(B) The total energy $E$ of a satellite orbiting at a height $h$ is given by $E = -\frac{GMm}{2(R+h)}$.We know that the acceleration due to gravity on

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) The total energy $E$ of a satellite orbiting at a height $h$ is given by $E = -\frac{GMm}{2(R+h)}$.We know that the acceleration due to gravity on the Earth's surface is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.Given that the height $h = R$,the orbital radius is $r = R + h = R + R = 2R$.Substituting $GM = gR^2$ and $r = 2R$ into the energy formula:$E = -\frac{(gR^2)m}{2(2R)}$$E = -\frac{gR^2m}{4R}$$E = -\frac{mgR}{4}$.

Class 11 Physics Gravitation Energy consideration in Planetary and Satellite motion

Medium

$A$ satellite of mass $m$ is orbiting the Earth (of radius $R$) at a height equal to the radius of the Earth. The total energy of the satellite in terms of $g$ (acceleration due to gravity on Earth's surface) is

A
$-\frac{mgR}{3}$
B
$-\frac{mgR}{4}$
C
$-\frac{mgR}{2}$
D
$-\frac{mgR}{5}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Gravitation

Subtopic

Energy consideration in Planetary and Satellite motion

Two identical satellites are at heights $R$ and $7R$ from the Earth's surface. Which of the following statements is incorrect? ($R =$ radius of the Earth)

Difficult

View Solution

In a satellite,if the time of revolution is $T$,then the kinetic energy $(K.E.)$ is proportional to:

Difficult

View Solution

For a satellite,if the time of revolution is $T$,then its $K.E.$ is proportional to

Medium

View Solution

$A$ 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

DifficultNEET 2016

View Solution

If $R$ is the radius of orbit of a satellite,then the kinetic energy of the satellite is

EasyAP EAMCET 2021

View Solution

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

$A$ satellite of mass $m$ is orbiting the Earth (of radius $R$) at a height equal to the radius of the Earth. The total energy of the satellite in terms of $g$ (acceleration due to gravity on Earth's surface) is

Similar Questions

Two identical satellites are at heights $R$ and $7R$ from the Earth's surface. Which of the following statements is incorrect? ($R =$ radius of the Earth)

In a satellite,if the time of revolution is $T$,then the kinetic energy $(K.E.)$ is proportional to:

For a satellite,if the time of revolution is $T$,then its $K.E.$ is proportional to

$A$ 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

If $R$ is the radius of orbit of a satellite,then the kinetic energy of the satellite is

Vedclass Test Series

Exam Paper Generator

Online Exam Module