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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.

Text Solution

Verified by Experts

The correct Answer is:
`(R_(e)k^(2))/(1-k^(2))`
We know `v_(e)=sqrt((2GM)/(R_(e)))`
`GM=(R_(e)v_(e)^(2))/2....(1)`
From energy conservation
`U_(i)+K_(i)=U_(f)+K_(f)`
`-(GMm)/(R_(e))+1/2m(kv_(e))^(2)=-(GMm)/((R_(e)+h))+0`
`-(GM)/(R_(e))+(k^(2)v_(e)^(2))/2=(-GM)/(R_(e)+h).....(2)`
From `(1)` and `(2)`
-`(v_(e)^(2))/(R_(e))+(k^(2)v_(e)^(2))/2=-(R_(e)v_(e)^(2))/(2(R_(e)+h))implies1-k^(2)=(R_(e))/((R_(e)+h))`
`(R_(e)+h)/(R_(e))=1/(1-k^(2))impliesh=(k^(2)R_(e))/(1-k^(2))`
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