Question

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

पर्याय

• 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, $$R_m=\frac{R_e}{4}=\frac{6400000}{4}=1600000m$$

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

$$g^{\prime }=\frac{GM}{(R_e+R_m{)}^2}=\frac{GM}{{R_e}^2(1+\frac{R_m}{R_e}{)}^2}$$

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

$$\Rightarrow g^{\prime }=6.4m/s^2$$