Question

Calculate the escape velocity of a body from the surface of earth. [Average density of earth $= 5.5 \times {10^3}\,kg{m^{ - 3}}$ , $G = 6.67 \times {10^{11}}\,N{m^2}\,k{g^{ - 2}}$ , radius of earth $R = 6.4 \times {10^6}\,m$]

Answer 1

Escape velocity is defined as a lowest velocity that is used by a body must to escape the gravitational pull of the earth. At first we will calculate the volume of earth and then the mass of the earth. Then we will use the formula of escape velocity to calculate the escape velocity of a body from the surface of earth.

Formula used:
The formula used for calculating the volume of earth is given below;
$V = \dfrac{4}{3}\pi {R^3}$
Here, $V$ is the volume of the earth and $R$ is the radius of earth.
The formula used for calculating the mass of earth is given below;
$M = \rho V$
Here, $M$ is the mass of earth, $\rho$ is the density of earth and $V$ is the volume of earth.
Also, the formula used for calculating the escape velocity is given below;
${V_{esc}} = \sqrt {\dfrac{{2GM}}{R}}$
Here, ${V_{esc}}$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of earth and $R$ is the radius of earth.

Complete step by step answer:
Consider a body that will be placed on the surface of earth. The given terms in the questions are given below;
The density of earth, $\rho = 5.5 \times {10^3}\,kg{m^{ - 3}}$
The gravitational constant, $G = 6.67 \times {10^{11}}\,N{m^2}\,k{g^{ - 2}}$
Also, the radius of earth, $R = 6.4 \times {10^6}\,m$
Now, the volume of the earth is calculated as shown below;
$V = \dfrac{4}{3}\pi {R^3}$
$\Rightarrow \,V = \dfrac{4}{3} \times 3.14 \times {\left( {6.4 \times {{10}^6}} \right)^3}$
$\Rightarrow \,V = \dfrac{4}{3} \times 3.14 \times 262.14 \times {10^{18}}$
$\Rightarrow \,V = 1094.74 \times {10^{18}}$
Now the mass of the earth can be calculated as shown below;
$M = \rho V$
$\Rightarrow \,M = 5.5 \times {10^3} \times 1094.74 \times {10^{18}}$
$\Rightarrow \,M = 6021.07 \times {10^{21}}$
Now, the escape velocity of a body on the surface of earth is calculated below;
${V_{esc}} = \sqrt {\dfrac{{2GM}}{R}}$
$\Rightarrow \,{V_{esc}} = \sqrt {\dfrac{{2 \times 6.67 \times {{10}^{ - 11}} \times 6021.07 \times {{10}^{21}}}}{{6.4 \times {{10}^6}}}}$
$\Rightarrow \,{V_{esc}} = 11219.75\,m{s^{ - 1}}$
$\therefore \,{V_{esc}} = 11.219\,km\,{s^{ - 1}}$

Therefore, the escape velocity of a body on the surface of earth is $11.219\,km\,{s^{ - 1}}$.

Note: Here, we can also change the radius of earth into $kilometer$ rather than changing the escape velocity into $kilometer$. Also, it is important to calculate the volume of earth so that we can calculate the mass of the earth. Using this value of mass we have calculated the escape velocity of the body on the surface of earth.