Question

The escape speed for a planet is ${{v}_{0}}$. What is the escape speed corresponding to another planet of twice the radius and half the density?

Answer 1

Hint: As a first step, one could read the question well and hence note down important points on the given two planets. Then you could recall the standard expression for the escape velocity of a planet. Then make necessary changes as per the given question and hence take the ratio and hence get the answer.

Formula used:
Escape velocity,
${{v}_{0}}=\sqrt{\dfrac{2GM}{R}}$


Complete step-by-step solution:
In the question, we are given the escape speed for a certain planet to be ${{v}_{0}}$. We are supposed to find the escape speed of another planet that has twice the radius and half the other planet’s density.
Let us recall the standard expression for escape velocity of a planet to be given by,
${{v}_{0}}=\sqrt{\dfrac{2GM}{R}}$…………………………….. (1)
Now let us express the mass in terms of density and volume.
$M=\rho V=\rho \left( \dfrac{4}{3}\pi {{R}^{3}} \right)$…………………………………… (2)
Now we could substitute the equation (2) in equation (1) to get,
${{v}_{0}}=\sqrt{\dfrac{2G\left( \rho \left( \dfrac{4}{3}\pi {{R}^{3}} \right) \right)}{R}}$
That is, taking all the constants in the equation into account, we could say that,
${{v}_{0}}\propto \sqrt{\rho {{R}^{2}}}$……………………………………… (3)
This could be an expression for the escape velocity of the first planet. Now for the second planet,
${{v}_{0}}'\propto \sqrt{\rho 'R{{'}^{2}}}$
From the given conditions of the question,
${{v}_{0}}'\propto \sqrt{\left( \dfrac{\rho }{2} \right){{\left( 2R \right)}^{2}}}$………………………………….. (4)
Now taking the ratio of (3) and (4), we get,
$\dfrac{{{v}_{0}}'}{{{v}_{0}}}=\dfrac{\sqrt{\left( \dfrac{\rho }{2} \right){{\left( 2R \right)}^{2}}}}{\sqrt{\rho {{R}^{2}}}}$
$\Rightarrow \dfrac{{{v}_{0}}'}{{{v}_{0}}}=\dfrac{\sqrt{2}}{1}$
$\therefore {{v}_{0}}'=\sqrt{2}{{v}_{0}}$
Therefore, we found the escape velocity of the second planet to be $\sqrt{2}$times that of the escape velocity of the first planet.

Note: Now, let us recall the definition of escape velocity. Escape velocity is known to be the minimum speed that is required for free propulsion in order to escape from the gravitational influence of a certain massive body and hence eventually attaining an infinite distance from it.