Question

The acceleration of the moon just before it strikes the earth in the previous question is

Options

• 10 m s⁻²

• 0⋅0027 m s⁻²

• 6⋅4 m s⁻²

• 5⋅0 m s⁻²

Answer 1

6⋅4 m s⁻²       

According to the previous question, we have :
Radius of the moon, \[R_m = \frac{R_e}{4} = \frac{6400000}{4} = 1600000 m\]

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

\[g' = \frac{GM}{( R_e + R_m )^2} = \frac{GM}{{R_e}^2 (1 + \frac{R_m}{R_e} )^2}\]

\[ \Rightarrow g' = \frac{g}{(1 + \frac{R_m}{R_e} )^2} = \frac{10}{(1 + \frac{1}{4} )^2} = \frac{10 \times 16}{25}\]

\[ \Rightarrow g' = 6 . 4 m/ s^2\]