Minimum energy required to launch a satellite of mass m from surface of earth of mass M and radius R

English

Gujarati

Hindi

7.Gravitation

medium

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ from the surface of the earth is:

A

$\frac{2 G m M}{3 R}$

B

$\frac{G m M}{2 R}$

C

$\frac{G m M}{3 R}$

D

$\frac{5 G m M}{6 R}$

(NEET-2024)

Solution

Apply energy conservation,

$U_I+K_I=U_f+K_f$

$\Rightarrow \quad-\frac{G M m}{R}+K_I=-\frac{G M m}{3 R}+\frac{1}{2} m v^2$

$\Rightarrow \quad-\frac{G M m}{R}+K_I=-\frac{G M m}{3 R}+\frac{1}{2} \times m \times \frac{G M}{3 R}$

$\Rightarrow \quad K_I=-\frac{1}{6} \frac{G M m}{R}+\frac{G M m}{R}$

$K_I=\frac{5}{6} \frac{G M m}{R}$

Standard 11

Physics

Share

Similar Questions

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)

hard

A shell of mass $M$ and radius $R$ has a point mass m placed at a distance $r$ from its centre. The gravitational potential energy $U \,(r)$ vs $r$ will be

medium

Two masses $m_1\, \& m_2$ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separation distance of $d$ is :

hard

A satellite of mass $m$, initially at rest on the earth, is launched into a circular orbit at a height equal to the radius of the earth. The minimum energy required is

hard

Two identical satellites $A$ and $B$ are circulating round the earth at the height of $R$ and $2R$ respectively, (where $R$ is radius of the earth). The ratio of kinetic energy of $A$ to that of $B$ is

medium