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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
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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
\[g' = \frac{GM}{( R_e + R_m )^2} = \frac{GM}{{R_e}^2 (1 + \frac{R_m}{R_e} )^2}\]
\[ \Rightarrow g' = \frac{g}{(1 + \frac{R_m}{R_e} )^2} = \frac{10}{(1 + \frac{1}{4} )^2} = \frac{10 \times 16}{25}\]
\[ \Rightarrow g' = 6 . 4 m/ s^2\]
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