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)

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Solution

The centripetal acceleration is given by $a = \frac{v^{2}}{6}$
This expression explicity depends on Moon's speed which i s non trivial . we can work with the formula $w^{2} R_{m} = a_{m}$
Since,
$a_{m} \to$ centrifugal acceleration of the moon due to earth's gravity
$w \to$ angular velocity
$R_{m} = 60 R = 60 \times 6.4 \times 10^{6} = 384 \times 10^{6} m$ is the distance between earth and the moon, which is $60$ times the radius of the earth.
$w = \frac{2 π}{T}$
angular velocity $w = \frac{2 π}{T}$ .
and $T = 27.3 d a y s = 27.3 \times 24 \times 60 \times 60$
Seconds= $2.358 \times 10^{6} sec$
putting these values in the formula for acceleration
$a_{m} = \frac{(4 π^{2}) (384 \times 10^{6})}{{(2.358 \times 10^{6})}^{2}} = 0.00272 m s^{- 2}$

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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 6.4×106 m)

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)