Question

A body of mass m kg.starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the earth’s surface [R-Radius of earth, M-Mass of earth, G-Gravitational constant]

• A. $\frac { 1 } { 2 } \frac { G M m } { R }$
• B. $\frac { 1 } { 6 } \frac { G M m } { R }$
• C. $\frac { 2 } { 3 } \frac { G M m } { R }$
• D. $\frac { 1 } { 3 } \frac { G M m } { R }$

Answer 1

Answer: (B)

$\frac { 1 } { 6 } \frac { G M m } { R }$

Answer 2

When body starts falling toward earth’s surface its potential energy decreases so kinetic energy increases.

Increase in kinetic energy = Decrease in potential energy

Final kinetic energy – Initial kinetic energy = Initial potential energy – Final potential energy

Final kinetic energy – 0 $= \left( - \frac { G M m } { r _ { 1 } } \right) - \left( - \frac { G M m } { r _ { 2 } } \right)$

$\therefore$ Final kinetic energy $= \left( \frac { - G M m } { R + h _ { 1 } } \right) - \left( - \frac { G M m } { R + h _ { 2 } } \right)$ $= \left( - \frac { G M m } { R + 2 R } \right) - \left( - \frac { G M m } { R + R } \right)$ $= \frac { 1 } { 6 } \frac { G M m } { R }$.