A rocket is launched vertically from the surface of earth with an initial velocity `v`. How far above the surface of earth it will go? Neglect the air resistance.

Options

A. `(R_(E)v^(2))/(gR_(E)-v^(2))`
B. `(R_(E)v^(2))/(gR_(E)+v^(2))`
C. `(R_(E)v^(2))/(2gR_(E)-v^(2))`
D. `(R_(E)v^(2))/(2gR_(E)+v^(2))`

Answer

C

Explanation

( c) Let the rocket reaches a height h from the surface of earth.
Total energy at the surface of the earth is
`E_(s)=(1)/(2)mv^(2)-(GM_(E)m)/(R_(E))`
where m and `M_(E)` are the masses of rocket and earth respectively.
At highest point, the velocity of the rocket becomes zero.
`:.` Total energy at the highest point is
`E_(h)=-(GM_(E)m)/((R_(E)+h))`
According to law of conservation of energy,
`E_(s)=E_(h)`
`:.(1)/(2)mv^(2)-(GM_(E)m)/(R_(E))=-(GM_(E)m)/(R_(E)+h)`
`(1)/(2)v^(2)-(GM_(E))/(R_(E))=-(GM_(E))/(R_(E)+h)`
`=(gR_(E)^(2))/(R_(E))-(gR_(E)^(2))/(R_(E)+h)( :. G=(GM_(E))/(R_(E)^(2)))`
`=gR_(E)(1-(R_(E))/(R_(E)+h))=gR_(E)((h)/(R_(E)+h))`
`v^(2)(R_(E)+h)=2gR_(E)h`
`v^(2)R_(E)=2gR_(E)h-v^(2)h`
`R_(E)v^(2)=h(2gR_(E)-v^(2))`
`h=(R_(E)v^(2))/(2gR_(E)-v^(2))`