The Moon orbits the Earth once in 27.3 days in an almost circular orbit. Calculate the centripetal acceleration experienced by the Moon? (Radius of the Earth is `6.4 xx 10^6 m` ).

Explanation

The centripetal acceleration is given by ` a = (v^2)/(r )`. This expression explicitly depends on Moon.s speed which is nontrivial. We can work with the formula
`omega^2 R_m = a_m`
`a_m` is centripetal acceleration of the Moon due to Earth.s gravity. `omega ` is angular velocity.
`R_m` is the distance between Earth and the Moon, which is 60 times the radius of the Earth.
`R_m = 60R = 60 xx 6.4 xx 10^6 = 384xx 10^6 m`
As we know the angular velocity `omega = (2pi)/(T)` and T = 27.3 days = `27.3 xx 24 xx 60 xx 60` second = `2.358 xx 10^6` sec
By substituting these values in the formula for acceleration
`a_m = ((4pi^2)(384 xx 10^6) )/((2.358 xx 10^6 )^2) = 0.00272 ms^(-2)`
The centripetal acceleration of moon towards the earth is `0.00272 ms^(-2)`