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}*) : ___________________
Period (

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

**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
: ___________________
Period (

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:

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 *P _{shuttle}* .
You will need a calulator that calculates square roots.

What is the answer in 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? _______________
*Reason:*

Will it be closer than the orbit of the Moon? _______________
*Reason:*

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:*

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:*