Question

What is the dimensional formula of escape velocity?

Answer 1

To solve this question we will have to understand escape velocity.Escape velocity is the minimum velocity required by a body to be projected to overcome the gravitational pull of earth. It is the minimum velocity required by an object to escape the gravitational field, that is, escape the land without falling back, ignoring the losses due to the atmosphere. We will substitute the dimensions of each physical quantity into the formula ${U_i} = - \dfrac{{GMm}}{R}$ to find the dimensions of escape velocity.

Complete answer:
We can calculate escape velocity by conservation of energy, the initial potential energy on the surface of earth due to gravity is,
${U_i} = - \dfrac{{GMm}}{R}$
The initial kinetic energy on the surface of earth is,
${K_i} = \dfrac{1}{2}m{V_E}^2$
The final potential energy and the final kinetic energy in the interstellar space is zero
${U_f} = {K_f} = 0$
Applying the law of conservation of energy.
${U_i} + {K_i} = {U_f} + {K_f}$
$\Rightarrow - \dfrac{{GMm}}{R} + \dfrac{1}{2}m{V_E}^2 = 0$
$\Rightarrow {V_E} = \sqrt {\dfrac{{2GM}}{R}}$
Where, $G$ is Gravitational constant $6.67408 \times {10^{ - 11}}{\text{ }}{{\text{m}}^{\text{3}}}{\text{k}}{{\text{g}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 2}}}}$, ${V_E}$ is escape velocity, $M$ is the mass of the earth and $R$ is the radius of earth.

So, we have the formula for escape velocity as,
${V_E} = \sqrt {\dfrac{{2GM}}{R}}$.
Now, we will substitute the dimensions of each physical quantity and find the dimensional formula of escape velocity.
Dimensional formula of the earth’s mass $= {M^1}{L^0}{T^0}$
Dimensional formula of universal gravitational constant $= {M^{ - 1}}{L^3}{T^{ - 2}}$
Dimensional formula of the radius of the earth $= {M^0}{L^1}{T^0}$
So, we have, ${V_E} = \sqrt {\dfrac{{2GM}}{R}}$
Substituting the dimensions of G, M and R into the formula to find the dimensions of escape velocity, we get,
$\Rightarrow {V_E} = \sqrt {\dfrac{{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]\left[ {{M^1}{L^0}{T^0}} \right]}}{{\left[ {{M^0}{L^1}{T^0}} \right]}}}$

Simplifying the calculations using the laws of exponents ${a^x} \times {a^y} = {a^{x + y}}$, we get,
$\Rightarrow {V_E} = \sqrt {\dfrac{{\left[ {{M^{ - 1 + 1}}{L^{3 + 0}}{T^{ - 2 + 0}}} \right]}}{{\left[ {{M^0}{L^1}{T^0}} \right]}}}$
$\Rightarrow {V_E} = \sqrt {\dfrac{{\left[ {{M^0}{L^3}{T^{ - 2}}} \right]}}{{\left[ {{M^0}{L^1}{T^0}} \right]}}}$
Using the law of logarithm $\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}$, we get,
$\Rightarrow {V_E} = \sqrt {\left[ {{M^{0 - 0}}{L^{3 - 1}}{T^{ - 2 - 0}}} \right]}$
Simplifying the calculations, we get,
$\Rightarrow {V_E} = \sqrt {\left[ {{M^0}{L^2}{T^{ - 2}}} \right]}$
Taking the square root,
$\therefore {V_E} = {M^0}{L^1}{T^{ - 1}}$
So, the dimension of escape velocity of Earth is ${V_E} = {M^0}{L^1}{T^{ - 1}}$.

Therefore, the dimensional formula of escape speed is ${M^0}{L^1}{T^{ - 1}}$.**

Note: We have seen rockets going into space leaving the earth. We also noticed that a very huge kick-start is required to leave the surface of the earth because of the strong gravitational field. This is because of the escape velocity. There exists a relationship between escape velocity and orbital velocity. The relationship between the escape velocity and orbital velocity is proportional in nature. Escape velocity refers to the minimum velocity needed to overcome the gravitational pull of the mass to fly to infinite space. Orbital velocity is a velocity that is required to rotate around a massive body. This means if the orbital velocity increases, the escape velocity also increases and if orbital velocity decreases, the escape velocity also decreases.