Answer:
Acceleration, [tex]a=2.22\times 10^{-3}\ m/s^2[/tex]
Explanation:
It is given that,
Time period of revolution of the moon, [tex]T=2060\times 10^3\ s[/tex]
If the distance from the center of the moon to the surface of the planet is, [tex]h=235\times 10^6\ m[/tex]
The radius of the planet, [tex]r=3.9\times 10^6\ m[/tex]
Let a is the moon's radial acceleration. Mathematically, it is given by :
[tex]a=R\times \omega^2[/tex], R is the radius of orbit
Since, [tex]\omega=\dfrac{2\pi}{T}[/tex]
The radius of orbit is,
[tex]R=r+h[/tex]
[tex]R=3.9\times 10^6\ m+235\times 10^6\ m=238900000\ m[/tex]
So, [tex]a=\dfrac{4\pi^2 R}{T^2}[/tex]
[tex]a=\dfrac{4\pi^2 \times 238900000}{(2060\times 10^3)^2}[/tex]
[tex]a=2.22\times 10^{-3}\ m/s^2[/tex]
Hence, this is the required solution for the radial acceleration of the moon.
