Binding energy of a revolving satellite at he | vedclass

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

Question

(B) The binding energy $(BE)$ of a satellite is defined as the minimum energy required to remove the satellite from its orbit to infinity.For a satell

Vedclass · Accepted Answer

$-7 	imes 10^8 \ J$

Vedclass · Answer

(B) The binding energy $(BE)$ of a satellite is defined as the minimum energy required to remove the satellite from its orbit to infinity.For a satellite of mass $m$ at a distance $r = R + h$ from the center of the Earth,the potential energy is $U = -\frac{GMm}{r}$.The total energy of the satellite is $E = K + U = -\frac{GMm}{2r}$.The binding energy is the negative of the total energy: $BE = -E = \frac{GMm}{2r}$.Comparing $BE$ and $U$,we see that $U = -2 	imes BE$.Given $BE = 3.5 	imes 10^8 \ J$,the potential energy is $U = -2 	imes (3.5 	imes 10^8 \ J) = -7 	imes 10^8 \ J$.

Binding energy of a revolving satellite at height $h$ is $3.5 \times 10^8 \ J$. Its potential energy is

Similar Questions

The total energy of a circularly orbiting satellite is

$A$ 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

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

What is the binding energy of a satellite? Write its equation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module