Question

What is the escape velocity to the orbital velocity ratio?
(A) $\sqrt{2}$
(B) ${\displaystyle \frac{1}{\sqrt{2}}}$
(C) $2$
(D) ${\displaystyle \frac{1}{2}}$

Answer 1

Hint: We are asked to find the ratio of escape velocity and orbital velocity. So we will derive the equations of the velocities and then find their ratio.
Formula Used
$K.E.={\displaystyle \frac{1}{2}}m{v}^2$
Where, $K.E.$ is the kinetic energy of a body, $m$ is the mass of the body and $v$ is the velocity of the body.
$F_G={\displaystyle \frac{GMm}{r^2}}$
Where, $F_G$ is the gravitational force, $G$ is the universal gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the body and $r$ is the radius of the planet.
$F_C={\displaystyle \frac{m{v}^2}{r}}$
Where, $F_C$ is the centripetal force on a body, $m$ is the mass of the body, $v$ is the velocity of the body and $r$ is the radius of motion.

Step By Step Solution
For finding the escape velocity of a planet, we should equate the kinetic energy of a body to the gravitational force on the body multiplied by the radius of the planet.
${\displaystyle \frac{1}{2}}m{v_e}^2={\displaystyle \frac{GMm}{r^2}}\times r$
After further evaluation, we get
$v_e=\sqrt{{\displaystyle \frac{2GM}{r}}}\cdot \cdot \cdot \cdot (1)$
Now,
For the orbital velocity of a planet, we can equate the centripetal force on a body and the gravitational force on the body.
${\displaystyle \frac{m{v_o}^2}{r}}={\displaystyle \frac{GMm}{r^2}}$
After further evaluation, we get
$v_o=\sqrt{{\displaystyle \frac{GM}{r}}}\cdot \cdot \cdot \cdot (2)$
Now,
Applying ${\displaystyle \frac{Equation(1)}{Equation(2)}}$ , we get
${\displaystyle \frac{v_e}{v_o}}=\sqrt{2}$

Hence, the answer is (A).

Note: Here we were asked to find the ratio of escape velocity to the orbital velocity of a planet, thus we got the ratio to be $\sqrt{2}$. But if the question was to find the ratio between orbital velocity to the escape velocity, then the answer will get reversed, that is the ratio becomes ${\displaystyle \frac{1}{\sqrt{2}}}$.