Question

The radius of the moon is roughly 2000km. The acceleration of gravity at the surface of the moon is about \(\frac{g}{6}\), where g is the acceleration of gravity at the surface of the earth. What is the velocity of escape for the moon?(Take g=10ms^-2)

(a) 2.58 kms^-1

(b) 4.58 kms^-1

(c) 6.28 kms^-1

(d) 12.28 kms^-1

This question was addressed to me during an interview.

I need to ask this question from Newton’s Law of Cooling and Escape Velocity in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (a) 2.58 kms^-1

The explanation is: Let R be radius of the earth and r be the variable distance from Newton’s law \(a = \frac{dv}{dt} = \frac{k}{r^2}\)  when r=R a=-g due to retardation of a body -gR^2=k

substituting the value of k back we get \(\frac{dv}{dt} = \frac{-gR^2}{r^2} = \frac{dr}{dt}  \frac{dv}{dr} = v \frac{dv}{dr}\) solving DE for v

we get \(\int v \,dv = \int \frac{-gR^2}{r^2}  dr + c \rightarrow v^2 = \frac{2gR^2}{r} + C\) to find c we use at r=R

v=ve thus we get \(v_e^2 – \frac{2gR^2}{R} = C\)… where 2c=C=constant substituting value of c we get \(v^2=\frac{2gR^2}{r} + v_e^2 – 2gR\)…..if r>>R \(\frac{2gR^2}{r} = 0\) and particle to get escape from earth v≥0 –> ve^2 – 2gR≥0 –> \(v_e=\sqrt{2gR}\) to find ve

from the moon g becomes \(\frac{g}{6}\), R=2000km=2×10^6 m, g=10ms^-2

                         therefore \(v_e = \sqrt{2*\frac{10}{6}(2×10^6)} = 2.58kms^{-1}.\).