Question

What is the relation between escape and orbital velocity?

Answer 1

In order to obtain the relation between the orbital velocity and escape velocity we first need to know the expression for orbital velocity as well as the escape velocity with common parameters. Assume that the satellites revolve close to the surface of the Earth. Further accordingly by taking the ration between the two we will obtain the required expression.

Formula used:
${{v}_{o}}=\sqrt{\dfrac{g{{R}^{2}}}{R+h}}$
${{v}_{e}}=\sqrt{2gR}$

Complete answer:
Orbital velocity is the velocity required to put the satellite into its orbit around the Earth.
The orbital velocity for a satellite at a height ‘h’ from the surface of the Earth is given by
${{v}_{o}}=\sqrt{\dfrac{g{{R}^{2}}}{R+h}}$
Where ‘R’ is the radius of the Earth and ‘g’ is acceleration due to gravity.
For a satellite moving just above the Earth’s surface h=0. Hence the expression for orbital velocity becomes,
$\begin{aligned} & {{v}_{o}}=\sqrt{\dfrac{g{{R}^{2}}}{R+h}} \\\ & \because h=0 \\\ & \therefore {{v}_{o}}=\sqrt{\dfrac{g{{R}^{2}}}{R}}=\sqrt{gR}....(1) \\\ \end{aligned}$
The escape velocity for a body to escape the force of gravity of the Earth is given by
${{v}_{e}}=\sqrt{2gR}.....(2)$
Taking ratio of equation 1 and 2 we get,
$\begin{aligned} & \dfrac{{{v}_{o}}}{{{v}_{e}}}=\dfrac{\sqrt{gR}}{\sqrt{2gR}} \\\ & \Rightarrow \dfrac{{{v}_{o}}}{{{v}_{e}}}=\dfrac{1}{\sqrt{2}} \\\ & \therefore {{v}_{e}}=\sqrt{2}{{v}_{o}} \\\ \end{aligned}$
Hence from the above equation, we can imply that the escape velocity is greater than the orbital velocity.

Note: It is to be noted that the escape velocity is independent of the mass of the body. This basically means that if we want an ant or an elephant to escape the gravitational potential, they both must be projected with the same velocity. The same holds true for orbital velocity as well, that is they both are dependent on the mass and the radius of the planet having different proportionality.