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 = Me80Me80
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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