Question

the radius of the moon is one fourth the radius of the earth and the mass of is 196 of the mass of the earth if escape speed from the surface of the Earth is 11 kilometre per second then find the escape velocity from the surface of the moon

Answer 1

Answer:

2.245 km/s

Explanation:

Escape velocity on the surface of a planet is the minimum velocity given to a body to make it free from the gravitational field of that planet.

Let vₓ be the escape velocity of the body on the surface of a planet and the mass of the body to be projected be m.

Now, conserving energy at the surface of the planet and infinity is :

$\frac{1}{2} m(v_{x})^{2} - \frac{GMm}{R} = 0$

[$T = \frac{GMm}{R}$ is the binding energy of the system ]

[Binding Energy of the system of two bodies is the amount of minimum energy needed to separate the body to a large distance.]

∴ vₓ = $\sqrt{\frac{2GM}{R} }$

∴ vₓ = $\sqrt{2gR}$

Hence vₓ of moon = $\sqrt{2(g/6)(R/4)}$

[∵ g of moon is around 1/6th g of Earth]

⇒ vₓ = $\frac{\sqrt{2gR} }{\sqrt{24} }$

⇒ vₓ = $\frac{11}{\sqrt{24} }$ km/s

∴ vₓ = 2.245 km/s [Approximate]

Escape velocity of Earth is 11.2 km/s and Hence escape velocity of moon is 2.38 km/s, to be more precise.

Thanks !