The period of the moon revolving under the gravitational force of the earth is $27.3$ days. Find the distance of the moon from the center of the earth if the mass of the earth is $5.97 \;*\;10^{24}kg$.

Given,

Time $(T) = 27.3$ days = $27.3 \; * \; 24 \; * \; 60 \; * \; 60  = 2358720 \; Sec$

Mass of earth $(M) = 5.97 \; *\; 10^{24}\; kg$

Height of the Moon from the center of the earth $(h) = \;?$

We know that,

$h = $ $\left ( \frac{T^2 \; * \; R^2\; * \;g}{4\; \pi^2} \right )^{1/3}$ $ - \;R$

$h =$ $ \left ( \frac{(2358720)^2 \; * \; {6400000}^2\; * \; 9.8}{4\; \pi^2} \right )^{1/3}$ $ - \; 6400000 = 3.77 * 10^8 \; m$

∴ Height of the Moon from the center of the earth $(h) = \;3.77 * 10^8 \; m$

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Calculate the points along a line joining the centers of earth and moon where there is no gravitational force. ($M_e = 6 * 10^{24}\;kg, M_m = 7.4 * 10^{22}\;kg, d = 3.8 * 10^8\;m$).

Given, (need figure)

Mass of the earth $(M_e) = 6 * 10^{24}\;kg$

Mass of the Moon $(M_m) = 7.4 * 10^{22}\;kg$

Distance between centers of earth and the moon $(d) = 3.8 * 10^8\;m$

Let $P$ be the point at which gravitational force between earth and the moon is zero, then

$\frac{G\;*\;M_e\;*\;1}{x^2} = \frac{G\;*\;M_m\;*\;1}{(d-x)^2}$

$\frac{6\;*\;10^{24}}{x^2} = \frac{7.4\;*\;10^{22}}{(d\;-\;x)^2}$

$\frac{600}{x^2} = \frac{7.4}{(d-x)^2}$

$\left ( \frac{24.5}{x} \right )^2 = \left ( \frac{2.72}{d-x} \right )^2$

$\frac{24.5}{x} = \frac{2.72}{(d-x)}$

$27.22\;*\;x = 24.5\;*\;d$

$x = \frac{24.5\;*\;3.8\;*\;10^8}{27.22} = 3.42\;*\;10^8\;m$

∴ The distance from the earth center is $3.42\;*\;10^8\;m$

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An earth satellite moves in a circular orbit with a speed of $6.2\;km/s$. Find the time of one revolution and its centripetal acceleration.

Given,

Orbital velocity $v_o = 6.2\;km/s = 6200\;m/s$

Radius of earth $(R) = 6400000\;m$

Time Period $(T) = \; ?$

Centripetal Acceleration $(a) = \;?$

We have,

$ v_{0} = R \sqrt { \frac{g}{R+h}} $

or, $6200 = 6400000\;*\; \sqrt{\frac{10}{6400000 \; + \; h}}$

or, $9.38 \; *\; 10^{-7} = \sqrt{\frac{10}{6400000 + h}}$

or, $h = 10655301 - 6400000$

h = $4255301\;m$

Now, 

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

or, $2\pi \; \sqrt{\frac{(4255301 + 6400000)^3}{10 * 6400000}}$

or, $2 \; \pi \;*\;1718.57 = 10792.65\;Sec = 2.99\;hrs$

∴ Time of revolution = $2.99\;hrs$

Again,

Centripetal Acceleration $a = \frac{v_0^2}{h + R} = \frac{(6200)^2}{4255301 + 6400000} = 3.6\;m/s^2$

∴ Centripetal Acceleration = $3.6\;m/s^2$

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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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Taking the earth to be the uniform sphere of radius $6400 \; km$, calculate the total energy needed to raise a satellite of mass $1000 \; kg$ to a height of $600 \; km$ above the ground and to set it into a circular orbit at that altitude.

Given;

Radius of earth (R) = 6400 km = 6400000 m

Mass of satellite (m) = 1000 kg

Height of satellite (h) = 600 km = 600000 m

Total Energy needed (E) = ?

Energy needed = Increase in Potential Energy + Kinetic Energy at Orbit

= $-$$\frac{G\;M\;m}{r}$$\;-\;$$\frac{-\;G\;M\;m}{R}$$\;+\;$$\frac{1}{2}$$mv^2$

= $-$$\frac{G\;M\;m}{r}$$\;-\;$$\frac{-\;G\;M\;m}{R}$$\;+\;$$\frac{1}{2}$$m$$ \frac{G\;M}{r}$                               [∵ r = R+h]

= $g\;R^2\;m[$$\frac{1}{R}$$\;-\;$$ \frac{1}{2(R+h)}$$]$     =     $g\;m[R\;-\;$$\frac{R^2}{2(R+h)}$$]$

= $1000\;*\;10\;[6400000 \;-\; $$\frac{(6400000)^2}{2(6400000 \;+\; 600000)}]$

= $3.47\;*\;10^{10}\;J$

∴ The total Energy needed (E) = $3.47\;*\;10^{10}\;J$

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