Question

the mass of the jupiter is 318 times that of the earth and the radius of jupiter is also11.2 times that of the earth.If escape velocity of a body on the surface is 11.2 km/s; then what will be the escape velocity of the body from the surface of the Jupiter? ​

Answer 1

Answer:-

Let the mass of the earth be M & earth's radius be R.

We know that,

Escape velocity (V) = √2GM/R

Given:

Escape velocity on earth (Vₑ) = 11.2 km/s

⟹ √2GM/R = 11.2 km/s -- equation (1)

And,

Mass of Jupiter is 318 times that of the earth.

⟹ Mass of Jupiter = 318 * M

⟹ Mass of Jupiter = 318M

Radius of Jupiter is 11.2 times that of the earth.

⟹ Radius of Jupiter = 11.2R

Hence,

⟹ Escape velocity of the body on Jupiter (Vⱼ) = √2G318M/11.2R

⟹ Vⱼ = (√2GM/R) * √(318/11.2)

Substitute the value of √2GM/R from equation (1).

⟹ Vⱼ = 11.2 * √318/√11.2

⟹ Vⱼ = (√11.2) * (√11.2) * √318 / √11.2

⟹ Vⱼ = √(11.2 * 318)

⟹ Vⱼ ≈ 59.5 km/s

∴ The escape velocity of the object of the body on Jupiter will be approx. 59.5 km/s.