Question

The escape velocity '$v$' of a body depends upon (1) the acceleration due to gravity (g) of the planet and (2) the radius (R) of the planet. Use the method of dimensions to obtain a relation between $v, g$ and $R$.

Answer 1

Hint:This question is based on the concept of dimensional analysis. In this question, firstly we will apply the concept of dimensional analysis and compare the dimensions of $v, g$ and $R$. On finding the dimensions, we will arrive at the final expression which gives the relation between $v, g$ and $R$.

Complete step by step answer:
The following are the uses of dimensional analysis:
- To check the correctness of a physical relation
- To convert the value of a physical quantity from one system to another.
- To derive relations between various physical quantities.
- To find the dimensions of dimensional constants

Escape speed is the minimum speed with which a mass should be projected from the Earth’s surface in order to escape Earth’s gravitational field. Escape speed, also known as escape velocity can also be defined as: The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.We know the formula for escape velocity,
\[{v_e} = \sqrt {\dfrac{{2GM}}{R}} = \sqrt {gR} \]

Hence, we can write that,
\[[v] = {[g]^a}{[R]^b}......(1)\]
\[\Rightarrow [{L^{ - 1}}{T^{ - 1}}] = {[{L^1}{T^{ - 2}}]^a}{[{L^1}]^b}\]
On further simplifying this, we get,
\[[{L^{ - 1}}{T^{ - 1}}] = [{L^{a + b}}][{T^{ - 2a}}]\]
On comparing the power,
\[a + b = 1\]
And \[ - 2a = - 1\]
On solving the above two equations, we get,
\[a = \dfrac{1}{2}\] and \[b = \dfrac{1}{2}\]
On putting the above values in equation (1),
\[\therefore {v_e} = \sqrt {gR} \]

Therefore, the relation between $v, g$ and $R$ is given as ${v_e} = \sqrt {gR}$.

Note:The escape velocity of celestial bodies like planets and their natural satellites (the moon for us) is the minimum velocity that has to be achieved by an object, to escape the gravitational sphere of influence (pull or force or attraction) of the celestial body. At this velocity, the sum of the gravitational potential energy and kinetic energy of the system will be equal to zero.