LEARNATHON
III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

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Theory:

Formula of orbital velocity:
The orbital velocity is calculated using the formula,
 
v=GM(R+h)
 
Where,
\(G\) is the Gravitational constant (\(6.673 \times 10^{–11}\) \(Nm^2kg^{-2}\)),
\(M\) is the mass of the Earth (\(5.972 \times 10^{24}\) \(kg\)),
\(R\) is the radius of the Earth (\(6371\ km\)),
\(h\) is the height of the satellite from the Earth's surface
 
Example:
Calculate the orbital velocity of a satellite orbiting at an altitude of \(500\ km\) above Earth?
 
\(G\) \(=\) \(6.673 \times 10^{–11}\) \(Nm^2kg^{-2}\)
 
\(M\) \(=\) \(5.972 \times 10^{24}\) \(kg\)
 
\(R\) \(=\) \(6731000\ m\)
 
\(h\) \(=\) \(500000\ m\)
 
The formula is given as
 
v=GM(R+h)
 
On substituting the given values, we get
 
v=6.673×1011×5.972×1024(6731000+500000)=39.851156×1013(7231000)=7613ms1=7.613kms1
 
Thus, the orbital velocity of a satellite is \(7.613\ kms^{-1}\).
Time period of a satellite:
The time taken by a satellite to complete one revolution around the Earth is called the time period (\(T\)).
Time period=DistancecoveredOrbitalvelocityT=2πrv
 
On substituting the values of \(v\), we get
 
T=2π(R+h)GM(R+h)
 
Example:
Find the orbital period of the satellite at an orbital height of \(500\ km\).
 
\(h\) \(=\) \(500\ km\)
 
\(R\) \(=\) \(6731\ km\)
 
\(v\) \(=\) \(7.613 kms{-1}\)
 
The formula is given as
 
T=2π(R+h)GM(R+h)
 
On substituting the given values, we get
 
T=2×227×(6731+500)7.613=5.6677×103s=5667.7sT95min
 
Thus, the orbital period of the satellite is \(95\ minutes\).