A satellite of mass is orbiting the earth (of radius R) at a height from its surface. The total energy of the satellite in terms of , the value of acceleration due to gravity at the earth’s surface, is [2016]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

Q.
A satellite of mass $m$ is orbiting the earth (of radius R) at a height $h$ from its surface. The total energy of the satellite in terms of $g_{0}$ , the value of acceleration due to gravity at the earth’s surface, is [2016]

1 $\frac{m g_{0} R^{2}}{2 (R + h)}$

2 $- \frac{m g_{0} R^{2}}{2 (R + h)}$

3 $\frac{2 m g_{0} R^{2}}{R + h}$

4 $- \frac{2 m g_{0} R^{2}}{R + h}$

Ans.
(2)

Total energy of satellite at height $h$ from the earth's surface,

   $E = P E + K E = - \frac{G M m}{(R + h)} + \frac{1}{2} m v^{2}$ ...(i)

Also, $\frac{m v^{2}}{(R + h)} = \frac{G M m}{{(R + h)}^{2}}$

or,    $v^{2} = \frac{G M}{R + h}$ ...(ii)

From eqns. (i) and (ii),

   $E = - \frac{G M m}{(R + h)} + \frac{1}{2} \frac{G M m}{(R + h)} = - \frac{1}{2} \frac{G M m}{(R + h)}$

   $= - \frac{1}{2} \frac{G M}{R^{2}} \times \frac{m R^{2}}{(R + h)} = - \frac{m g_{0} R^{2}}{2 (R + h)}$

Solution

Q.
A satellite of mass $m$ is orbiting the earth (of radius R) at a height $h$ from its surface. The total energy of the satellite in terms of $g_{0}$ , the value of acceleration due to gravity at the earth’s surface, is [2016]

1 $\frac{m g_{0} R^{2}}{2 (R + h)}$

2 $- \frac{m g_{0} R^{2}}{2 (R + h)}$

3 $\frac{2 m g_{0} R^{2}}{R + h}$

4 $- \frac{2 m g_{0} R^{2}}{R + h}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g0, the value of acceleration due to gravity at the earth’s surface, is [2016]

1 mg0R22(R+h)

2 -mg0R22(R+h)

3 2mg0R2R+h

4 -2mg0R2R+h

Ans. (2) Total energy of satellite at height h from the earth's surface, E=PE+KE=-GMm(R+h)+12mv2 ...(i) Also, mv2(R+h)=GMm(R+h)2 or, v2=GMR+h ...(ii) From eqns. (i) and (ii), E=-GMm(R+h)+12GMm(R+h)=-12GMm(R+h) =-12GMR2×mR2(R+h)=-mg0R22(R+h)

Want Us to Email you About Special Offers & Updates?

Q.
A satellite of mass $m$ is orbiting the earth (of radius R) at a height $h$ from its surface. The total energy of the satellite in terms of $g_{0}$ , the value of acceleration due to gravity at the earth’s surface, is [2016]

1 $\frac{m g_{0} R^{2}}{2 (R + h)}$

2 $- \frac{m g_{0} R^{2}}{2 (R + h)}$

3 $\frac{2 m g_{0} R^{2}}{R + h}$

4 $- \frac{2 m g_{0} R^{2}}{R + h}$