Question 102: Gravitation

Question

A planet having mass $$9 \mathrm{Me}$$ and radius $$4 \mathrm{R}_{\mathrm{e}}$$, where $$\mathrm{Me}$$ and $$\mathrm{Re}$$ are mass and radius of earth respectively, has escape velocity in $$\mathrm{km} / \mathrm{s}$$ given by:
(Given escape velocity on earth $$\mathrm{V}_{\mathrm{e}}=11.2 	imes 10^{3} \mathrm{~m} / \mathrm{s}$$ )

Accepted Answer

The escape velocity on a planet is given by the following formula:
$$v_{esc} = \sqrt{\frac{2GM}{R}}$$
where $v_{esc}$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.
For Earth, we are given that $v_e = 11.2 	imes 10^3$ m/s. We know that the mass of the planet in question is $9M_e$ and its radius is $4R_e$. Let's find the escape velocity of this planet:
$$v_{esc} = \sqrt{\frac{2G(9M_e)}{4R_e}}$$
Divide both sides by the Earth's escape velocity formula:
$$\frac{v_{esc}}{v_e} = \frac{\sqrt{\frac{2G(9M_e)}{4R_e}}}{\sqrt{\frac{2GM_e}{R_e}}}$$
Simplify:
$$\frac{v_{esc}}{11.2 	imes 10^3 	ext{ m/s}} = {\sqrt{\frac{9}{4}}}$$
$$\frac{v_{esc}}{11.2 	imes 10^3 	ext{ m/s}} = \frac{3}{2}$$
Now, solve for $v_{esc}$:
$$v_{esc} = 11.2 	imes 10^3 	ext{ m/s} 	imes \frac{3}{2}$$
$$v_{esc} = 16.8 	imes 10^3 	ext{ m/s}$$
Converting this to km/s, we get:
$$v_{esc} = 16.8 	ext{ km/s}$$

Answer

33.6

Answer

11.2

Answer

16.8

Answer

67.2

Gravitation

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