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If g be the acceleration due to gravity ...

If g be the acceleration due to gravity of the earth's surface, the gain is the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

A

`1/2mgR`

B

`2mg`

C

`mgR`

D

`1/4mgR`

Text Solution

Verified by Experts

The correct Answer is:
A
`U_(t)=-GMm//R`
`=`Initial potential energy of the system
`U_(f)=-GMm//2R=` Final `PE` of the system
`:. /_\U=U_(f)-U_(t)`
`=-GMm[1/(2R)-1/R]=(GMm)/(2R)` ……..i
But `g=GM//R^(2)`
`dK=-dU+W_("air friction")`
`:. GM=gR^(2)`.........ii
From eqn i and ii `/_\U=(gR^(2)m)/(2R)=(gRm)/2`
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