Question

A space station is at height equal to the radius of the earth. If ${{v}_{E}}$ is the escape velocity on the surface of the earth the same on the space station is -----times ${{v}_{E}}$.
$\begin{aligned} & A.\text{ }\dfrac{1}{2} \\\ & B.\text{ }\dfrac{1}{4} \\\ & C.\text{ }\dfrac{1}{\sqrt{2}} \\\ & D.\text{ }\dfrac{1}{\sqrt{3}} \\\ \end{aligned}$

Answer 1

The definition of the escape velocity states that it is the speed at which an object travels to break free from the either the planet’s gravity and leave without any development of the propulsion the equation of escape velocity is obtained by equating the kinetic energy of an object with the mass m and travelling with a velocity of v and the gravitational potential energy of the same object.

Formula used:
${{v}_{E}}=\sqrt{\dfrac{2GM}{r}}$

Complete answer:
As we know that formula for escape velocity is,
${{v}_{E}}=\sqrt{\dfrac{2GM}{r}}$
Where, ${{v}_{E}}$= escape velocity.
G = universal gravitational constant
M = mass of the planet.
r = distance from the center of the object to the center of the earth.
On the surface of the earth the distance from the center of the object is very small as compared to the radius of the earth.
Therefore we can rewrite equation (1) as
${{v}_{E}}=\sqrt{\dfrac{2GM}{R}}.....\left( 2 \right)$,
Where, R = radius of the earth.
Now if the object is at Height h from surface of the earth, then the formula for escape velocity can be written as,
${{v}_{E}}=\sqrt{\dfrac{2GM}{R+h}}.....\left( 3 \right)$
Now in the case of the space station the height h will be equal to the radius of the earth.
$\therefore h=R$
Substituting value of h in the equation (3)
$\begin{aligned} & {{v}_{E}}'=\sqrt{\dfrac{2GM}{R+R}} \\\ & {{v}_{E}}'=\sqrt{\dfrac{2GM}{2R}} \\\ \end{aligned}$
Taking$\dfrac{1}{\sqrt{2}}$ common from the equation
${{v}_{E}}'=\dfrac{1}{\sqrt{2}}\sqrt{\dfrac{2GM}{R}}...\left( 4 \right)$
Now substitute value of the equation (3) in the equation (4)
${{v}_{E}}'=\dfrac{1}{\sqrt{2}}{{v}_{E}}$

Therefore option (C) $\dfrac{1}{\sqrt{2}}$ is correct.

Note:
As an approach to solve this question first we write equation for the escape velocity on the surface of the earth then we find the equation of the escape velocity of the space station and by comparing both the equation we found that escape velocity of the space station is $\dfrac{1}{\sqrt{2}}$times of the escape velocity on the surface of the earth.