Question

If the mass of the planet is eight times the mass of the Earth and it's radius is twice of the Earth,what will be the escape velocity for that planet?​

Answer 1

Answer :

• The escape velocity of the planet X is 4 times of the escape velocity of Earth.

Explanation :

Given :

• Mass of the planet X = 8 times the mass of the Earth.
• Radius of the planet X = 2 times the radius of the Earth.

To find :

• The escape velocity of that planet with respect to the earth.

Knowledge required :

Formula for escape velocity of a planet :

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline{\boxed{\sf{v_{e} = \sqrt{\dfrac{2GM}{R}}}}}$

Where,

• $\sf{v_{e}}$ = Escape velocity of the planet.
• M = Mass of the planet.
• R = Radius of the planet.
• G = Universal Gravitational constant.

Solution :

Let the mass of the Earth be m kg and the radius of the Earth be r m.

So,

The mass of the planet X will be 8m kg and radius of the planet X will be 2r m.

Now, Let's find out the escape velocity of both the planets.

• Escape velocity of the Earth :

By using the formula for escape velocity of a planet and substituting the values in it, we get :

$:\implies \sf{v_{e} = \sqrt{\dfrac{2GM}{R}}} \\ \\ :\implies \sf{v_{E} = \sqrt{\dfrac{2Gm}{r}}} \\ \\\boxed{\therefore \sf{v_{E} = \sqrt{\dfrac{2Gm}{r}}}}$

Hence the escape velocity of the Earth is .

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf{v_{E} = \sqrt{\dfrac{2Gm}{r}}}$

• Escape velocity of the planet X :

By using the formula for escape velocity of a planet and substituting the values in it, we get :

$:\implies \sf{v_{X} = \sqrt{\dfrac{2GM}{R}}} \\ \\ :\implies \sf{v_{X} = \sqrt{\dfrac{2G \times 8m}{2r}}} \\ \\ :\implies \sf{v_{X} = \sqrt{\dfrac{8Gm}{r}}} \\ \\ \boxed{\therefore \sf{v_{X} = \sqrt{\dfrac{8Gm}{r}}}}$

Hence the escape velocity of the planet X is .

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf{v_{E} = \sqrt{\dfrac{8Gm}{r}}}$

Now let's compare the escape velocity of both the planets.

$:\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{\sqrt{\dfrac{2Gm}{r}}}{\sqrt{\dfrac{8Gm}{r}}}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{\dfrac{2 \times G \times m}{r}}{\dfrac{8 \times G \times m}{r}}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{2 \times G \times m}{r} \times \dfrac{r}{8 \times G \times m}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{\not{2} \times \not{G} \times \not{m}}{\not{r}} \times \dfrac{\not{r}}{\not{8} \times \not{G} \times \not{m}}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{1}{4}} \\ \\ :\implies \sf{4v_{E} = v_{X}} \\ \\ \boxed{\therefore \sf{4v_{E} = v_{X}}}$

Hence the escape velocity of the planet X is 4 times of the escape velocity of Earth.

Answer 2

Answer :

The escape velocity of the planet X is 4 times of the escape velocity of Earth.

Explanation :

Given :

Mass of the planet X = 8 times the mass of the Earth.

Radius of the planet X = 2 times the radius of the Earth.

To find :

The escape velocity of that planet with respect to the earth.

Knowledge required :

Formula for escape velocity of a planet :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\underline{\boxed{\sf{v_{e} = \sqrt{\dfrac{2GM}{R}}}}}$

Where,

$\sf{v_{e}}v$

= Escape velocity of the planet.

M = Mass of the planet.

R = Radius of the planet.

G = Universal Gravitational constant.

Solution :

Let the mass of the Earth be m kg and the radius of the Earth be r m.

So,

The mass of the planet X will be 8m kg and radius of the planet X will be 2r m.

Now, Let's find out the escape velocity of both the planets.

Escape velocity of the Earth :

By using the formula for escape velocity of a planet and substituting the values in it, we get :

$\begin{gathered}:\implies \sf{v_{e} = \sqrt{\dfrac{2GM}{R}}} \\ \\ :\implies \sf{v_{E} = \sqrt{\dfrac{2Gm}{r}}} \\ \\\boxed{\therefore \sf{v_{E} = \sqrt{\dfrac{2Gm}{r}}}} \\ \\ \end{gathered}$

Hence the escape velocity of the Earth is .

⠀⠀⠀⠀⠀⠀⠀⠀

$\begin{gathered}\sf{v_{E} = \sqrt{\dfrac{2Gm}{r}}} \\ \\ \end{gathered}$

Escape velocity of the planet X :

By using the formula for escape velocity of a planet and substituting the values in it, we get :

$\begin{gathered}:\implies \sf{v_{X} = \sqrt{\dfrac{2GM}{R}}} \\ \\ :\implies \sf{v_{X} = \sqrt{\dfrac{2G \times 8m}{2r}}} \\ \\ :\implies \sf{v_{X} = \sqrt{\dfrac{8Gm}{r}}} \\ \\ \boxed{\therefore \sf{v_{X} = \sqrt{\dfrac{8Gm}{r}}}} \\ \\ \end{gathered}$

Hence the escape velocity of the planet X is .

⠀⠀⠀⠀⠀⠀⠀⠀

$\begin{gathered}\sf{v_{E} = \sqrt{\dfrac{8Gm}{r}}} \\ \\ \end{gathered} </p><p>v$

Now let's compare the escape velocity of both the planets.

$\begin{gathered}:\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{\sqrt{\dfrac{2Gm}{r}}}{\sqrt{\dfrac{8Gm}{r}}}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{\dfrac{2 \times G \times m}{r}}{\dfrac{8 \times G \times m}{r}}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{2 \times G \times m}{r} \times \dfrac{r}{8 \times G \times m}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{\not{2} \times \not{G} \times \not{m}}{\not{r}} \times \dfrac{\not{r}}{\not{8} \times \not{G} \times \not{m}}} \\ \\ :\implies \sf{\dfrac{v_{E}}{v_{X}} = \dfrac{1}{4}} \\ \\ :\implies \sf{4v_{E} = v_{X}} \\ \\ \boxed{\therefore \sf{4v_{E} = v_{X}}} \\ \\ \end{gathered}$

Hence the escape velocity of the planet X is 4 times of the escape velocity of Earth.