Question

If ${{v}_{e}}$ is escape velocity and ${{v}_{o}}$ is the orbital velocity of a satellite for orbit close to the earth’s surface, then these are related by:
A. ${{v}_{o}}={{v}_{e}}$
B. ${{v}_{e}}=\sqrt{2{{v}_{o}}}$
C. ${{v}_{e}}=\sqrt{2}{{v}_{o}}$
D. ${{v}_{o}}=\sqrt{2}{{v}_{e}}$

Answer 1

Hint: For this question we will find the value of escape velocity such that when the particle reaches infinity it has non-negative total energy(zero for the equivalency case). And then we will find the value of orbital velocity close to the surface of earth by balancing the gravitational force with the centripetal force. And then we will compare the two.

Formula used:
Kinetic energy: $\dfrac{m{{v}^{2}}}{2}$
Gravitational potential energy: $-\dfrac{GMm}{\operatorname{R}}$
Centripetal force: \[\dfrac{m{{v}^{2}}}{\operatorname{R}}\]
Gravitational force: \[\dfrac{GMm}{{{\operatorname{R}}^{2}}}\]

Complete step-by-step answer:
First, let us find the escape velocity from the surface of the earth. We will take the energy of the particle at the surface of the earth which will be the sum of gravitational potential energy (we will take reference at infinity) and its kinetic energy. This must be greater than zero because when the object reaches infinity it will have zero potential energy, but its total energy remains constant.
So, $\dfrac{m{{v}_{e}}^{2}}{2}-\dfrac{GMm}{{{\operatorname{R}}_{e}}}\ge 0$
Here M is the mass of the earth and m is the mass of the object. \[{{\operatorname{R}}_{e}}\] is the radius of the earth as we are taking the point of launch as the surface of the earth. At escape velocity the particle just reaches infinity so we will take the condition where energy is equal to zero. We will replace the value of g here.
$\begin{align}
  & \dfrac{m{{v}_{e}}^{2}}{2}-\dfrac{GMm}{{{\operatorname{R}}_{e}}}=0 \\
 & \dfrac{m{{v}_{e}}^{2}}{2}=\dfrac{GMm}{{{\operatorname{R}}_{e}}}=mg{{\operatorname{R}}_{e}}\Rightarrow {{v}_{e}}=\sqrt{2g{{\operatorname{R}}_{e}}} \\
\end{align}$
Now, let us find the orbital velocity of a satellite orbiting close to the surface of earth. We will take the radius of the motion of the satellite to be the same as that of earth in our calculations. We will equate the force of gravity and the centripetal force to find the orbital velocity
\[\begin{align}
  & \dfrac{m{{v}_{o}}^{2}}{{{\operatorname{R}}_{e}}}=\dfrac{GMm}{{{\operatorname{R}}_{e}}^{2}}=mg \\
 & {{v}_{o}}=\sqrt{g{{\operatorname{R}}_{e}}} \\
\end{align}\]
So, we get that ${{v}_{e}}$ is $\sqrt{2}$ times ${{v}_{o}}$.

So, the correct answer is “Option C”.

Note: Special care must be taken to avoid calculation mistakes. Also take care that when we take the reference as infinity for gravitational potential energy, the gravitational potential energy near the surface of earth will be negative as potential energy is always taken with a reference. The common formula of mgh will not be used as the potential energy is considered as zero at the surface in that case.