Question

A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.17 hours. What is density of the planet

Answer 1

Explanation:

Lets consider

Circumference of orbit = T

as it is mentioned in the question that a satellite is in orbit that is very close to the surface of planet. so

circumference of orbit = circumference of planet

Time period = T

radius of planet = R

orbital velocity = V

gravitational constant = G

mass of planet = m

Solution:

Time period for a uniform circular motion of orbit is,

T = $\frac{2\pi R}{V}$

$T = \frac{2\pi R }{\sqrt{\frac{GM}{R} } }$

$T= 2\pi \sqrt{(\frac{R^3}{GM} )}$

$M = \frac{4}{3} \pi R^{3}p$

where p = density

$T = 2\pi \sqrt{\frac{R^{3} }{G\frac{4}{3}\pi R^{3} p } }$

$T = \sqrt{\frac{3\pi }{Gp} }$

T = 2.17 hours = 7812 sec

(7812)² = [( 3×3.14)/6.67×$10^{-11}$×ρ)]

ρ = 6.28/6.67×$10^{-11}$×6.10×$10^{-7}$

ρ = 6.28/40.687×$10^{-18}$

ρ = 0.1543×$10^{18}$kg/m³

ρ = 15.43×$10^{16}$kg/m³