GRAVITY or GRAVITATION

This will give you what you need on gravity for the typical first physics course, plus some additional things about elliptical orbits that are found in more advanced courses. Also there is a brief discussion of gravity waves which you can ignore.

In everyday life, the gravitational force on an object is almost exactly constant, mg, where m is the mass and g is the acceleration due to gravity (also known as the gravitational field). I say almost, because g varies with elevation, but the variation is so small that you are unlikely to need to bother with it. Space flight, of course, is another matter.

To understand the universe, we need to understand gravity. Gravitational force is the attraction of a mass to any other mass. To all other masses in the universe, in fact. This is why the universe consists of lumps of matter (stars, planets, etc.) There was the Big Bang 13.7 billion years ago or so, sending matter out in all directions, and it has been slowing down and forming lumps ever since. Of course gravitation is a theory, so we know it about as well as we know evolution. Both may need a little tweaking from time to time.

Big news of 1998: The slowing down part of the last paragraph above may not be true! There is evidence that the velocities are increasing, so maybe there is some kind of repulsive force. Einstein had one (the cosmological constant) in his original general theory of relativity, and he later called that his biggest blunder. But maybe it is correct after all. (Einstein’s general relativity came before Hubble discovered that the galaxies are moving apart, which in turn lead to the idea of the Big Bang. Einstein made the reasonable assumption that the universe is static.)

The Sun and its planets were formed from stuff spewed out from supernova explosions in stars, and there was a lot of matter swirling around it which formed into planets. This probably happened around lots of other stars (some of these planets have been discovered), so it is likely that there is life elsewhere in the universe. (My assumption is that if a planet can support life, life will develop and evolve, but what do I know?)

You are a mass. Your physics book is a mass. You are attracted to your physics book! (And your book is attracted to you- equal & opposite force.)

Newton's law of gravity describes the attraction between masses in a way very much like the electrical force between charges. One big difference is that there is no such thing as gravitational repulsion. (Of course one could argue that maybe there is repulsive mass out there, and it must be long gone from our world, so how would we find it?)

(You may not need to know this paragraph, so skip it.) Here is sort of the way Newton reasoned: masses attract each other, evidently, because look at the moon going around Earth and Earth around the Sun. By weight = mg, we know it is proportional to one mass, but this doesn't make sense if it is an attraction between two masses-- it must be proportional to Mm. Now how does it vary with distance? We know gravitational acceleration here on Earth, 9.8 m/s², and we also know the moon's acceleration toward Earth, v²/r. Using the radius of Earth and distance from Earth to moon, he found that these accelerations were inversely proportional to the square of the distances. (The radius of the moon's orbit is about 60 times the radius of Earth, and he found that the moon's
v² /r is 9.8/60² m/s² .)

Thus F is proportional to Mm/r². To make an equation out of it, we stick in a constant of proportionality:

F = GMm/r²

We call big G the universal gravitation constant, and we need to find G by measurement. This was first done by Cavendish, and with today's precision it is

G = 6.6726 x 10^-11 N m²/kg².

It turns out that for spheres, you use the center-to-center distance for r in F=GMm/r². If you want to calculate the attraction between you and your physics book, a reasonable approximation would be to use the center-to-center distance, but to tell you the truth, this is not exactly right, because of the significant variation in distance between the various parts. For calculating the force between you and Earth or Moon, your head and feet are both about the same distance away from the centers, so the sphere to sphere approximation is ok.

In everyday life, weight = mg, and g is a constant 9.8 N/kg. Equate mg to GMm/R², cancel m, plug in the radius of Earth, 6.371 x 10⁶ m, and you can solve for the mass of the Earth.

Satellite problems: One force acts on a satellite, the force of gravity. The f = ma equation is therefore
GMm/r² = mv²/r if the satellite is in circular orbit. Note that m, the mass of the satellite, cancels out. Also one factor of r. From this we find how v is related to the radius of orbit:
v=(GM/r)^1/2

If you have not studied energy, ignore the rest of this document.

Gravitational potential energy is mgy if g = constant, where y is the vertical distance above a reference point. For practical problem solving, use the lowest point of m's motion as the reference point, so grav PE is never negative in this system. For space flight or motion of other bodies in space, g is not constant, so we need a different approach. This time, we use r = ¥ as the reference "point", so grav PE is never positive. The way to determine the PE formula is to imagine m moving toward M from infinitely far away. PE = -(work done by gravity). This is the integral
PE = ò (GMm/r²)dr,
where the negative disappeared because the element of distance traveled is -dr. Then we find that
PE = -GMm/r. If we used the surface of the Earth as a reference point we would find
PE = GMm/r_Earth - GMm/r, which is more awkward.

For a non-circular orbit (elliptical, with the planet's center at one focus), one can use Kepler's laws, which were found empirically from the data about planetary orbits. From a fundamental physics point of view, it seems better to use Newtonian physics: Newton's 2^nd law, conservation of energy and conservation of angular momentum. If you have not had all of these concepts yet, quit this stuff.

The total energy of the satellite, kinetic plus potential, is
¹/₂mv² - GMm/r. This energy is negative for a satellite (elliptical path). For a circular orbit, GMm/r² = mv²/r, so if you replace the mv² in the total energy expression with GMm/r and do a little algebra, you will find that total energy E=-GMm/(2r). For an ellipse, E=-GMm/(2a), where a is the semi-major axis. E is always negative for orbits. If E is zero, it would escape along a parabolic path and its velocity would approach zero as the distance approaches infinity, and if greater than zero, a hyperbolic path. The angular momentum is a cross product, rx(mv), and its magnitude is rmvsinq (see below)

So the two conservation laws tell us that
1. ¹/₂mv₁² - GMm/r₁ = ¹/₂mv₂² - GMm/r₂ and
2. r₁mv₁sinq ₁ = r₂mv₂sinq ₂ or r₁² w ₁ = r₂² w ₂ because w is (vsinq )/r.

Notice that m cancels out of both, hence satellites of different masses can have the same orbit. From Newton's 2^nd law and the equations above, it can be shown that the orbit is an ellipse (see an advanced classical mechanics text for the derivation). Equation 2 is equivalent to equal areas in equal times "swept" out by r. Area rate is r²w /2.

Now ignore the rest of this article- you do not need to know it.

A special case: if r₁ and r₂ are the maximum and minimum, the angles are 90^o, and with a little ;-) algebra on equations 1 and 2 above, you can show that
3. v₁ = (GM/r₁')^1/2 where r₁' = r₁(r₁+r₂)/(2r₂), and
4. v₂ = (GM/r₂')^1/2 where r₂' = r₂(r₂+r₁)/(2r₁).
As always, don't use specialized equations like this on tests unless you derive them. Also with either equation 3 or 4 and with the info on ellipses below, you can easily ;) show that total energy E = -GMm/(2a) as mentioned earlier.

Now some facts about ellipses: An ellipse is not just any oval shape; within it are two points called foci (plural of focus), located on the major axis (the long way across). Call the length of the major axis 2a. By definition, the ellipse consists of points located where the sum of the distances to the foci = 2a. Call the distance across the short way (the minor axis) = 2b. Let s= distance from a focus to the center of the ellipse. The eccentricity e is defined as e=s/a. (A circle has e = 0, and a long, thin ellipse has e close to 1.) Call the distance from a focus to the nearest point of ellipse (the perigee) distance p.

(Note: Most people, when guessing where the foci are, put them closer to the center than they should be. I did that in the previous sketch even though I was aware of the tendency. If you measure the hypotenuse and the dotted a in this sketch, you will see that the picture is very close to being correct.)

Given any two of the above quantities, you can use the definition of ellipse and the Pythagorean theorem in the right triangle below to find the others. There is nothing else to it, so this is not worth spending much time on unless you suspect that it might be on a test. Here are the grubby little results all worked out for you so you can spend your time on something meaningful, like enjoying a good brew:

Given a and b
s = (a² - b²)^1/2, e = s/a, p = a - s

Given a and s
b = (a² - s²)^1/2, e = s/a, p = a - s

Given a and e
s = ae, b = (a² - s²)^1/2, p = a - s

Given a and p
s = a - p, b = (a² - s²)^1/2, e = s/a

Given b and e
a = b/(1-e²)^1/2, s = (a² - b²)^1/2 , p = a - s

Given b and s
a = (b² + s²)^{1/2 ,} e = s/a, p = a - s

Given b and p
a = (b² + p²)/(2p), s = a - p, e = s/a

Given s and e
a = s/e, b= (a² - s²)^1/2, p = a - s

Given s and p
a = s + p, b = (a² - s²)^1/2, e = s/a

Given e and p
a = p/(1-e), s = a - p, b = (a² - s²)^1/2

Another approach to equation 3 and 4 shown earlier is to write them in terms of (GM/a)^1/2 and eccentricity e. The maximum speed (at the perigee) is
(GM/a)^1/2 times [(1+e)/(1-e)]^1/2 and the minimum speed (at the apogee) is
(GM/a)^1/2 divided by [(1+e)/(1-e)]^1/2.

With the origin at the center of the ellipse, the equation of the ellipse is

x²/a² + y²/b² = 1

With the origin at the focus where the attracting body is, using polar coordinates r and q ,
r(1 + ecosq ) = a(1 - e²) where q is defined to be zero at the perigee. Incidentally, that same equation gives you a parabola if e = 1 and a hyperbola if e>1.

For the orbit case it can be shown that
r² w = [GMa(1-e²)]^1/2 .
This is twice the rate of area swept out by the radius line. From this and the area of an ellipse, p a²(1-e²)^1/2 , we calculate the period T, the area divided by the rate:
T = 2p a^3/2/(GM)^1/2, so T² = [4p ²/(GM)]a³.
The fact that T² is proportional to a³ was discovered by Kepler from data about the planets, but the underlying theory was not known at that time.

To draw an ellipse, tie loops on both ends of a string. Its length will be 2a. Put thumb tacks or drive nails in where you want the foci to be (less than 2a apart) and with the string loops held by the nails or tacks, use a pencil to hold the string taut as you draw.

One more interesting property of the ellipse: lines drawn from the foci to a point on the ellipse form equal angles to the ellipse. So if you take a strip of flexible mirror material and form an elliptical shape, light from one focus goes to the other. The ruby rod laser: A ruby rod with flat ends will lase if it is pumped up with a flash of light. So a flash tube is put at one focus and a ruby rod at the other, and voila- we have a laser.

Light reflections off parabolas and hyperbolas: Parallel rays toward a parabolic mirror along the axis reflect toward the focus. Rays through one focus of a hyperbola reflect off in a direction away from the other focus.

Einstein's general theory of relativity explains gravity in an entirely different way, and it gives correct results in the few cases where Newton's law is slightly off. It is very difficult, though. The heart of it is simple: in free fall, all physics is the same as if there is no gravity. Conversely in a gravitational field, physics is the same as having no gravity but being accelerated. From that simple beginning, Einstein showed that gravity is curved space, a very hard concept to understand. The math involved is complicated also. Yet there are some easy things you can do with it using easy math. But don't click on general relativity. Too much knowledge is bad for you. We need good janitors, burger turners, etc.

Gravity Waves

You may not be aware of how electromagnetic waves (radio, microwave, light, etc.) are produced, so let's look at that first. It is similar to generating a wave in a garden hose. Shake the end of it rapidly up and down, and you will generate a wave pulse which travels on the hose. Similarly if you shake an electric charge up and down, a wave pulse travels along its field lines, and if you shake a mass up and down, a pulse travels along the gravitational field.

In general it does not need to be an up and down shake; any acceleration will do. Thus the rotating Earth generates gravitational waves and the Earth in orbit about the sun likewise. These waves are much too weak to be of any significance. But pulsars are rapidly rotating neutron stars with frequencies around 1 Hz or so. These radiate significant amounts of gravity wave power. Also binary stars- stars that orbit each other. One pair was observed for a number of years and its orbital period is slowing down, indicating energy loss, and its loss rate is consistent with the predicted gravity wave power.

No one has detected gravity waves yet, but the above is evidence of their existence. Also, gravity waves are predicted by general relativity, and the other predictions of general relativity have panned out, so physicists are confident that they exist.

You will now be gravitationally attracted back to the main physics page (mechanics) due to the curvature of cyberspace.

Other main physics pages:

fluids, heat, electricity and magnetism
vibrations and waves
quantum

Or look up stuff alphabetically in my index.

Comments, questions: fredrick.gram @ tri-c.edu