Question

The moon orbits the Earth once in 27.3 days in an almost circular orbit. Calculate the centripetal acceleration experienced by the Earth? (Radius of the Earth is 6.4 $\times$ ${10^6}m$)

Answer 1

Hint Centripetal acceleration is defined as the property of the motion of a body that is traversing in a circular path. The acceleration is then directed radially towards the centre of the circle and has a magnitude equal to the square of the body’s speed along the curve is divided by the distance from the centre of the circle to the moving body.

Complete step by step answer
The centripetal acceleration is given by a = $\dfrac{{{v^2}}}{6}$
This expression explicitly depends on Moon’s speed which is not trivial we can work with the formula ${w^2}{R_m} = {a_m}$
Since here:
${a_m}$ is the centrifugal acceleration of the moon due to the earth’s gravity
w is the angular velocity
${R_m} = 60R = 60 \times 6.4 \times {10^6} = 384 \times {10^6}m$is the distance between earth and the moon, which is 60 times the radius of the earth.
$w = \dfrac{{2\pi }}{T}$
And the angular velocity w = $\dfrac{{2\pi }}{T}$
And T = 27.3 days = $27.3 \times 24 \times 60 \times 60$
Seconds = $2.358 \times {10^6}\sec$
Now putting the values in the formation for the acceleration is given as:
${a_m} = \dfrac{{(4{\pi ^2})(384 \times {{10}^6})}}{{{{(2.358 \times {{10}^6})}^2}}} = 0.00272m{s^{ - 2}}$ ${a_m} = \dfrac{{(4{\pi ^2})(384 \times {{10}^6})}}{{{{(2.358 \times {{10}^6})}^2}}} = 0.00272m{s^{ - 2}}$

Note We should know that centripetal force is caused by the centripetal accelerations. In the special case of the Earth’s circular motion around the Sun, or any satellite’s circular motion around any celestial body, the centripetal force causing motion is the result of the gravitational attraction between them.