Lets take a look at some of the other aspects of satellite behavior. To do so, we will use various combinations of Kepler's Third Law and Newton's Law of Universal Gravitation, and our recently acquired knowledge of centripetal motion.

First... remember this formula from section 8.1 (formula #11 on that page.)?

This relationship can be solved for the period T of any satellite of an object with mass m₁, at a distance r from the center of that mass.

1. and

2.

• This relationship should look familiar to you by now. It is the same as the period formulae we derived for the period of a pendulum, and for anything in centripetal motion. That took the form of
• 3. where r is the centripetal radius and a_c the centripetal acceleration.
• In the previous section (equation 10) , we determined that which says that
• 4. . If we substitute this value for a_c into formula #3 above, you arrive at formula #2 again!
• In orbital systems, the centripetal acceleration can also be regarded as the field strength g of the system at a distance r from the center of mass of the central body in the system.
• Note also that m₁ here is the mass of the central body in the orbital system, NOT the mass of the satellite. The mass of the satellite itself is apparently irrelevant!

So.. note that for any given orbital radius r, there is only one value of period T that is valid, that value calculated by formula #2 above. If a satellite has a constant orbital radius and therefore covers a specific circumference in period T, the satellite must have a specific velocity that works for that particular radius. By this I mean.. for any given orbital radius r, there appears to be only ONE velocity at which any satellite of any kind can travel. Lets explore this a little more.

In the section on centripetal motion in chapter 7 we learned that

5. . Substituting this value in to equation 4 above, we see that

6. . Multiply both sides by r, and take the square root to arrive at

7.

• This assures us again that at any orbital radius r, there is only one value of v that will work, regardless of the mass of the satellite at that radius.
• Remember.. m₁ here is the mass of the body at the center of the orbital system. When it comes to orbits, the central mass RULES. It's never about the mass of the satellite!
• Note here also that if you can observe both v and r, you can determine m₁, the mass at the center of the system.

You try:

• The radius of the moon's orbit around the Earth is 3.8 x 10⁸ meters. The mass of Earth is 5.98 x 10²⁴ kg. What is the velocity of the moon?
• Using formula #7, you should arrive at 1.0 x 10³ m/s. You should arrive at this same value using any other velocity formula.

Now.. one last formula for velocity of a satellite...

In formula #11 from the previous section, we saw that Kepler's constant for an orbital system could be expressed as

8. . Lets assign this constant to the symbol k for now such that

9. . Note that now

10.. If we substitute this value into velocity formula #7 above we see that

11..

• This lets us find the velocity of any satellite in an orbital system for which we already know the value of k, and r. k can be determined from the observations of other satellites.
• Conversely, it lets us also solve for r if we can observe velocity!

Weight and weightlessness.

Ok.. consider an astronaut in the space shuttle. You have all seen video of them moving around as if they have no weight at all. Yet, Universal Gravitation clearly indicates that a gravitational force of attraction (weight) exists. Also, you can calculate the acceleration rate g as

as we saw in the previous section.

The shuttle is at about 400 km above the surface of the Earth. Add that value to the radius of the Earth for "d" in this formula, and the mass of the Earth, and we find that the value of g at that altitude is -8.7 m/s/s.... only a little less than here on Earth. So what gives... why do those astronaut guys seem to be floating "weightlessly" ?

Simple..

• they still have weight, and they still are accelerating.
• What they don't have is something on Earth like a spring scale pushing back up on them, and what they are not doing is running into the Earth! They are simply falling.
• They also have a velocity as they accelerate toward the Earth and move around the orbit simultaneously.
• At those special velocities, (remember.. the only ones that work for a specific orbital radius) the surface of the Earth curves away from the falling objects as fast as the object approaches the surface!
• And that is what an orbit is all about!

Video: "The Apple and the Moon" from the Mechanical Universe Series.

You have seen that there you can calculate the value of g at any specific point from the center of the Earth. Imagine a collection of zillions of these points.

• The closer to the Earth, the higher the value of g, and the farther away the lower the value.
• This collection of all possible points with varying values of g is called the gravitational field of the Earth. All masses have them.
• g is the field strength and can also be calculated as F/m... the Force per unit mass on an object at a certain point in the field.
• At the Earth's surface, g is -9.8 N/kg.

Einstein's theory of gravity

Albert Einstein theorized that gravity itself is not a force, but a behavior of space. He imagined that a mass would warp space around itself, thus bringing any objects in that space closer. So far, this General Theory of Relativity has proven to produce correct results every time it is tested. Relativity is a whole new ball game that we will not enter into at this point, but you should know its basic premise.

Activity: Project NEAR and Eros!

Homework: set #3 Ch8 questions 9-15 page 194 set #4 ch8 probs 41-49, page 195-96

video: "The Apple and the Moon" 15 min.

back to ch 8.1