Question

A satellite is revolving round the earth with orbital speed $v _ { 0 }$. If it stops suddenly, the speed with which it will strike the surface of earth would be ($v _ { e } =$escape velocity of a particle on earth’s surface)

• A. $\frac { v _ { e } ^ { 2 } } { v _ { 0 } }$
• B. $v _ { 0 }$
• C. $\sqrt { v _ { e } ^ { 2 } - v _ { 0 } ^ { 2 } }$
• D. $\sqrt { v _ { e } ^ { 2 } - 2 v _ { 0 } ^ { 2 } }$

Answer 1

Answer: (D)

$\sqrt { v _ { e } ^ { 2 } - 2 v _ { 0 } ^ { 2 } }$

Answer 2

Applying conservation of mechanical energy between A and B point

$- \frac { G M m } { r } = \frac { 1 } { 2 } m v ^ { 2 } + \left( - \frac { G M m } { R } \right)$; $\frac { 1 } { 2 } m v ^ { 2 } = \frac { G M m } { R } - \frac { G M m } { r }$

$= v _ { e } ^ { 2 } - 2 v _ { 0 } ^ { 2 }$ $\Rightarrow v = \sqrt { v _ { e } ^ { 2 } - 2 v _ { 0 } ^ { 2 } }$

[As escape velocity $v _ { 0 } = \sqrt { \frac { G m } { r } }$]