Question

Escape velocity of an atmospheric particle which is $1000km$ above the earth’s surface, is (radius of earth is $6400km$ and $g = 10m/{s^2}$ )
(A) $6.5km/s$
(B) $8km/s$
(C) $10km/s$
(D) $11.2km/s$

Answer 1

Hint To solve this question, we have to use the definition of the escape velocity. Then we need to deduce the expression for the escape velocity using the energy conservation principle. Then on substituting the values which are given in the question, we get the value of the escape velocity.

Formula Used: The formula which is used in solving this question is given by
$U = - \dfrac{{Gm{M_e}}}{{{R_e} + h}}$ , here $U$ is the gravitational potential energy of an object of mass $m$ , which is situated at a height of $h$ above the surface of the earth having mass ${M_e}$ and radius ${R_e}$ .

Complete step by step answer
We know that the escape velocity of an object is defined as the minimum velocity required to be provided to that object so that it gets free from the gravitational influence of the earth. Let the escape velocity of the atmospheric particle be ${v_e}$ . Now, we know that the gravitational potential energy of an object located at some height above the earth’s surface is given by
$U = - \dfrac{{Gm{M_e}}}{{{R_e} + h}}$ (i)
When at the same location, the particle is given an escape velocity of ${v_e}$ , then the kinetic energy imparted to the particle is given by
$K = \dfrac{1}{2}m{v_e}^2$ (ii)
So the total mechanical energy of the particle at the location becomes
$E = K + U$
From (i) and (ii) we have
$E = \dfrac{1}{2}m{v_e}^2 - \dfrac{{Gm{M_e}}}{{{R_e} + h}}$ (iii)
Now, as we know that the gravitational potential energy is taken with respect to the infinity, where it is taken as zero. So, as we are providing the particle the minimum velocity, so the kinetic energy imparted is just sufficient to make the particle reach to infinity. Thus at infinity, the kinetic energy of the particle must become zero. As the potential energy is also zero at infinity, so the total energy of the particle at infinity becomes
${E_\infty } = K + U = 0$ (iv)
As no other force apart from the gravitational force acts on the particle, so from the energy conservation principle we get
$E = {E_\infty }$
From (iii) and (iv)
$\dfrac{1}{2}m{v_e}^2 - \dfrac{{Gm{M_e}}}{{\left( {{R_e} + h} \right)}} = 0$
$\Rightarrow \dfrac{1}{2}m{v_e}^2 = \dfrac{{Gm{M_e}}}{{\left( {{R_e} + h} \right)}}$
Cancelling $m$ from both the sides, we get
$\Rightarrow \dfrac{1}{2}{v_e}^2 = \dfrac{{G{M_e}}}{{\left( {{R_e} + h} \right)}}$
Multiplying by $2$ and taking square root both the sides, we get
${v_e} = \sqrt {\dfrac{{2G{M_e}}}{{\left( {{R_e} + h} \right)}}}$
Multiplying and dividing by ${R_e}^2$ inside the root
${v_e} = \sqrt {\dfrac{{2G{M_e}{R_e}^2}}{{{R_e}^2\left( {{R_e} + h} \right)}}}$ (v)
Now, we know that the acceleration due to gravity is given by
$g = \dfrac{{G{M_e}}}{{{R_e}^2}}$
Putting in (v) we get
${v_e} = \sqrt {\dfrac{{2g{R_e}^2}}{{\left( {{R_e} + h} \right)}}}$ (vi)
According to the question, we have $g = 10m/{s^2}$ ,
${R_e} = 6400km = 6.4 \times {10^6}m$ , and
$h = 1000km = {10^6}m$
Substituting these in (vi) we get
${v_e} = \sqrt {\dfrac{{2 \times 10 \times {{\left( {6.4 \times {{10}^6}} \right)}^2}}}{{\left( {6.4 \times {{10}^6} + {{10}^6}} \right)}}}$
On solving we get
${v_e} = 10521.54m/s$
This can be approximated as
${v_e} \approx 10km/s$
Thus, the escape velocity of the atmospheric particle is approximately equal to $10km/s$ .
Hence, the correct answer is option (C).

Note
We should not use the popular formula for the escape velocity, which is given by
${v_e} = \sqrt {2g{R_e}}$
This is an approximation for the escape velocity when the height of the particle above the earth’s surface is negligibly small compared to the radius of the earth. But in this question, we can clearly see that the height is comparable to the radius. Hence, we had to derive the exact expression.