Question

An artificial satellite is in a circular orbit d=730.0 km above the surface of a planet of radius r=4.55×103 km. The period of revolution of the satellite around the planet is T=2.15 hours. What is the average density of the planet? kg/m³

Answer 1

Gravitational constant = G = 6.67 x 10^-11 N.m²/kg²

Mass of the planet = M

Mass of the satellite = m

Radius of the planet = R_p = 4.55 x 10³ km = 4.55 x 10⁶ m

Distance of the satellite above the surface of the planet = H = 730 km = 730 x 10³ m = 0.73 x 10⁶ m

Radius of orbit of the satellite = R

R = R_p + H

R = 4.55x10⁶ + 0.73x10⁶

R = 5.28 x 10⁶ m

Orbital period of the satellite = T = 2.15 hours = 2.15 x (3600) sec = 7740 sec

Orbital speed of the satellite = V

VT = 2$\pi$R

V(7740) = 2$\pi$(5.28x10⁶)

V = 4286.204 m/s

The centripetal force for the circular motion of the satellite is provided by the gravitational force of the planet on the satellite.

$\frac{mV^{2}}{R} = \frac{GMm}{R^{2}}$

$V^{2} = \frac{GM}{R}$

$M = \frac{V^{2}R}{G}$

$M = \frac{(4286.204)^{2}(5.28\ast 10^{6})}{6.67\ast 10^{-11}}$

M = 1.454 x 10²⁴ kg

Volume of the planet = V₀

V₀ = 4$\pi$R_p³/3

V₀ = 4$\pi$(4.55x10⁶)³/3

V₀ = 3.945 x 10²⁰ m³

Average density of the planet = $\rho$

M = $\rho$V₀

1.454x10²⁴ = $\rho$(3.945x10²⁰)

$\rho$ = 3685.68 kg/m³

Average density of the planet = 3685.68 kg/m³