|Astronomy 110||PRINT Name   __________________________|
|Fall 2005   Section 006|
Homework 4 : Orbits
(Due Thursday, Oct 6, 2005)
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 :
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
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:
(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.