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 = [tex]\frac{2\pi R}{V}[/tex]
[tex]T = \frac{2\pi R }{\sqrt{\frac{GM}{R} } }[/tex]
[tex]T= 2\pi \sqrt{(\frac{R^3}{GM} )}[/tex]
[tex]M = \frac{4}{3} \pi R^{3}p[/tex]
where p = density
[tex]T = 2\pi \sqrt{\frac{R^{3} }{G\frac{4}{3}\pi R^{3} p } }[/tex]
[tex]T = \sqrt{\frac{3\pi }{Gp} }[/tex]
T = 2.17 hours = 7812 sec
(7812)² = [( 3×3.14)/6.67×[tex]10^{-11}[/tex]×ρ)]
ρ = 6.28/6.67×[tex]10^{-11}[/tex]×6.10×[tex]10^{-7}[/tex]
ρ = 6.28/40.687×[tex]10^{-18}[/tex]
ρ = 0.1543×[tex]10^{18}[/tex]kg/m³
ρ = 15.43×[tex]10^{16}[/tex]kg/m³
