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A satellite of mass m is revolving aroun...

A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:

A

`(mgR)/4`

B

`(mgR)/2`

C

mgR

D

`2mgR`

Text Solution

Verified by Experts

The correct Answer is:
A
`(mv^(2))/(r)=(GMn)/(r^(2))" "," "(1)/(2)mv^(2)=(GMm)/(2r)`
But r=2R `:.(1)/(2)mv^(2)=(GMm)/(4R)`
`KE=(gR^(2)m)/(4R)," "KE=(mgR)/(4)`
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