Question

The total energy and kinetic energy of an Earth’s satellite are respectively
A. positive and negative
B. negative and positive
C. positive and positive
D. negative and negative
E. zero and positive.

Answer 1

The sum of kinetic energy and gravitational potential energy is required to solve the given problem.

Complete step-by-step answer:
Let us consider the total energy of the earth’s satellite.
The total energy of the Earth’s satellite is given as,
Total energy= kinetic energy + potential energy ……………… (1)
Now we will find the equation for the kinetic energy of an Earth’ satellite.
Kinetic energy, $K = \dfrac{1}{2}m{v_o}^2$ …………… (2)
Where ${v_o}$is orbital velocity of a satellite.
The speed with which a body revolves around the earth or any other celestial body in a given orbit is called its orbital velocity.
Consider a satellite of mass ‘m’ revolving around the earth with orbital velocity ${v_o}$ in a circular orbit of radius ‘r’. Let ‘M’ be the mass and ‘R’ be the radius of the earth. The centripetal force necessary for the satellite to revolve around the earth is provided by earth’s gravitational force on the satellite.
Therefore, centripetal force = Gravitational force
$\dfrac{{m{v_o}^2}}{r} = \dfrac{{GMm}}{{{r^2}}}$
${v_o} = \sqrt {\dfrac{{GM}}{r}}$
If ‘h’ is the height of the satellite from the surface from the surface of the earth, then $r = R + h$
$\therefore {v_o} = \sqrt {\dfrac{{GM}}{{R + h}}}$ ……………. (3)
For satellite close to the earth, $h \ll R$,so that $R + h \approx R$
${v_o} = \sqrt {\dfrac{{GM}}{R}}$
${v_o} = \sqrt {gR}$ (since $g = \dfrac{{GM}}{{{R^2}}}$)
Now we can substitute the value of ${v_o}$that is equation (3) in equation (2),
We get, $K = \dfrac{1}{2}m{v_o}^2$
$K = \dfrac{1}{2}m\left( {\dfrac{{GM}}{{R + h}}} \right) = \dfrac{{GMm}}{{2(R + h)}}$ …………… (4)
From the above equation we can say kinetic energy of an orbiting satellite is positive.
We know that, gravitational potential energy ${U}$ = $- \dfrac{{GMm}}{{(R + h)}}$
Now, let us substitute the value of kinetic energy and potential energy in equation (1),
Total energy = K + U
$E = \dfrac{{GMm}}{{2(R + h)}}$ $- \dfrac{{GMm}}{{(R + h)}}$
$E = - \dfrac{{GMm}}{{2(R + h)}}$
From this above equation we can say that the total energy of an earth’s satellite is negative.
Hence, option (B) negative and positive is the correct answer.

Note: The total energy of the satellite is negative because it is revolving round the earth at a finite distance from it. Only objects which escape from the gravitational field of the earth have positive energy.