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In this section we will use the results of the previous sections to develop a simple mathematical model of planetary motion around the sun. We will see that a simple model can provide some very accurate results concerning planetary motion. This is good – most scientists (me included) like mathematical models of the universe to be as simple as possible. For one thing, the results of simple models are usually easier to comprehend than the results from complex models. Another advantage of simple models is that they have less onerous data requirements than complex models.
However, we are going to make a major simplification in our consideration of planetary motion. We are going to assume that the orbits of the planets around the sun are circular. The true orbits are in fact elliptical, but the maths of elliptical orbits is a little more complex than for circular orbits, and mathematical complexity is not what I’m seeking in these articles.
Nevertheless, the important results that we are going to obtain with our “circular approximation” are in fact the same as those that would be obtained if we considered elliptic orbits. (Mathematically inclined readers will already know that a circle is a special case of an ellipse with coincident foci, so our approximation is not all that silly.)
The planetary system that we will consider is shown in the following diagram:
The object around which the orbit occurs need not be the sun, of course, but it can be shown (and in fact is obvious from common sense) that the orbiting body needs to be much less massive than the body around which the orbit occurs.
[As a digression, strictly, both bodies rotate about a point determined by the centre of mass of the two bodies. But if one body is much more massive, the centre of mass lies very close to the massive body and it becomes a good approximation to assume that the massive body is stationary and the smaller body rotates in the manner that we are assuming].
From our discussion of gravitation, we know that the sun exerts a gravitation force on the orbiting body given by:
We also know from our discussion in the first article that if an object is to undergo circular motion, then it must experience a centripetal force equal to
Therefore, if our body in the figure above is to orbit the sun, the gravitational attraction to the sun must equal the required centripetal force:
We can turn this equation into a useful form by noting that the angular velocity is related to the time period for the orbit (i.e. the time required for the orbiting body to complete one rotation). One orbit corresponds to travelling through an angle of 2π radians. Thus, if the time for one complete orbit is T:
Substituting this into the previous equation, and cancelling the (identical) gravitational and inertial masses m on either side of this equation, we can write down an expression for the time period of the orbit:
This is a very interesting result, for a number of reasons:
1. The result is independent of the mass m of the orbiting body. This is because the gravitational and inertial masses are identical and we cancelled them out.
2. The time period is proportional to the radius of the orbit to the power 3/2. This result in fact is known as Kepler’s third law, and the same result is obtained if we assume an elliptic orbit.
Example Calculation - Time Periods
OK, so let’s try out this formula. The mass of the sun is 2 1030 kg, and the earth orbits the sun at a distance of about 93,000,000 miles. So what is the time period of the orbit? We need to convert the radius of the orbit into metres (multiply miles by 1,610) and substitute into the above formula. We find that:
T = 3.15 107 seconds = 0.99 years.
That is, the earth completes an orbit of the sun in one year. We know of course that this result is true, and so we give a thumbs up to our simple theory for getting it right!
Note that we would not normally use the formula in this way – the time period of the earth’s orbit is something we know very well, and so we might instead use the above formula to estimate the mass M of the sun. Clearly if we substituted a time period of one year into our formula, we would obtain a value for M of 2 1030 kg.
Having said that, let’s compute the time period of the moon’s orbit. For this, the radius of the orbit is 239,000 miles, and the mass of the earth (since the moon orbits around the earth) is about 6 1024 kg. We find that:
T = 2.37 106 seconds = 27.4 days.
So, our theory tells us that the moon orbits the earth in a little less than one calendar month. Again, this is another result that we know to be correct. (The time period of the moon around the earth, 27.4 days, is known as the “sidereal month”).
Example Calculation - Geostationary Satellite
Let’s try and obtain a result that is of some practical use. Let us compute the radius of the orbit of a geostationary satellite – i.e. a satellite that appear to us from earth to remain in the same location in the sky. We therefore require the time period of the satellite’s orbit to be one day, since the time period must match the rotational time period of the earth. Let the height of the satellite’s orbit above the earth be a. The radius of its orbit, measured from the centre of the earth, is therefore d + a, if we use the symbol d to represent the radius of the earth. Therefore, in our formula above, we have
T = 1 day = 86,400 seconds.
We can substitute these numbers into our formula, and we find that:
d + a ~ 4.23 107 m
Therefore the geostationary orbit corresponds to a height a above the surface of the earth equal to about 3.6 106 metres, or 22 miles.
Kepler’s Laws for a Circular Orbit
Finally, we will take a look at Kepler’s laws, developed by Johannes Kepler in the 16th century. These laws relate to the motion of planets, and were developed by Kepler on the basis of astronomical observations made by Tycho Brahe, a Danish astronomer. (Kepler himself was German).
There are three laws, and these are:
1. The planets orbit the sun in ellipses, with the sun at one focus of the ellipse.
2. For a given planet, the areas swept by the straight line joining the centre of the planet and the centre of the Sun are equal for equal time intervals.
3. The ratio of the cube of a planet's mean distance, R, from the Sun to the square of the time period of its orbit, T, is a constant.
Taking these in reverse order, we have already derived Kepler’s third law, at least for a circular orbit. We have shown that
and hence by rearrangement,
which is the result that we need.
For a circular orbit, the second law automatically holds, since the velocity of a body in its orbit is a constant. This is because the radius r of the orbit is constant, and since angular momentum is a constant for the orbiting system, it follows that v is a constant.
For an elliptic orbit, the velocity is not constant at all points in the orbit, and the second law is more difficult to appreciate. We begin our consideration of the second law for an ellipse by considering Kepler's first law for an ellipse. Although mathematically more complex, it can be shown that Kepler's third law also holds for an elliptical orbit. We will not consider the third law further in this article.
Kepler's First Law
The following diagram illustrates Kepler’s first law - an elliptical orbit.
It can be seen that the elliptic orbit is not about the centre of the ellipse, but about a point located near one edge of the ellipse. This point is called a “focus” of the ellipse. Any ellipse has two foci, and a circle is a special case of an ellipse in which the foci are coincident. The foci play a special role in defining the mathematical properties of an ellipse, but this is not of interest to us here.
Now, in this system, angular momentum is conserved. The angular momentum of the system is conserved. If the sun is assumed to be stationary, the angular momentum J is
J = mvr
where v is the velocity of the mass m when the radius of the orbit is r. Thus, as r decreases (the mass gets closer in the orbit to the sun), v must increase to conserve angular momentum.
Kepler's Second Law
The second law is illustrated in the following diagram:
Because of the variation in velocity of the orbiting body, the angle traced out by the body in a given time varies depending on whereabouts in the orbit the body is. However, the velocity variation is such that the areas PABP and PCDP in the diagram above are the same, even though the angle swept in each part of these segments (θ1 and θ2) are different.
It can also be shown that Kepler’s second law follows directly from the conservation of angular momentum in this two body orbiting system. The angular momentum J can be written as
where we have set the velocity v = ωr and noted that the angular velocity ω is simply the rate at which the angle BPA in the above diagram is swept out. But the rate at which the area A of the orbit is swept out is just
which is of course a constant because J is a constant. This is Kepler’s second law for an elliptical orbit.