Answer:
Radius of the planet, [tex]r=5.8\times 10^6\ m[/tex]
Explanation:
It is given that,
Mass of the satellite, m = 20 kg
Radius of the circular orbit, [tex]r=8\times 10^6\ m[/tex]
Time period of the motion of satellite, [tex]T=2.4\ h=8640\ s[/tex]
The acceleration on the surface of the planet is, [tex]a=8\ m/s^2[/tex]
The relation between the time period of the satellite and its radius is given by third law of Kepler as :
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
M is the mass of planet
[tex]M=\dfrac{4\pi^2r^3}{T^2G}[/tex]
[tex]M=\dfrac{4\pi^2\times (8\times 10^6)^3}{(8640)^2\times 6.67\times 10^{-11}}[/tex]
[tex]M=4.059\times 10^{24}\ Kg[/tex]
The acceleration on the surface of planet is given by :
[tex]a=\dfrac{GM}{r^2}[/tex]
[tex]r=\sqrt{\dfrac{GM}{a}}[/tex]
[tex]r=\sqrt{\dfrac{6.67\times 10^{-11}\times 4.059\times 10^{24}}{8}}[/tex]
[tex]r=5.8\times 10^6\ m[/tex]
So, the radius of the planet is [tex]5.8\times 10^6\ m[/tex]. Hence, this is the required solution.
