The total energy of a satellite in a circular orbit at a distance $(R+h)$ from the centre of the Earth varies as [ $R$ is the radius of the Earth and $h$ is the height of the orbit from Earth's surface].
- A$-\frac{1}{(R+h)}$
- B$\frac{1}{(R+h)^2}$
- C$-\frac{1}{(R+h)^2}$
- D$\frac{1}{(R+h)}$
Explore More
Similar Questions
The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statement about the curves $A, B$ and $C$.
Medium
View SolutionWhat is the binding energy of a satellite? Write its equation.
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 SolutionThe minimum energy required to launch a satellite of mass $m$ from the surface of the Earth of mass $M$ and radius $R$ into a circular orbit at an altitude of $2R$ from the surface of the Earth is:
MediumNEET 2024
View SolutionThe additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$,to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2$ $(R_2 > R_1)$ is:
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo