Satellites and Parking Orbits, Energy of a satellite

Satellites are bodies, natural or artificial which move in orbits around the moon or planets. Artificial  satellites are made by man.
Consider a satellite of mass m moving round the earth of mass M in an orbit as shown below:

satellite orbit

If R is the radius of the earth, r is the radius of the orbit and v is the velocity with which the satellite is moving,

We have that centripetal force (due to the satellite) = gravitational force (due to the earth)


Hence, v=gr−−√ ——-(9)

To get the time for one complete revolution, i.e, period, we use


Therefore, the period T is given as:

T=length of circumferencespeed=2πRv

But v=gr−−√∴T=2πRgR√

By squaring both sides, we have,


Hence, T=2πRg−−√ ——-(10)

Since R = 6400km and g = 9.8m/s2, therefore, T = 5077.58s = 84.6mins. This is the period for an artificial satellite in the orbit above the earth.

If the period of the satellite in its orbit is equal to the period of the earth about its axis, which is 24hours, the satellite will stay at the same place above the earth as the earth rotates. Such orbit is called ‘’Parking Orbit’’.

Now, mv2R=GMmR2

But GM=gr2(fromGMr=gr=Vg)

We have, mv2R=mgrR2

Now, multiplying through by R/m, we have:


Again, v=2πRT


By simplification,

R=gT2R24π2−−−−−√3 ——-(11) Radius of orbit

Hence, the length the satellite is above the earth surface l=Rorbit−Rearth


  1. State the relationship between radius of the orbit and the radius of the earth.
  2. What is a satellite?
  3. What is the value of g at the orbit?

Energy of the Satellite

Supposing a satellite of mass m is revolving in an orbit of radius r round the earth of mass M with  a velocity v,then it will have both kinetic and potential energies.

Recall that mv2r=GMmR2=mg∴mv2r=GMmr2

Hence, v2=GMr

Now, k.e=12mV2

Putting in the value of vabove, we have:


Hence, k.e=GMm2r ——-(12)

Also, p.e.=mgh=−GMmr

∴ Total energy k.e+p.e=GMm2r+−GMmr

Hence, total energy in orbit =−GMm2r ——-(13)

Worked Examples

  1. Find the satellite orbit above the earth’s surface if the radius of that orbit is 6.5 × 106m.



  1. Determine the speed of revolution of a satellite round an orbit revolving at 85mins per revolution if the satellite is at 110km above the earth’s surface.



Now, T = 85mins = 5100s,

R = radius of the orbit =110km+6400km=6510=6,510,000mSpeed=2π×65100005100=8020.3m/s

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