Question

Gravitational acceleration on the surface of a planet is $\dfrac{\sqrt{6}}{11}g$, where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\dfrac{2}{3}$ times that of the earth. If the escape speed on the surface of earth is taken to be $11km\,{{s}^{-1}}$ what will be the escape speed on the surface of this planet?

Answer 1

The gravitational force is defined as the force which acts between two masses. It is always attractive.Escape speed of a moving body is the minimum speed required to escape from the gravitational field of a planet and move into space.We use the relation of gravitational acceleration and escape velocity to obtain the escape velocity on the planet.
Formula used:
Acceleration due to gravity on the surface of a planet is ${{g}_{p}}=\dfrac{G{{M}_{P}}}{{{R}_{p}}^{2}}$ and,the escape velocity from the surface of the planet is ${{v}_{p}}=\sqrt{\dfrac{2G{{M}_{p}}}{{{R}_{p}}}}$

Complete answer:
Acceleration due to gravity is the acceleration of the object due to gravitational force of a planet.
Acceleration due to gravity on a planet of mass ${{M}_{p}}$ and radius ${{R}_{p}}$ is given by relation
${{g}_{p}}=\dfrac{G{{M}_{P}}}{{{R}_{p}}^{2}}$ where G is the gravitational constant
The escape velocity of an object from the surface of the planet is
${{v}_{p}}=\sqrt{\dfrac{2G{{M}_{p}}}{{{R}_{p}}}}$
Mass density is defined as volumetric mass and given by
$\rho =\dfrac{m}{V}$ where m and V denotes mass and volume of the material respectively.
Assuming a planet to be spherical, its mass density can be given as
${{\rho }_{p}}=\dfrac{{{M}_{p}}}{\dfrac{4}{3}\pi {{R}_{p}}^{3}}$
Substituting this in (1) and rearranging, we get
${{R}_{p}}=\dfrac{3{{g}_{p}}}{4\pi G{{\rho }_{p}}}$
Now we substitute this value in (2) and get
${{v}_{p}}=\sqrt{\dfrac{3g_{p}^{2}}{2\pi G{{\rho }_{p}}}}$
Similarly acceleration due to gravity of earth is given by’
${{v}_{e}}=\sqrt{\dfrac{3g_{e}^{2}}{2\pi G{{\rho }_{e}}}}$
Taking ratio of ${{v}_{p}}$ and ${{v}_{e}}$ and substituting the given values; we get
$\dfrac{{{v}_{p}}}{{{v}_{e}}}=\sqrt{\dfrac{{{({{g}_{p}}/{{g}_{e}})}^{2}}}{{{\rho }_{p}}/{{\rho }_{e}}}}=\sqrt{\dfrac{{{(\sqrt{6}/11)}^{2}}}{2/3}}$
$\Rightarrow {{v}_{p}}=\sqrt{\dfrac{(6/121)}{2/3}}\,{{v}_{e}}=\dfrac{3}{11}\times 11km/s$
$\Rightarrow {{v}_{p}}=3km/s$
We conclude that escape velocity on the planet is 3 km /s.

Note:
Escape velocity is the minimum speed required to get out from the gravitational field of the planet. Escape velocity is independent of mass of the object. It depends on the radius of the planet and acceleration due to gravitational force of the planet.
Escape velocity on the surface of the earth is 11.2 km/s.