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Question: A spherical planet far out in space has a mass \(M_{0}\) and diameter \(D_{0}.\) A particle of mass ...

A spherical planet far out in space has a mass M0M_{0} and diameter D0.D_{0}. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

A

GM0/D02GM_{0}/D_{0}^{2}

B

4mGM0/D024mGM_{0}/D_{0}^{2}

C

4GM0/D024GM_{0}/D_{0}^{2}

D

GmM0/D02GmM_{0}/D_{0}^{2}

Answer

4GM0/D024GM_{0}/D_{0}^{2}

Explanation

Solution

We know g=GMR2=GM(D/2)2=4GMD2g = \frac{GM}{R^{2}} = \frac{GM}{(D/2)^{2}} = \frac{4GM}{D^{2}}

If mass of the planet =M0= M_{0} and diameter of the planet =D0= D_{0}. Then g=4GM0D02g = \frac{4GM_{0}}{D_{0}^{2}}.