Uniform Circular Motion.

Labs written for the CARA Space Explorers, Fall 1993

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This is meant to be handed out to the students.*

Recall the distances involved in circles. We will usually deal with the radius (R) and will use the units of kilometers. If you walked all the way along the edge of the circle, you would walk around its circumference (C). It too will be measured in kilometers in this class. These are related by the famous equation:

C = 2R

Here's an example of how to use this relationship. The Earth's radius is 6,378 km. What is its circumference? (Recall that =3.14) Plugging into above we get: C(Earth) = 2 x 6,378 km = 6.28 x 6,378 km = 40,050 km

When something moves from point A to point B, a line drawn from the center of the circle (called a radial) sweeps out an angle w. In this class we will use degrees as our units of angle. Recall that a full circle has 360° . The relationship between the distance (s) along the edge of the circle between points A and B is simply the proportion:

s/C = w/360 °

Here's an example of how to use this equation. Since there are 24 hours in a day, that means the Earth turns 1/24th of its circumference every hour. How many degrees does it turn in one hour? Using the above equation we get: 1/24 = w/360° . Multiplying both sides by 360° we have, 360° x 1/24 = w. Or, w=360° /24 = 15° .

When anything rotates or moves in uniform circular motion, we then need to take the above relationship one step further. Suppose we have a satellite moving at a speed v, where v is the distance it covers in km/hour.

If the satellite is moving at a speed v along the circle travels a distance s in a time t.

v=s/t is just the definition of v.

It also sweeps out an angle w in the same time. Now define = w/t. This funny looking number is called the angular velocity of the satellite. In this class will have the units degrees per second. It's related to v by a simple relation:

= 57.3(v/r)

When a satellite is moving at a constant speed in its orbit, then v/r is constant too, so that is constant as well. When this angular speed stays constant, then we have uniform circular motion. One important way to write this equation is gotten by multiplying both sides by r and dividing both by 57.3. Then we get:

v = r / 57.3

Let's apply this. Suppose the Space Shuttle completes one orbit in 1.5 hours. Also, assume it orbits at an altitude of 500 km over the Earth. That means the radius of its orbit is 6,378 km (the radius of the Earth) + 500 km = 6,878 km.

Since one orbit takes it through a total of 360° , then the Shuttle's angular velocity is = 360° / 1.5 hours = 240° per hour. Plugging this and the radius into the above equation, we get the speed of the Shuttle:

v = 6,878 x 240/57.3 km/hour = 28,808 km/hour

That's very fast!

1. Chicago is about 40 km from its southern border to its northern border. Suppose you drew a great circle, centered on the Earth's core, going north and south, all the way around the Earth and cutting through Chicago. How many degrees of the circle would the city take up in the north-south direction?

2. You are driving at an angular speed along the Earth of 1.5 degrees per hour. Do you think you might get a traffic ticket if the regular speed limit is 65 miles per hour? (Recall that 1 mile is equal to 1.62 kilometer.)

3. A jet liner travels at about 1,000 km/hr. Assume it travels very close to the Earth (in other words, the radius of its orbit is 6,378 km). How long will it take to complete one "orbit" of the Earth? (Hint: Just think about how to do this problem. You won't simply be able to get the answer by plugging into one of the equations above.)

Challenge Problem: 4. It turns out that the period of a satellite's orbit will be very important to us. It is defined as the length of time it takes for a satellite to complete one orbit. Derive a simple equation for the period (P) of a satellites orbit in terms of its speed in its orbit (V) and the radius or its orbit (R). (Hint: Begin by using the circumference of the orbit.)

**Important Disclaimers and
Caveats**

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