Orbital motion problem

Here’s a problem I ran up against while teaching a student some physics, which allows us to see how Kepler’s three laws work together. An object – the Earth – is orbiting the Sun with a period of one year (3.16\times10^7 s). At aphelion, its distance from the Sun is 1.522\times10^{11} m, and its angular speed is 1.92\times10^{-7} rad/s. If we know the mass of the Sun is about 1.989\times10^{30} kg, can we calculate the angular speed of the Earth at perihelion? Without Google? Below is a picture of the situation, greatly exaggerated of course.

Earth at aphelion
Earth located at its aphelion position, with the Sun at one focus.

We will assume we know Kepler’s three laws:

1) Planets travel around the Sun in elliptical orbits, with the Sun at one of the foci.

2) The line connecting the planet and Sun sweeps out area at a constant rate – equal areas in equal times.

3) The square of the planet’s period is proportional to the cube of its orbit’s semimajor axis, and the constant of proportionality is set by the Sun.

This problem really hinges on the second law, so let’s start there. In any give amount of time, the line connecting the Earth and the Sun will sweep out a certain area. Double the area, double the time, and so forth. This can be expressed as A=\alpha t, or A/t = \alpha, where \alpha is some constant of proportionality. It is not straightforward to calculate these areas, since they involve transcendental equations. This dilemma formed the basis for the so-called “Kepler Problem” of 17th and 18th Centuries, the which partially motivated the development of the calculus. There is one area and time that can be calculated exactly, which will provide the exact value of \alpha, though. The area of the entire ellipse is \pi ab, where a and b are the semimajor and semiminor axes of the ellipse, respectively. The time it takes for one orbit is the period, T. Hence, \alpha = {\pi ab}/{T}.

It’s still not easy to calculate the area of a sector of the ellipse, since both the angle and the radius is constantly changing. However, we can get a little traction by looking at infinitesimal motions.

Infinitesimal sector of ellipse
An infinitesimal sector of the ellipse.

In the infinitesimal, the tiny sector traversed by our planet, ds, is approximately a straight line. Therefore, our sector is approximately a triangle, with two sides equal to r, and one equal to ds. We know from circle geometry (sure, an ellipse is not a circle, but don’t worry about that right now!) that the size of the arc is equal to the angle times the radius. Thus, ds=r d\theta. The area of a triangle is equal to half the base times the height. Here, the base is r, and the height is, well, ds=r d\theta. So, the area of the sector, dA is equal to \frac{1}{2} r r d\theta = \frac{1}{2} r^2 d\theta. If we just divide both sides by the tiny amount of time that transpired, dt, then we have Kepler’s second law expressed in the small:

\frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt}

Since \frac{d\theta}{dt} is just the angular speed \omega,

\frac{dA}{dt} = \frac{1}{2} r^2 \omega

And, since it doesn’t matter how small the area or time are for Kepler, we know that this value is always constant. In other words, as the radius goes down, the angular speed goes up. I our problem, we have the angular speed at aphelion, and need to find the angular speed at perihelion. Since \frac{dA}{dt} is constant,

r_\alpha^2 \omega_\alpha = r_\pi^2 \omega_\pi

or

\omega_\pi = \omega_\alpha (\frac{r_\alpha^2}{r_\pi^2})

So, all we need to find is the perihelion distance, r_\pi. This brings us to Kepler’s first law, and a little geometry.

The ellipse has fascinating relationships within it, and it is incredible that the universe is constructed such that orbits follow this geometry. Here, we will only scratch the surface. The ellipse is defined as that shape the sum of whose distance from two points is always constant. In other words, take a point on an ellipse, and draw lines to the two foci. The total length of those lines is the same, no matter where on the ellipse you started.

How to draw an ellipse
The generator of the ellipse comes from the two foci, f1 and f2. f1P1 + P1f2 = f1P2 + P2f2.

One unique way to draw the two lines is to start at one focus, draw straight through the other focus, and then back to that other focus. Now, it should be obvious that the total length of the line is equal to the long diameter of the ellipse, the diameter that passes through both foci. Half of that diameter is the semimajor axis, a. Therefore, the total length of line from focus to ellipse to the other focus will always be equal to twice the semimajor axis, or 2a.

Well, the diameter can also be broken into two other lengths. If we draw the line from the Earth at aphelion to the Sun, and then keep going, we will hit the Earth’s perihelion position. The major axis is thus equal to the perihelion distance plus the aphelion distance, or 2a = r_\alpha + r_\pi. Therefore, if we knew the length of the semimajor axis, a, then we could calculate the perihelion distance, r_\pi, and thus the angular speed at perihelion! But, how do we determine the length of the semimajor axis? Kepler’s third law.

Geometry of the Ellipse
Some primary measures of the ellipse. Semimajor axis = a, Semiminor axis = b, distance from center to focus = c.

Kepler’s third law states that the square of the period is proportional to the cube of the semimajor axis, and the proportionality is a constant for any orbital system. Some math voodoo shows that the constant is really only a function of the mass of the central body, in this case, the Sun. In other words, T^2 = (\frac{4\pi^2}{GM})a^3, where M is the mass of the Sun. So, we can calculate the semimajor axis:

a = ((\frac{GM}{4\pi^2})T^2)^{1/3}

Let’s put it all together now.

Kepler’s second law: \omega_\pi = \omega_\alpha (\frac{r_\alpha}{r_\pi})^2

Kepler’s first law: r_\pi = 2a - r_\alpha

Kepler’s third law: a = ((\frac{GM}{4\pi^2})T^2)^{1/3}

Or, cowboy style: \omega_\pi = \omega_\alpha (\frac{r_\alpha}{2((\frac{GM}{4\pi^2})T^2)^{1/3} - r_\alpha})^2

First, use Kepler’s third law to get the semimajor axis:
a=((\frac{GM}{4\pi^2})T^2)^{\frac{1}{3}}
= (\frac{(6.673\times10^{-11}\frac{m^3}{kg\dot s^2})(1.989\times10^{30} kg)}{4\pi^2})(3.16\times10^7 s)^2)^{\frac{1}{3}}
= 1.497\times10^{11} m

Second, use that vaule in Kepler’s first to get the distance to the perihelion:
r_\pi = 2a-r_\alpha
= 2(1.497\times10^{11} m)-(1.522\times10^{11} m)
= 1.472\times10^{11} m

Last, put that value into Kepler’s second law to finally get the perihelion angular speed:
\omega_\pi = \omega_\alpha (\frac{r_\alpha}{r_\pi})^2
= (1.94\times10^{-7} \frac{rad}{s})(\frac{1.522\times10^{11} m}{1.472\times10^{11} m})^2
= 2.06\times10^{-7} \frac{rad}{s}

Just to check with NASA, this angular speed would give a perihelion speed of v = r\omega = (1.522\times10^{11} m)(2.06\times10^{-7} \frac{rad}{s} = 30313 \frac{m}{s}, which is pretty close to NASA’s value of 30290 \frac{m}{s}. I mean, it’s less than 1% different!

Let’s recap. If we have the aphelion angular speed, we can calculate the perihelion angular speed through Kepler’s three laws. The third law gives us the semimajor axis from the period, the first law gives us the perihelion distance, and the second finally gives us the angular speed. How is that for some Wednesday morning physics?

2 thoughts on “Orbital motion problem”

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