Question

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

Answer 1

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
$K.E. = \dfrac{1}{2}m{v^2}$
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.
${F_G} = \dfrac{{GMm}}{{{r^2}}}$
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.
${F_C} = \dfrac{{m{v^2}}}{r}$
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.
$\dfrac{1}{2}m{v_e}^2 = \dfrac{{GMm}}{{{r^2}}} \times r$
After further evaluation, we get
${v_e} = \sqrt {\dfrac{{2GM}}{r}} \cdot \cdot \cdot \cdot (1)$
Now,
For the orbital velocity of a planet, we can equate the centripetal force on a body and the gravitational force on the body.
$\dfrac{{m{v_o}^2}}{r} = \dfrac{{GMm}}{{{r^2}}}$
After further evaluation, we get
${v_o} = \sqrt {\dfrac{{GM}}{r}} \cdot \cdot \cdot \cdot (2)$
Now,
Applying $\dfrac{{Equation(1)}}{{Equation(2)}}$ , we get
$\dfrac{{{v_e}}}{{{v_o}}} = \sqrt 2$

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 $\dfrac{1}{{\sqrt 2 }}$.