A $200 \; kg$. satellite is lifted to an orbit of $2.2\;*\;10^4\;km$ radius. If the radius and mass of the earth are $6.37\;*\;10^6\;m$. and $5.98\;*\;10^{24};kg$ respectively, how much additional potential energy is required to lift the satellite?

 Given,

Mass of the satellite (m) = 200 kg

Radius of the orbit (r) = $2.2 * 10^7\; m$

Radius of the earth (R) = $6.37 * 10^6\;m$

Mass of the earth (M) = $5.98 * 10^{24}\; kg$

Additional potential energy = ?

We have, 

Additional P.E = P.E on the orbit - P.E at the earth's surface

                        = $-\frac{GMm}{r}$ $-$  ($-\frac{GMm}{R})$

                        =  $\frac{GMm}{R} - \frac{GMm}{r}$

                        = $GMm(\frac{1}{R}- \frac{1}{r})$

        
                = $6.7 * 10^-{11} * 5.98 * 10^{24} * 200 \; (\frac{1}{6.37 * 10^{6}} - \frac{1}{2.2 * 10^7})$

                        = $4.47 * 10^7\; J$

$\therefore$ Required Additional P.E. is $4.47 * 10^7\;J$.

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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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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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What is Gravity?:

The Natural force that tends to cause physical things to move towards each other. 

First, [Newton's Law of Universal Gravitation] is a mathematical in its expression.... Second, it is not exact. Einstein had to modify it.... There is always an edge of mystery, always a place where we have some fiddling around to do yet.... But the most impressive fact is that gravity is simple.... It is simple, and therefore it is beautiful.... Finally, comes the universality of the gravitational law and the fact that it extends over such enormous distances....

~ Richard P. Feynman.

Sir Issac Newton:                                          Albert Einstein:

                            

$F = G\frac{m_1 m_2}{r^2}$	$G\mu \nu = 8\pi GT\mu \nu$
» It is a force.	» It is a  distortion of space and time.
» It depends on mass and distance.	» It depends on Energy.

» Newton's stated that the force of gravity is always attractive. It affects everything with mass, work instantaneously at a distance and has an infinite range. "Every mass attracts every other mass in the universe (across the empty space)", Newton had imagined. [1]

» But in Einstein's model, gravity is not a force. It is a warping of space-time. Space is really curved, and as a result objects are deflected from a straight path in a way that looks like a force. According to Einstein: "Mass tells space how to bend, and space tells mass how to move".