Question

The acceleration of the moon just before it strikes the earth in the previous question is
(a) 10 m s⁻²
(b) 0⋅0027 m s⁻²
(c) 6⋅4 m s⁻²
(d) 5⋅0 m s⁻²

Answer 1

(c) 6⋅4 m s⁻²

According to the previous question, we have:
Radius of the moon, $R_m=\frac{R_e}{4}=\frac{6400000}{4}=1600000\mathrm{m}$
So, when the Moon is just about to hit the surface of the Earth, its centre of mass is at a distance of (R_e + R_m) from the centre of the Earth.

Acceleration of the Moon just before hitting the surface of the earth is given by

$$\begin{gathered} g\text{'}=\frac{GM}{(R_e+R_m{)}^2}=\frac{GM}{{R_e}^2(1+{\displaystyle \frac{R_m}{R_e}}{)}^2} \\ \Rightarrow g\text{'}=\frac{g}{(1+{\displaystyle \frac{R_m}{R_e}}{)}^2}=\frac{10}{(1+{\displaystyle \frac{1}{4}}{)}^2}=\frac{10\times 16}{25} \\ \Rightarrow g\text{'}=6.4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$