Question

The radius of a planet is r and a satellite revolves around it in a circle of radius 3r. The time period of revolution of satellite is t. Acceleration due to gravity of the planet at its surface will be:

Answer 1

Answer:

Time period of revolution:

T=2πR2v

Hence, the expression for orbital velocity is,

v=2πR2T

But, the expression for orbital velocity is,

v=GmR2−−−√

By equating the expressions of velocity:

2πR2T=GmR2−−−√(2πR2T)2=GmR2Gm=R2(2πR2T)2

Acceleration due to gravity on the surface:

g=GmR12  =R2(2πR2T)2R12   =R2(2πR2TR1)2

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