Calculating the Earth-Moon distance

This post follows up my previous post on geostationary satellites.  Long before we were launching satellites (even non-geostationary ones), our natural satellite, the Moon, was orbiting the Earth.  As the moon goes around the earth, its phase (shape) changes, and in fact, the word "month" derives from "moonth", the time taken for the moon to go from new to full to new again.  This time is the time taken for one complete orbit around the Earth - the different phases of the moon are a result of us seeing a different amount of the lit half of the moon (I once based a very neat science lesson on this principle - in fact I used it in my interview lesson  and subsequently got the job).

One of my photographs of the moon, taken through a telescope.
The darkening at the bottom of the image is the edge of the
telescope's field of view

We can use physics, and our knowledge of the mass of the Earth, the value of pi and the time the Moon takes to complete one orbit, to work out how far it is from the Moon to the Earth.

Back to the two key equations that we'll need, which are the force on a body moving in a circular path:

where

And Newton's Law of Gravity

Equating the two, and rearranging to find r, gives us

This is the same equation used for geostationary satellites, and describes the basic relationship between the distance between two bodies (a planet and a moon, for example, or a star and a planet).  This gives it great power as it can be used in many different situations.

Turning to the current situation, then:

Calculation of the Earth-Moon distance:

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, which for the Moon is 27.32166 days, which is 2,360,591 seconds.

Plugging the numbers into the formula above gives the distance as 383,201 km

However, this is not the distance to the Moon from the Earth's surface.  Newton's law of gravity gives the distance between the centres of gravity of the two bodies.  I'm ignoring the radius of the Moon (which is perhaps an oversight on my part, you decide) but we must subtract the radius of the Earth from this value, to give the orbital height.  Radius of Earth = 6378.1 km, so the distance to the Moon is calculated as  = 376,823 km, or, if you prefer, (at 1.61 km to the mile), 234,147 miles.

Previously, I've learned that the Moon is about a quarter of a million miles away, so I'm glad the method I've used shows a figure which is 'about right' without any checking.  Looking at other sources, it looks like my figures are close enough, considering the assumptions I've made.  One key assumption I've made is to suggest that the moon travels in a circular orbit, and it doesn't.  It has an elliptical orbit, which means the distance from Earth to Moon changes during the orbit - so I've calculated an average distance.  Still, my figure is pretty close, and not a million miles away (and next time somebody reliably informs you that their opinion is not a million miles away, you can tell them that not even the Moon is that far away).

Other astronomy related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?

Calculating (and explaining) escape velocity
Measuring g using a pendulum

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:

where:
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).

Other astronomy related articles you may find interesting:

Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?

Calculating (and explaining) escape velocity