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 ).

The Moon orbits the Earth once in 27.3 days in an almost circular orbi

1.	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 ).
Answer» Solution :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)`