Question

A satellite is revolving round the earth with orbital speed \[{{{v}}_0}\]. If it stops suddenly, find the speed with which it will strike the surface of earth (\[{{{v}}_{{e}}} = \] escape velocity of a particle on earth’s surface).

Answer 1

Hint: Let v be the velocity of the satellite which is revolving round the earth with some orbital speed. We know that the energy of the satellite, ${{E = }}\dfrac{{{{ - G}}{{{M}}_{{e}}}{{m}}}}{{{R}}}$. As the orbiting satellite is very close to be acted upon the gravity of the earth and centripetal force keeps it rotating round the earth. Finally, equate the energy of the satellite with centripetal force.

Complete step by step solution:
Given: \[{{{v}}_{{e}}} = \] escape velocity of a particle on earth’s surface
\[{{{v}}_0} = \] Orbital speed of the satellite
Let us consider that v be the velocity with which the satellite strikes the surface of the earth.
An orbiting satellite is close enough to be acted upon by Earth's gravity. This force is constantly pulling the satellite in towards the centre of the earth and the force is named as centripetal force which causes a centripetal acceleration.
This energy of the satellite is equivalent to the centripetal force,
$\Rightarrow \dfrac{{{{G}}{{{M}}_{{e}}}{{m}}}}{{{{{R}}^{{2}}}}}{{ = }}\dfrac{{{{m}}{{{v}}_{{0}}}^{{2}}}}{{{R}}}$
Multiplying both sides with R, we get
$\Rightarrow \dfrac{{{{G}}{{{M}}_{{e}}}{{m}}}}{{{R}}}{{ = }}\dfrac{{{{m}}{{{v}}_{{0}}}^{{2}}}}{1}$
Total energy of the satellite is given by,
$\Rightarrow {{E = }}\dfrac{{{{ - G}}{{{M}}_{{e}}}{{m}}}}{{{R}}}$
Thus, energy on the surface of the earth is given by conservation of mechanical energy
$\Rightarrow {{{E}}_{{T}}}{{ = }}\dfrac{{{{ - G}}{{{M}}_{{e}}}{{m}}}}{{{R}}}{{ + }}\dfrac{{{1}}}{{{2}}}{{m}}{{{v}}^{{2}}}$
Where ${{{E}}_{{T}}}$is the total energy
Now, total energy of the satellite
$\Rightarrow \dfrac{{{{ - G}}{{{M}}_{{e}}}{{m}}}}{{{R}}}{{ = m}}{{{v}}_{{0}}}^{{2}}$
$
\Rightarrow \dfrac{1}{2}m{v^2} - \dfrac{{G{M_e}m}}{R} = - m{v_0}^2 \\
 \Rightarrow {{{v}}^{{2}}}{{ = }}{{{v}}_{{e}}}^{{2}}{{ - 2}}{{{v}}_{{0}}}^{{2}} \\
  \therefore {{v = }}\sqrt {{{{v}}_{{e}}}^{{2}}{{ - 2}}{{{v}}_{{0}}}^{{2}}} $

Thus, the speed with which the satellite will strike surface of earth is ${{v = }}\sqrt {{{{v}}_{{e}}}^{{2}}{{ - 2}}{{{v}}_{{0}}}^{{2}}} $.

Note: According to the law of conservation of total mechanical energy the total mechanical energy in a system (the sum of the potential and kinetic energy) remains constant as long as the only forces acting are conservative forces. In short, the total energy of a satellite is just the sum of its gravitational potential energy and kinetic energy.