ m

$R$ = radius of the orbit = $r + d$ = $2.05\times10^{6} + 0.31\times10^{6} = 2.36\times10^{6} m$

$M$ = mass of the planet

$T$ = Time period of the planet = 1.15 h = 1.15 x 3600 = 4140 s

Using Kepler's third law

$T^{2} = \frac{4\pi^{2} R^{3} }{GM} \\4140^{2} = \frac{4(3.14)^{2} (2.36\times10^{6})^{3} }{(6.67\times10^{-11})M}\\M = 4.53\times10^{23} kg$

Volume of the planet is given as

$V = \frac{4\pi r^{3} }{3} \\V = \frac{4(3.14) (2.05\times10^{6})^{3} }{3}\\V = 3.61\times10^{19} m^{3}$

Average density of the planet is given as

$\rho = \frac{M}{V} \\\rho = \frac{4.53\times10^{23}}{3.61\times10^{19}}\\\rho = 12548.5 kgm^{-3}$