Let the kinetic energy of a satellite be x,th | vedclass

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

Question

(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $K.E. = \frac{GMm}{2r}$.Given $

Vedclass · Accepted Answer

$x^{-3/2}$

Vedclass · Answer

(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $K.E. = \frac{GMm}{2r}$.Given $K.E. = x$,we have $x = \frac{GMm}{2r}$.This implies $r = \frac{GMm}{2x}$.Since $GMm/2$ is a constant,we can write $r \propto \frac{1}{x}$.According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$,i.e.,$T^2 \propto r^3$.Substituting $r \propto x^{-1}$ into the relation,we get $T^2 \propto (x^{-1})^3 = x^{-3}$.Taking the square root on both sides,we get $T \propto (x^{-3})^{1/2} = x^{-3/2}$.Thus,the time of revolution $T$ is proportional to $x^{-3/2}$.

Let the kinetic energy of a satellite be $x$,then its time of revolution $T$ is proportional to ..............

Similar Questions

$A$ 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

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

The energy required to take a satellite to a height $h$ above the Earth's surface (radius of Earth $R = 6.4 \times 10^3 \, km$) is $E_1$,and the kinetic energy required for the satellite to be in a circular orbit at this height is $E_2$. The value of $h$ for which $E_1 = E_2$ is:

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 $C$.

$A$ satellite is orbiting the Earth. If its orbital radius is reduced to half of its initial value,then the percentage change in its total energy is: (in $\%$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module