Question

What should be the new radius of the earth, in order to reduce the escape velocity to half of the present value without changing the mass of the earth?

Answer 1

Hint: First of all you should know what is meant by escape velocity. We can define escape velocity as the minimum velocity required to escape from the gravitational pull of the earth. Since it refers to minimum velocity it depends on the gravitational constant, mass of the earth and radius of the earth.

Complete step-by-step solution:
The escape velocity of a body is given by the formula,
\[{{V}_{esc}}=\sqrt{\dfrac{2GM}{R}}\]…………..(1)
Also given that the the escape velocity then reduces to half of the present value without changing the mass of the earth,
Hence this can be written as,
$V_{esc}^{'}=\dfrac{{{V}_{esc}}}{2}$………………..(2)
Substituting equation (1) in (2) we get,
\[{{V}_{esc}}^{'}=\dfrac{\sqrt{\dfrac{2GM}{R}}}{2}\]
\[{{V}_{esc}}^{'}=\sqrt{\dfrac{2GM}{4R}}\]
In this relation the numerator ‘G’ is the gravitational constant, ‘M’ is the mass of the earth.
That is, both of them are constant quantities.
\[{{V}_{esc}}^{'}=k\sqrt{\dfrac{1}{4R}}\]
Where k is another constant.
Therefore we can say that when the radius of the earth becomes 4 times greater than the initial then the escape velocity will reduce to its half value.

Note:We must also remember the value of gravitational constant as it is an important quantity in gravitation. That is, its value is $6.67\times {{10}^{-11}}{}^{N{{m}^{2}}}/{}_{k{{g}^{2}}}$. And the escape velocity has the same unit of velocity which is metre per second. The planet Pluto has the least escape velocity, whereas mercury has the highest escape velocity since it is the fastest planet.