Question

Obtain the value of g from the motion of moon assuming that it’s period of rotation round the earth is 27days 8hours and the radius of its orbit is 60.1 times the radius of the earth

Answer 1

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Answer 2

Answer:

The value of acceleration due to gravity is $g=9.785m/s^{2}$

Explanation:

Given that,

The period of rotation of moon round the earth is 27days,8 hours. and

The radius of its orbit is 60.1 times the radius of earth.

$= > r_{m}=60.1R_{e}=60.1\times64\times10^{5}m=3.84\times10^{8}m$

We shall convert the period of rotation of moon into seconds as shown

$T=(27\times24+8)\times60\times60=2361600sec$

Now ,thee time period of rotation is given by the relation as follows

$T=2\pi \sqrt{\frac{r_{m}^{3}}{gR^{2}} }$

Squaring it on both sides,we get

$T^{2}=4\pi^{2}\frac{r_{m}^{3}}{gR^{2}}= > g=\frac{4\pi ^{2}r_{m}^{3}}{T^{2}R^{2}}$

Substituting,the known values in above relation,we get

$g=\frac{4(9.87)(3.84\times10^{8})^{3}}{2361600^{2}\times(64\times10^{5})^{2}} \\=9.785m/s^{2}$

Therefore,the value of acceleration due to gravity is found to be $g=9.785m/s^{2}$

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