# Homework Help: 2 Gravitation Problems

1. Apr 20, 2007

### ttback

There is a set of two problems, I can solve problem 1, can't solve problem 2. But in order for you to help me solving problem 2, I think you need to use the answer for problem 1.

1. The problem statement, all variables and given/known data
P1. A satellite hovers over a certain spot on the equator of (rotating) Earth.
What is the altitude of its orbit(called a geosynchronous orbit)

2. Relevant equations
We use Kepler's Law of Period to get r, which is the distance between the satellite and the center of Earth.
T^2 = (4*pi^2 / G * Me) r^3

T = Time of one day = 24 h = 86400 s
Me = 5.98 x 10^24 kg
G = 6.67 x 10^-11 (some crazy unit)
r = That is the unknown we want

3. The attempt at a solution

Altitude = r - Re
r = 4.225 x 10^7
Re = 6.37 x 10^6 m

Altitude = r - Re = 3.58 x 10^7 m

================================================================
P2. Assume that the satellite of P1 is in orbit at the longitude of Chicago. You are in Chicago (latitude 47.5 degree) and want to pick up the signals the satellite broadcasts. In what direction should you point antenna?

The situation in the problem, as I interpret, is that the satellite is now on the same longitude as Chicago, same latitude as equator, and still hovering over one fixed point on the equator.

If so, the antenna must turn to the south and makes a angle with the ground to point at the satellite.

Now I know the satellite's altitude from P1, I can also find out the distance between Chicago and the Equator,(from the latitude) then I can calculate the angle using Arctan(Altitude/Distance).

But my question is, isn't Earth round? So the distance between Chicago and that Equator was measured by a curved path? Then we wouldn't really have a triangle, right?

Can we still use Arctan() to solve this problem? Or maybe I am just overthinking because the curve can be ignored?

2. Apr 20, 2007

### ttback

never mind, i think I know how to do it.

Law of Cosine to find the central angle to the arc between Chicago and Equator.

Law of Sine to solve for the chord length between them

Then we can get the answer.

3. Apr 20, 2007

### rootX

radius of the earth is touching both equator and Chicago, and using latitudes and longitudes thing you can find the angle between the line connecting equator and Chicago, from the center.

4. Apr 20, 2007

### rootX

yea, me had the same thing.

5. Apr 20, 2007

### ttback

P2. Assume that the satellite of P1 is in orbit at the longitude of Chicago. You are in Chicago (latitude 47.5 degree) and want to pick up the signals the satellite broadcasts. In what direction should you point antenna?

The situation in the problem, as I interpret, is that the satellite is now on the same longitude as Chicago, same latitude as equator, and still hovering over one fixed point on the equator.

If so, the antenna must turn to the south and makes a angle with the ground to point at the satellite.

The angle, is the angle the antenna makes with the tangent line to Earth's surface.

It is annoying to explain the geometry, so I just drew it on the file diagram.jpg.

Now I know the satellite's altitude from P1, it is H.

We want the arc radian between chicago and equator,

Using the formular here: http://www.krysstal.com/sphertrig.html

The Cosine rule gives us the central angle.

Note: There can be other ways to get the central angle, cuz this is a geometry problem, full of hidden paths, but what the heck, I need to finish this before I lose interest.

Now, we have an Isoscele, we know Earth radius, central angle, and the two end angles. So we can find the chord length by law of sine.

Using Arc(H/chord), we got a2.

Now we have to find a1.

Using geometry, a1= 90-[(180-Central Angle)/2)]

#### Attached Files:

• ###### diagram.JPG
File size:
14.7 KB
Views:
303
6. Apr 22, 2007