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

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 (*d _{moon}*) :

Is this the sidereal period or the synodic period? **Sidereal***Reason:*

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 (*d _{shuttle}*) = 300 km + radius of Earth
:

To solve for *P _{shuttle}*,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:
**

**
P ^{2}_{shuttle}/P^{2}_{moon}=d^{3}_{shuttle}/d^{3}_{moon}**

Notice that the constant

Use this equation to solve for the value of *P _{shuttle}* .
You will need a calulator that calculates square roots.

**
P _{shuttle}=P_{moon}*(d^{3}_{shuttle}/d^{3}_{moon})^{1/2}
**

**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__*Reason:*

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__*Reason:*

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.
*Reason:*

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?
*Reason:*

No. Although the equation P^{2}=K x d^{3} 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.