Question

the acceleration of moon with respect to the earth is 0.0027 metre per second square and the acceleration of an apple falling on earth surface is about 10 metre per second square assume that the radius of the moon is one fourth of the earth radius if the moon is a stopped for an instant and then released it will fall towards the earth the acceleration of moon just before it strikes the earth is

Answer 1

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Answer 2

Answer: The moon’s acceleration is $\bold{6.4 m / s^{2}}$

Solution:

The acceleration of falling moon is to be calculated through the gravitational force we know to act, which is as follows –

Given is the acceleration of moon with respect to earth = $0.0027 \mathrm{m} / \mathrm{s}^{2}$

Acceleration of apple falling on Earth = 10 $\mathrm{m} / \mathrm{s}^{2}$

As we know that force of gravitation acts on object falling on earth surface. We have – F = $\frac{G m 1 m 2}{r^{2}}$

$\frac{G M m}{r^{2}} = 0.0027$

Also $\frac{G M m}{R^{2}}=10$  

Equating both together we have $\bold{\frac{G M m}{r^{2}}=\frac{G M m}{R^{2}}}$

$\begin{array}{l}{\frac{G M m}{r^{2}}=\frac{G M m}{(r+0.25 r)^{2}}} \\ {\mathbf{F}=\mathbf{m} \mathbf{a}}\end{array}$

$\begin{aligned} a &=\frac{r^{2}}{(r+0.25 r)^{2}} \\ a &=\frac{10}{1.5625} \end{aligned}$

a = $\bold{6.4 m / s^{2}}$