Determine the acceleration of the moon with respect to the earth using the following data: the distance between the earth and the moon is 3.85 × 10^5 km , and the time taken by the moon to complete one revolution around

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Quantrex Academy · Accepted Answer

Correct answer: B. 2.73 × 10^ -3 m/s^2

The distance between the earth and the moon is

r = 3.85 \times 10^5 \text{ km} = 3.85 \times 10^8 \text{ m}

.

The time period of revolution of the moon is

T = 27.3 \text{ days} = 27.3 \times 24 \times 60 \times 60 \text{ s} \approx 2.36 \times 10^6 \text{ s}

.

The centripetal acceleration of the moon with respect to the earth is given by:

a = \omega^2 r = \left( \dfrac{2\pi}{T} \right)^2 r = \dfrac{4\pi^2 r}{T^2}

Substituting the values:

a = \dfrac{4 \times \pi^2 \times 3.85 \times 10^8}{(2.36 \times 10^6)^2}

a = \dfrac{4 \times 9.87 \times 3.85 \times 10^8}{5.57 \times 10^{12}}

a = \dfrac{152 \times 10^8}{5.57 \times 10^{12}}

a \approx 27.3 \times 10^{-4} \text{ m/s}^2

a = 2.73 \times 10^{-3} \text{ m/s}^2

Determine the acceleration of the moon with respect to the earth...

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