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A particle is fired vertically from the surface of the earth with a velocity `kupsilon_(e)`, where `upsilon_(e)` is the escape velocity and `klt1`. Neglecting air resistance and assuming earth's radius as `R_(e)`. Calculated the height to which it will rise from the surface of the earth.

A

`(R )/(1 - k^(2))`

B

`(R )/(k^(2))`

C

`(1-k^(2))/(R )`

D

`(k^(2))/(R )`

Text Solution

Verified by Experts

The correct Answer is:
A
According to law of conservation of mechanical energy,
`(1)/(2)m(k upsilon_(e))^(2) + ((-GM m)/(R )) = - (GM m)/((R + h))`
or `(1)/(2) m (ksqrt(2gR))^(2) = (GM m)/(R ) - (GM)/((R + h)) = (GM mh)/(R(R + h))`
or `(1)/(2) mk^(2) xx 2g R = (mg Rh)/(R + h) = (mg R(r - R))/(r )`
[As `h = r - R]`
or `k^(2) = (r - R)/(r ) = 1- (R )/(r )` or `r = (R )/(1 - k^(2))`
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