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A projectile is fired vertically upwards...

A projectile is fired vertically upwards from the surface of the earth with a velocity `Kv_(e)`, where `v_(e)` is the escape velocity and `Klt1`.If `R` is the radius of the earth, the maximum height to which it will rise measured from the centre of the earth will be (neglect air resistance)

A

`(R_(E))/(1-k^(2))`

B

`(R_(E))/(k^(2))`

C

`(1-k^(2))/(R_(E))`

D

`(k^(2))/(R_(E))`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) Let a projectile is fired vertically upwards from the earth's surface with a velocity v and reaches a height h. Energy of the projectile at the surface of the earth is
`E_(i)=(1)/(2)mv^(2)-(GM_(E)m)/(R_(E))`
Energy of the projectile at a height h
`E_(f)= -(GM_(E)m)/(R_(E)+h)` (`:.` At height h velocity of the projectile is zero.)
According to law of conservation of energy,
`E_(i)=E_(f)`
`(1)/(2)mv^(2)-(GM_(E)m)/(R_(E))=-(GM_(E)m)/(R_(E)+h)`
`(1)/(2)mv^(2)=GM_(E)m[(1)/(R_(E))-(1)/(R_(E)+h)]=(GM_(E)mh)/((R_(E))(R_(E)+h))`
`(1)/(2)mv^(2)=(mghR_(E))/(R_(E)+h)" "( :. g=(GM_(E))/(R_(E)^(2)))`
As per question,
`v=kv_(e)=ksqrt(2gR_(E))" "( :. v_(e)=sqrt(2gR_(E)))`
and `h=r-R_(E)`
`:. (1)/(2)mk^(2)2gR_(E)=(mg(r-R_(E))R_(E))/( r)`
`k^(2)=(r-R_(E))/(r) implies (1-k^(2))r=R_(E)`
`implies r=(R_(E))/(1-k^(2))`
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