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 = `\frac {Me}{80}`
Similarly, Let the radius of the earth is = Re, then
Radius of the moon = Rm = `\frac {Re}{4}`

To Find:

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

Solution:

We know the formula for escape velocity from the earth Vâ‚‘ and moon Vₘ are:

From the earth:

Vâ‚‘ = `\sqrt frac {2 G Me}{Re}`  ----------(1)

And from the moon:

Vₘ = `\sqrt frac {2 G Mm}{Rm}`

by putting values

Vₘ = `\sqrt frac {2 G frac {Me}{80}}{frac {Re}{4}}`

Vₘ = `\sqrt frac {2 ｘ4 ｘG ｘMe}{80ｘRe}`

Vₘ = `\sqrt frac {2 ｘ G ｘMe}{20ｘRe}`   [dividing the numerator and denominator by 4]

Vₘ = `\sqrt frac {2G frac{Me}{Re}}{20}`

Vₘ = `\ frac {sqrt frac{2GMe}{Re}}{sqrt{20}}`

Putting the value from equation (1)

Vₘ = `\ frac {Vₑ}{sqrt{20}}`

Escape veolcity from the earth = Vâ‚‘ = 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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