# The Acceleration of the Moon Just before It Strikes the Earth in the Previous Question - Physics

MCQ

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

• 10 m s−2

• 0⋅0027 m s−2

• 6⋅4 m s−2

• 5⋅0 m s−2

#### Solution

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 (Re + Rm) from the centre of the Earth.

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

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

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

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

Concept: Acceleration Due to Gravity of the Earth
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 11 Gravitation
MCQ | Q 2 | Page 224