Question

A mass of $6 \times {10^{24}}\;{\text{kg}}$(equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is $3 \times {10^8}\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$. What should be the radius of the sphere.

Answer 1

In this question, the concept of the escape velocity will be used, that is, it is the minimum velocity required by the object to escape from the given object gravitational influence. It depends on the mass and the radius of the object. The value of universal gravitational constant $G$ is $6.67 \times {10^{ - 11}}\;{\text{Nk}}{{\text{g}}^{ - 2}}{{\text{m}}^2}$.

Complete step by step answer:
In this question, we have given, the mass of the spherical body as $6 \times {10^{24}}\;{\text{kg}}$ which is equal to the mass of the earth and the escape velocity from its surface is $3 \times {10^8}\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$. In this problem, we need to calculate the radius of the sphere.

As we know that the escape velocity is the minimum velocity required by the object to escape from the given object gravitational influence.

In this question we will use the escape velocity formula as,
$\Rightarrow V_e^2 = \dfrac{{2GM}}{R}$
Further rearrange the formula we get,
$\Rightarrow R = \dfrac{{2GM}}{{V_e^2}}$
As we know here, $G$ is the universal Gravitational constant, $M$ is mass, ${V_e}$ is escape velocity, and $R$ is the radius of the sphere. As we know that the value of universal gravitational constant $G$ is $6.67 \times {10^{ - 11}}\;{\text{Nk}}{{\text{g}}^{ - 2}}{{\text{m}}^2}$.

Now we substitute the value of $m,G$ and ${V_e}$ in above equation as,
$\Rightarrow R = \dfrac{{2\left( {6.67 \times {{10}^{ - 11}}} \right)\left( {6 \times {{10}^{24}}} \right)}}{{{{\left( {3 \times {{10}^8}} \right)}^2}}}$
On simplification of the above expression we get,
$\Rightarrow R = 8.89 \times {10^{ - 3}}\;{\text{m}}$
Now, we convert the radius from meter to millimeter as,
$\Rightarrow R = \left( {8.89 \times {{10}^{ - 3}}\;{\text{m}}} \right)\left( {\dfrac{{1000\;{\text{mm}}}}{{1\;{\text{m}}}}} \right)$
After simplification we get,
$\therefore R = 8.89\;{\text{mm}}$
Therefore, the radius of sphere is $8.89\;{\text{mm}}$

Note: As we know that the escape velocity from the earth is $1.1186 \times {10^3}\;{\text{m/s}}$, but here the escape velocity is much greater that is $3 \times {10^8}\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$, so the radius of the given sphere object is very small that is $8.89\;{\text{mm}}$ while the radius of the Earth is $6371\;{\text{km}}$.