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Let the minimum external work done in sh...

Let the minimum external work done in shifted a particle form centre of earth to earth's surface be `W_(1)` and that form surface of earth to infinity be `W_(2)` . Then `(W_(1))/(W_(2))` is equl to

A

`1:1`

B

`1:2`

C

`2:1`

D

`1:3`

Text Solution

Verified by Experts

The correct Answer is:
B
`(W_(1))/(W_(1)) =(-(GM)/(R)-((3)/(2)(GM)/(R)))/(0- (-(GM )/(R)))=1/2 `
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