Question

maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth surface is

Answer 1

Answer:

H = R/3

Step by step explanation:

The body reaches maximum height when the Kinetic Energy of the body becomes zero.

Therefore,

$- \frac { G M m } { R } + \frac { 1 } { 2 } m v ^ { 2 } = - \frac { G M m } { R + H }$

v= 50% of the escape velocity. Hence,

$v = \frac { \sqrt { \frac { 2 G M } { R } } } { 2 } = \sqrt { \frac { G M } { 2 R } }\\ \\- \frac { G M m } { R } + \frac { 1 } { 2 } m \frac { G M } { 2 R } = - \frac { G M m } { R + H }\\ \\- \frac { 3 G M m } { 4 R } = - \frac { G M m } { R + H }\\ \\H = \frac { R } { 3 }$

Where R is the radius of the Earth.