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 $6400 \; km$, and the value of $g$ at the surface of earth $10\;m/s^2$, calculate the total energy needed to raise a satellite of mass $2000 \; kg$ to a height of $800 \; km$ 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) = 2000 kg

Height of satellite (h) = 800 km = 800000 m

Total Energy needed (E) = ?

We have,

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

$\left [ -\;\frac{G\;M\;m}{r}\;-\;\left (\frac{-\;G\;M\;m}{R}\right )\; \right ]$

= $-$$\frac{G\;M\;m}{r}$$\;-\;$$\frac{-\;G\;M\;m}{R}$$\;+\;$$\frac{1}{2}$$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\;-\;$$\frac{R^2}{2(R+h)}$$]$

= $2000\;*\;10\;[6400000 \;-\; $$\frac{(6400000)^2}{2(6400000 \;+\; 800000)}]$

= $7.12\;*\;10^{10}\;J$

∴ The total Energy needed (E) = $7.12\;*\;10^{10}\;J$

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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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