Question

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Give that escape velocity from the earth is$11{km}/{s}\;$, the escape velocity from the surface of the planets is

& A.\,100{km}/{s}\; \\\ & B.\,110{km}/{s}\; \\\ & C.\,120{km}/{s}\; \\\ & D.\,130{km}/{s}\; \\\ \end{aligned}$$

Answer 1

The formula for computing the escape velocity of the planet should be used to solve this problem. The escape velocity of the planet gives the relation between the gravitational constant, the mass of the planet and the radius of the planet, so, using this formula, we can compute the value of the escape velocity from the surface of the planets.
Formula used:
${{v}_{e}}=\sqrt{\dfrac{2G{{M}_{e}}}{{{R}_{e}}}}$

Complete step by step answer:
From the given information, we have the data as follows.
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. The escape velocity from the earth is$11{km}/{s}\;$.
The mass of the planet, ${{M}_{p}}=10{{M}_{e}}$
Where ${{M}_{e}}$ is the mass of the earth.
The radius of the planet, $10{{R}_{p}}={{R}_{e}}$
Where ${{R}_{e}}$ is the radius of the earth.
The escape velocity of the earth gives the relation between the gravitational constant, the mass of the earth and the radius of the earth. The mathematical representation of the same is given as follows.
${{v}_{e}}=\sqrt{\dfrac{2G{{M}_{e}}}{{{R}_{e}}}}$
The escape velocity of the planet gives the relation between the gravitational constant, the mass of the planet and the radius of the planet.
${{v}_{p}}=\sqrt{\dfrac{2G{{M}_{p}}}{{{R}_{p}}}}$
Represent the above equation in terms of the escape velocity of the earth.

& {{v}_{p}}=\sqrt{\dfrac{2G{{M}_{p}}}{{{R}_{p}}}} \\\ & \Rightarrow {{v}_{p}}=\sqrt{\dfrac{100\times 2G{{M}_{e}}}{{{R}_{e}}}} \\\ & \therefore {{v}_{p}}=\sqrt{100}{{v}_{e}} \\\ \end{aligned}$$ Substitute the value of the escape velocity of the earth. $$\begin{aligned} & {{v}_{p}}=10{{v}_{e}} \\\ & \Rightarrow {{v}_{p}}=10\times 11 \\\ & \therefore {{v}_{p}}=110\,{km}/{s}\; \\\ \end{aligned}$$ $$\therefore $$ The value of the escape velocity from the surface of the planets is, $$110\,{km}/{s}\;$$ **So, the correct answer is “Option B”.** **Note:** The escape velocity of the planet is same as the escape velocity of the earth, the difference being, if we substitute the values of the mass and the radius of the earth, then, the escape velocity formula gives the escape velocity of object from the surface of the earth.