Question

The orbital radius of the moon around the earth is $3.8 \times {10^8}$ meter and its time period is 27.3 days. The centripetal acceleration of the moon will be
(A) $- 2.4 \times {10^{ - 3}}m/{s^2}$
(B) $11.2m/{s^2}$
(C) $2.7 \times {10^{ - 3}}m/{s^2}$
(D) $9.8m/{s^2}$

Answer 1

Hint The centripetal acceleration, as known by us is defined as the property of the motion of a body which is traversing in a circular path. The acceleration is signified by the radially directed towards the centre of the circle and the magnitude will be similar to the square of the body’s speed along the curve which is divided by the distance from the centre of the circle to the body which is moving.

Complete step by step answer:
We know that centripetal acceleration is defined as net centripetal force / mass.
The net centripetal force is given as: $\dfrac{{GMm}}{{{R^2}}} - m{(\omega )^2}R$
The value of $\omega$is given as: $\dfrac{2\pi }{27.3\times 24\times 3600}$
Here the value of R = $3.8 \times {10^8}$M and M is $5.92 \times {10^2}4$
The centripetal acceleration is given as:
$\dfrac{{\dfrac{{GMm}}{{{R^2}}} - m{{(\omega )}^2}R}}{m}$
We have to put the values in the above expression to get the expression as:
$\dfrac{{6.67 \times {{10}^{ - 11}} \times 5.92 \times {{10}^2}4}}{{{{(3.8 \times {{10}^8})}^2}}} - {\left( {\dfrac{{2\pi }}{{27.3 \times 24 \times 3600}}} \right)^2} \times 3.8 \times {10^8}$
The value we get after the evaluation is given as: $2.7 \times {10^{ - 3}}m/{s^2}$.

Hence the correct answer is option C.

Note: We should know that any summation of the force which is the cause behind the uniform circular motion is known as the centripetal force. As we know that Newton's second law of motion gives us an idea that the net force is defined as the mass times acceleration. But we should remember that for any uniform circular motion the acceleration of a body is the centripetal acceleration.