Obtain the value of $'g'$ from the motion of moon assuming that its period of rotation around the earth is $27$ days $8$ hours and the radius of its orbit is $60.1$ times the radius of the earth.

Given,

T = 27 days 8 hours = (27 * 24 + 8) hours = 656 hours = 2361600 Sec

Radius of Earth (R) = $6.36 * 10^6\;m$

Radius of orbit $(r) = 60.1 * R = 60.1 * 6.36 * 10^6$

Value of $g$ = ?

We have,

$T = $ $ \frac{2\;\pi\;r}{R}  \sqrt{ \frac{r}{g}}$

By Solving,

$g$ = $\frac{4\; \pi^2\;r^3}{T^2 \; R^2}$ = $\frac{4 \; \pi^2 \; * \; (60.1)^3 \; R^3}{T^2 \; R^2}$ = $\frac{4 \; \pi^2 \; * \; (60.1)^3 \; * \; 6.36\; * 10^6}{2361600}$ = $9.77 \; m/s^2$

∴ The required value of $g$ from the motion of the moon is $9.8\;m/s^2$

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What is the period of revolution of a satellite of mass m that orbits the earth in a circular path of radius $7880 \; km$ about $1500 \; km$ above the surface of the earth?

Given,

Radius of the orbit $(r) = 7880\;km = 7880\; * \;10^3\;m$

Height $(h) = 1500\;km = 1500\;*\;10^3\;m$

Period of revolution $(T) = \;?$

Radius of earth $(R) = r - h = (7880 - 1500)\; * \; 6380\;*\;10^3\;m$

We have,

$T = 2 \pi \; * \; \sqrt{ \frac{(R\;+\;h)^{3}}{(g\;*\;R)^2}}$

or, $ T = 2 \pi \; * \; \sqrt{ \frac{(6380000\;+\;1500000)^3}{9.8\;*\;(6380000)^2}} = 6955.29\;Sec = 1.91\; hrs$

∴ The time period of the revolution = $1.91\;hrs$.

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