There is only one main force acting on a satellite when it is in orbit, and
that is the gravitational force exerted on the satellite by the Earth. This force is
constantly pulling the satellite towards the centre of the Earth.
A satellite doesn't fall straight down to the Earth because of its velocity. Throughout a
satellites orbit there is a perfect balance between the gravitational force due to the Earth,
and the centripetal force necessary to maintain the orbit of the satellite.
The formula for centripetal force is: F = (mv^{2})/r
The formula for the gravitational force between two bodies of mass M and m is (GMm)/r^{2}
The most common type of satellite orbit is the geostationary orbit. This is described in more
detail below, but is a type of orbit where the satellite is over the same point of Earth
always. It moves around the Earth at the same angular speed that the Earth rotates on its
axis.
We can use our formulae above to work out characteristics of the orbit.
(mv^{2}/r) = (GMm)/r^{2}
=> v^{2}/r = (GM)/r^{2}
Now, v = (2πr)/T.
=> (((2πr)/T)^{2})/r = (GM)/r^{2}
=> (4π^{2}r)/T^{2} = (GM)/r^{2}
=> r^{3} = (GMT^{2})/4π^{2
}We know that T is one day, since this is the period of the Earth. This is 8.64 x 10^{4}
seconds.
We also know that M is the mass of the Earth, which is 6 x 10^{24} kg.
Lastly, we know that G (Newton's Gravitational Constant) is 6.67 x 10^{-11} m^{3}/kg.s^{2}
So we can work out r.
r^{3} = 7.57 x 10^{22}
Therefore, r = 4.23 x 10^{7} = 42,300 km.
So the orbital radius required for a geostationary, or geosynchronous orbit is 42,300km.
Since the radius of the Earth is 6378 km the height of the geostationary orbit above the
Earth's surface is ~36000 km.
There are many different types of orbits used for satellite
telecommunications, the geostationary orbit described above is just one of them. Outlined
below are the most commonly used satellite orbits. The orbits are sometimes described by their
inclination - this is the angle between the orbital plane and the equatorial plane. |