Question

Two identical satellites are at heights R and $7R$ from the earth’s surface. Then which of the following statements is correct:( $R = $ Radius of the earth)
A. Ratio of total energy of both is $5$
B. Ratio of kinetic energy of both is $6$
C. Ratio of potential energy of both is $9$
D. Ratio of total energy of both is $4$

Answer 1

Hint: In order to solve this question we need to know the gravitational potential concept. So according to Newton, Earth attracts everything toward it with a force known as gravitational force, so by symmetry from the electrical field lines, gravitational field lines do exist in space around a planet or an object of some mass. Similarly, analogy from electrical potential, gravitational potential do exist and gravitational field lines are in the direction of decreasing gravitational potential.

Complete step by step answer:
Let the mass of earth be $M$. So the gravitational potential at a height $h$ measured from center of earth is given by,
${U_g} = - \dfrac{{GM}}{h}$
Here, $G$ is a gravitational constant.
Since satellite one is at height of $R$ from surface of earth, so its height from center of earth is,
${h_1} = R + R$
$\Rightarrow {h_1} = 2R$
Similarly, satellite two is at height of $7R$ from surface of earth, so its height from center of earth is,
${h_1} = 7R + R$
$\Rightarrow {h_1} = 8R$
So the gravitational potential for satellite one is given by,
${U_{G1}} = - \dfrac{{GM}}{{{h_1}}}$

Putting values we get, ${U_{G1}} = - \dfrac{{GM}}{{2R}}$
Also, the gravitational potential for satellite two is given by,
${U_{G1}} = - \dfrac{{GM}}{{{h_2}}}$
Putting values we get, ${U_{G2}} = - \dfrac{{GM}}{{8R}}$
So the ratio of potential energy of both satellites is,
\[\dfrac{{{U_{G1}}}}{{{U_{G2}}}} = \dfrac{{ - \dfrac{{GM}}{{2R}}}}{{ - \dfrac{{GM}}{{8R}}}}\]
\[\Rightarrow \dfrac{{{U_{G1}}}}{{{U_{G2}}}} = \dfrac{{8R}}{{2R}}\]
\[\Rightarrow \dfrac{{{U_{G1}}}}{{{U_{G2}}}} = 4\]
So the ratio of potential energy for both is $4$.

Kinetic energy of any satellite is given as,
${K_G} = \dfrac{{ - 1}}{2}{U_G}$
So for satellite one, kinetic energy is given as,
${K_{G1}} = - \dfrac{1}{2}{U_{G1}}$
And, for satellite two, kinetic energy is given as,
${K_{G2}} = - \dfrac{1}{2}{U_{G2}}$
So the ratio is,
$\dfrac{{{K_{G1}}}}{{{K_{G2}}}} = \dfrac{{ - \dfrac{1}{2}{U_{G1}}}}{{ - \dfrac{1}{2}{U_{G2}}}}$
$\Rightarrow \dfrac{{{K_{G1}}}}{{{K_{G2}}}} = \dfrac{{{U_{G1}}}}{{{U_{G2}}}}$
Putting values we get, $\dfrac{{{K_{G1}}}}{{{K_{G2}}}} = 4$
So, the ratio of kinetic energy for both is $4$.

Since total energy of any satellite is given as,
$T = \dfrac{1}{2}{U_G}$
So for satellite one, total energy is given as,
${T_{G1}} = \dfrac{1}{2}{U_{G1}}$
And, for satellite two, total energy is given as,
${T_{G2}} = \dfrac{1}{2}{U_{G2}}$
So the ratio is,
$\dfrac{{{T_{G1}}}}{{{T_{G2}}}} = \dfrac{{\dfrac{1}{2}{U_{G1}}}}{{\dfrac{1}{2}{U_{G2}}}}$
$\Rightarrow \dfrac{{{T_{G1}}}}{{{T_{G2}}}} = \dfrac{{{U_{G1}}}}{{{U_{G2}}}}$
Putting values we get,
$\therefore \dfrac{{{T_{G1}}}}{{{T_{G2}}}} = 4$
So, the ratio of total energy for both is $4$.

So the correct option is D.

Note: It should be remembered that here we have assumed that the motion is analyzed in inertial frame of reference, Inertial frame are those frame which moves with constant velocity and if this velocity is very less in comparison to light then Newton’s law would be valid otherwise it would prove to be invalid in relativistic case.