The mass of the moon is 1/80 of the mass of the earth and the corresponding radius is 1/4 of the earth. Calculate the escape velocity on the surface of the moon. (2.5 km s⁻¹)



Given:

Let the mass of the earth be = Me, then
Mass of the moon = Mm = Me80
Similarly, Let the radius of the earth is = Re, then
Radius of the moon = Rm = Re4

To Find:

Escape velocity at the surface of the moon = Vescape = ?

Solution:

We know the formula for escape velocity from the earth Vₑ and moon Vₘ are:

From the earth:

Vₑ = √2GMeRe  ----------(1)

And from the moon:

Vₘ = √2GMmRm

by putting values

Vₘ = √2GMe80Re4

Vₘ = √2ｘ4ｘGｘMe80ｘRe

Vₘ = √2ｘGｘMe20ｘRe   [dividing the numerator and denominator by 4]

Vₘ = √2GMeRe20

Vₘ =  √2GMeRe√20

Putting the value from equation (1)

Vₘ =  Vₑ√20

Escape veolcity from the earth = Vₑ = 11.2 km s⁻¹, so

Vₘ = `\ frac {11.2 km s⁻¹}{4.472}`

Vₘ = 2.503 km s⁻¹

or

Vₘ = 2.5 km s⁻¹

Hence the escape velocity from the moon according to the given conditions of the question is 2.5 km s⁻¹ .


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