Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$ is

  • A
    $\frac{G M m}{12 R^{2}}$
  • B
    $\frac{G M m}{3 R^{2}}$
  • C
    $\frac{ G M m }{8 R }$
  • D
    $\frac{ G M m }{6 R }$

Explore More

Similar Questions

$A$ rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V$. Assuming the radius of the earth to be $R$ and negligible air resistance,the maximum height attained by the rocket above the surface of the earth is

Difficult
View Solution

$A$ body (mass $m$) starts its motion from rest from a point distant $R_0$ $(R_0 > R)$ from the centre of the Earth. The velocity acquired by the body when it reaches the surface of the Earth will be ($G =$ universal constant of gravitation,$M =$ mass of Earth,$R =$ radius of Earth).

What is the minimum energy required to launch a satellite of mass $m$ from the surface of the Earth of mass $M$ and radius $R$ to an altitude of $2R$?

An object is propelled vertically to a maximum height of $4 R$ from the surface of a planet of radius $R$ and mass $M$. The speed of the object when it returns to the surface of the planet is

$A$ body of mass $m$ is released from a height of $2R$ above the Earth's surface. What will be its kinetic energy at a height of $R$ from the Earth's surface? (Where $R$ is the radius of the Earth)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo