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A FEW CONSTANTS DEFINED

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G=6.67E-11;
R_earth=6374*1000;
M_earth=5.9736E+24;
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<h1 style="text-align: center;">Satellite Orbit Calculations</h1>
Iyer Arvind Sundaram

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altKm=600
alt=altKm*1000
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The altitude of the satellite is set to be $alt=\sage{alt}$m from the surface of the earth
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R=R_earth+alt
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The radius of this orbit is 
R=R_{earth}+alt=\sage{R}m
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omega=sqrt(G*M_earth/R**3)

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\noindent
The centripetal force due to the rotation is to be balanced by the force due to gravity, hence
\begin{eqnarray*}
F_{centripetal}&=&m_{sat}\omega^2 R\\
F_{gravity}&=&G \dfrac{m_{sat}M_{earth}}{R^2}\\
F_{centripetal}&=&F_{gravity} \\
\implies m_{sat}\omega^2 R&=&G \dfrac{m_{sat}M_{earth}}{R^2}\\
\implies\omega&=&\sqrt{\dfrac{GM_{earth}}{R^3}}
&=&\sage{omega}rad/s
\end{eqnarray*}
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Time period of orbit is \[T=\dfrac{2\pi}{\omega}\]
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T=2*pi/omega
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omega;T
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Total revolutions per day is \[N=\dfrac{86400}{T}\]
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N=numerical_approx(86400/T,digits=5)
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Thus the satellite performs \sage{N} revolutions in a day
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