Question

Mass and radius of a planet are two times the value of earth. What is the value of escape velocity from the surface of this planet?

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

As we know that :

$\large{\boxed{\boxed{\sf{V_e \: = \: \sqrt{\dfrac{2GM}{R}}}}}}$

Where,

• M is mass of earth
• R is radius of earth
• G is gravitational constant

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Now, take case when a planet have mass and radius double that of earth.

Let,

• Mass of planet be M'. So, M' = 2M
• Radius of planet be R'. So, R' = 2R

Put Values in Escape Velocity Formula :

$\implies {\boxed{\sf{V_e ' \: = \: \sqrt{\dfrac{2GM'}{R'}}}}} \\ \\ \implies {\sf{V_e ' \: = \: \sqrt{\dfrac{2G \: \times \: 2M}{2R}}}} \\ \\ \implies {\sf{V_e ' \: = \: \sqrt{\dfrac{2GM}{R}}}} \\ \\ \implies {\sf{V_e ' \: = \: V_e}}$

So, escape velocity on that planet and that of earth are same

Answer 2

Answer:

Given:

Mass and radius of planet is twice than that of Earth.

To find:

Value of escape Velocity from surface of this planet.

Concept:

Escape Velocity is the minimum Velocity with which when an object is projected goes out of the gravitational pull of the planet. In other words , the object reaches infinity.

Calculation:

$v = \sqrt{ \dfrac{2Gm}{r} } .....(earth)$

For that planet ,

mass = 2m, radius = 2r

$v2 = \sqrt{ \dfrac{2G(2m)}{(2r)} } .....(planet)$

$= > v2 = \sqrt{ \dfrac{2Gm}{r} }$

$= > v2 = v$

As per data , escape Velocity from Earth is 11.2 km/sec. So the escape Velocity of the planet is :

$= > v2 = v = 11.2 \: km {s}^{ - 1}$

So final answer :

$\boxed{ \huge{ \green{ \sf{ \bold{v2 = 11.2 \: \frac{km}{s}}}}}}$