Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and the moon = 3.85 × 10^{5} km and the time taken by the moon to complete one revolution around the earth = 27.3 days.

#### Solution

Distance between the Earth and the Moon :

\[r = 3 . 85 \times {10}^5 \text{km }= 3 . 85 \times {10}^8 \text{m}\]

Time taken by the Moon to revolve around the Earth :

\[T = 27 . 3 \text { days }\]

\[ = 24 \times 3600 \times 27 . 3 s = 2 . 36 \times {10}^6 \text{s}\]

\[\text { Velocity of the Moon }: \]

\[\text{v} = \frac{2\pi r}{T}\]

\[ = \frac{2 \times 3 . 14 \times 3 . 85 \times {10}^8}{2 . 36 \times {10}^6} = 1025 . 42 \text{ m/s}\]

\[\text { Acceleration of the Moon }: \]

\[\text{a} = \frac{\text{v}^2}{\text{r}} = \frac{(1025 . 42 )^2}{2 . 36 \times {10}^6} = 0 . 00273 \text{ m/ s}^2\]

\[\Rightarrow a = 2 . 73 \times {10}^{- 3} \text{m/ s}^2\]