Sunday, 16 January 2011

Maths Problems 5: Geostationary Satellites

Now that I've shown how to estimate (or calculate) pi with a high degree of accuracy and precision, it's time to start using it!

Modern communications around the world rely on being able to bounce a signal of a satellite in orbit around the earth, so that it can be relayed beyond the horizon of the person sending it (or better still, over a specific area of the earth). The ability to bounce the signal off the satellite is very important, and relies on one important factor - knowing where the satellite is, and better still, having it stay directly above the same point on the earth, so that its apparent position in the sky is fixed.

In order to do this, the satellite needs to occupy a geostationary orbit, meaning that it goes around the earth at the same rate as the earth rotates on its axis - once every 24 hours. A geostationary satellite appears to stay in the same point in the sky because it's rotating at the same rate as the earth.

The question is - how can this be done? The answer is to put the satellite in orbit at a specific height above the earth, and above the earth's equator. The rest is physics. Well, it's not rocket science, is it?

Firstly, there are two key formulae to use - the first is Newton's universal law of gravity, because gravity is what holds a satellite in orbit, and the second is the laws of motion for an object moving in a circle (we'll be assuming that the satellite follows a circular orbit).

The first formula is Newton's universal law of gravitation, describing the force of attraction between two bodies.  It is:

G is the universal gravitational constant, 6.67300 × 10-11 m3 kg-1 s-2
M is the mass of the heavier body (in this case, the Earth)
m is the mass of the lighter body (in this case, the satellite)
and r is the distance between the centres of gravity of the two bodies.

The second formula is the equation which describes the force required to keep an object moving in a circular path.

In this formula, F is the force which points towards the centre of the circle, m is the mass of the object (in this case, the satellite), v is the angular velocity (how quickly it's moving in a circle), and r is the radius of the circular path being described by the body.

The first thing I'm going to do is unpack v in the second equation.  v, the angular velocity, is defined in the same way as all speeds as the distance travelled divided by the time taken.  For a circular path, the distance is the circumference of the circle, and the time is the time taken to complete one circle (or orbit), T.
We can plug v2 into the formula for circular motion, and we can also equate the two formulae, since the force that will hold the satellite in orbit is the force of gravity.

By cancelling and rearranging, we have one expression for the radius of a geostationary satellite held in orbit by the Earth's gravity.

One thing to notice here is that the mass of the satellite, m, does not appear in our final expression.  The period of rotation of a satellite depends only on the height of its orbit.  The time taken to complete an orbit is only affected by the height of the orbit - the mass of the satellite is not important.

Anyway, we know the time taken for one orbit has to be 24 hours, and all the other components of the formula are constants, so we can plug them in and calculate r.

G is the universal gravitational constant, 6.67300 × 10-11 m3 kg-1 s-2
M is the mass of  the Earth, 5.9742 × 1024 kg
T is the time to complete one orbit, 24 hrs = 86 400 s

Therefore, r = 42,243 km.

However, this is not the altitude of the satellite from the Earth's surface.  Remember that when I first gave Newton's law of gravity, the r was the distance between the centres of gravity of the two bodies.  The size of the satellite is small enough to be ignored, but we must subtract the radius of the Earth from this value, to give the orbital height.  Radius of Earth = 6378.1 km, so orbital radius = 35,865 km.

Later edits of this post will include some nice diagrams, but for now I'm happy to have posted the result!

NASA gives the height as approx 35,790 km using a more exact value of the length of one day, while Wikipedia has a similar figure and more information on geostationary orbits.

In my next post... I'm not sure.  Possibly more on orbits generally, including calculating a Moon - Earth distance, or a Sun - Earth distance, or a practical experiment to determine g (acceleration due to gravity on Earth).

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