Astronomy 110 PRINT Name   __________________________
Fall 2005   Section 006

Homework 4 : Orbits

(Due Thursday, Oct 6, 2005)

This homework asks you to compute some of the orbital properties of artificial satellites of the Earth by comparing with the Earth's natural satellite --- the Moon --- using Kepler's third law. Although Kepler formulated this law for the planets orbiting the Sun, it is also valid for any orbit situation where gravity is the force keeping the satellites in orbit. Comparing an unknown situation with a known one (in this case, the orbit of the Moon) is a powerful technique that makes the algebra simpler. However, you will have to be careful about units, e.g. all orbital radii in kilometers, say, and all orbital periods in days. It doesn't matter which units you choose, as long as they are consistent among the different orbits.

Start by writing down the parameters of the Moon's orbit --- its average size and its period. These are your known quantities. You can find these in the textbook (try the appendices). The quantities you want are the distance of the Moon from the center of its orbit (the center of the Earth) and the period of its orbit.

Write these in here :   (Remember the UNITS!)

Size of orbit  (dmoon)   :   3.844*10 5 km     Period   (Pmoon)   :   27.32 days

Is this the sidereal period or the synodic period? Sidereal

The time to complete one orbit is called the sidereal period. The synodic period is the time between phases, which is longer due to Earth's own motion around the Sun. See Fig 2.22 in our text.

Low Earth Orbit (Space Shuttle)     The first thing you'll calculate is the period of a satellite, such as the Space Shuttle, that (usually for reasons of economy) is in an orbit quite close to the surface of the Earth. Assume the average height of the satellite is 300 km above the surface of the Earth. Remember that it's distance from the center of the Earth that's important, however, so you'll need to add the 300 km to the radius of the Earth to get the size of the satellite orbit. This is also one of the known (or assumed) quantities.

Write this in here :   (Remember to use the SAME UNITS as for the Moon!)

Size   (dshuttle) = 300 km + radius of Earth   :   6678 km     Period   (Pshuttle)   :   unknown

To solve for Pshuttle,the method is to write down Kepler's third law, as an equality, twice, once for the Moon and once for the shuttle :

P2moon = K x d3moon        and          P2shuttle = K x d3shuttle


The trick is to divide one equation by the other so that you end up with one equation for what you want to know (the period of the space shuttle's orbit) in terms of the other three known quantities. Do that here:
The result from dividing the two equations is:


Notice that the constant K goes away so you don't need to worry about its value.

Use this equation to solve for the value of Pshuttle . You will need a calulator that calculates square roots. Remember the UNITS.
Using the equation above we can solve for the orbital period of the shuttle:


Using our units of days and km, our answer will be in days, which we need to convert to minutes. The equation gives 0.063 days:

0.063*24*60=90 min.

(Note: my equation takes the distances to the one-half power, this is equivalent to taking the square-root, but easier to write on a webpage.)

What is the answer in minutes? 90 minutes

Notice that this is less than a day. A satellite in an orbit with a period of one day is called a geostationary satellite. Such a satellite stays over the same spot on Earth all the time.

Do you expect that a geostationary satellite will orbit closer to the Earth than the low-Earth satellite you just calculated, or further away? Farther away

Our equation tells us that as the size of the orbit increases, the time to complete one orbit also increases. So since the shuttle takes much less than one day to complete its orbit, it must be closer to Earth than a geostationary satellite.

Will it be closer than the orbit of the Moon? Yes, closer than the Moon

Similarly to the last question, since the moon takes 27.32 days to orbit Earth, it must have a larger orbit than the geostationary satellite which takes 1 day to orbit.

Consider a satellite in a very low orbit, one just grazing the surface of the Earth (you'd have to clear away mountains and tall buildings). Explain why such a satellite doesn't orbit once in 24 hours, even though the ground just underneath it rotates once round in 24 hours.

We saw that the smaller an orbital distance the shorter the orbital period. We can use the equation for the shuttle's period to calculate the period of something orbiting on the surface of the Earth where the orbital distance is the Earth's radius d=6378 km. Plugging in this value gives a period of about 84 minutes. This has nothing to do with the fact that it takes the Earth 24 hours to rotate (it's not orbiting itself). There is only one place where the orbital period is 24 hours, that's where a geostationary satellite is located, which we said is out between the orbits of the shuttle and the Moon.

Can you use your equation to calculate the properties of the orbit of, say, Mars? that is, can you directly compare in this way the orbit of the Moon and the orbit of Mars?

No. Although the equation P2=K x d3 applies to all types of orbits, we used data from the Moon orbiting the Earth in all our calculations. That means that all our calculations assumed we were orbiting the same thing, the Earth. Since Mars orbits the Sun, and not the Earth, we would have to use information about objects orbiting the Sun to learn about Mars' orbit.