S-5.Waves & Photons
Optional: Quantum Physics
Q2. Atoms (and 6 more)
S-6.The X-ray Sun
S-7.The Sun's Energy
S-7A. The Black Hole at
our Galactic Center
of Atoms and Nuclei
26. Robert Goddard
27. Early Rocketry
29a. Looking Outwards
29b. Looking Earthwards
29c.Observing local space
29d. Useful Spaceflight
29e.Exploring far Space
30.To Space by Cannon?
Center of Gravity
Most people have at least an intuitive notion of the center of gravity (CG) of an object: it is the point on which the object can be perfectly balanced. Grab a broom at one end and the other end tries to drop down; grab it at its center of gravity, and it stays balanced, neither end tipping over.
If you have learned to balance a chair or a broom on the palm of your hand, you know the trick is to place your hand right below the center of gravity. Since your hand is not at the CG but below it, it must be constantly moved to keep that strategic position.
There also exists a precise mathematical definition--it has nothing to do with gravity, which is why many scientists and engineers prefer the term center of mass. However, it leads off the main subject and therefore we won't bother with it now. A lightweight stick with two balls of equal weight at its end obviously has its CG in the middle. When one ball is twice the weight of the other, the CG divides the distance between them by a ratio 1:2, in a way that makes it closer to the heavier mass (see figure). And similarly for other ratios.
Balls that push each other
Now imagine that instead of a lightweight stick the above two heavy balls have a spring between them, held compressed by a string. Even though the balls are separate, one can speak of their common center of gravity, on the line connecting their centers, 1/3 of the distance from the center of the heavier ball.
(The CG of the Earth-Moon system can be defined in the same manner. Since the ratio of masses of the two bodies is about 81:1, the CG is the point on the line between their centers dividing it by that ratio. It can be shown that--neglecting the pulls of the Sun and of other planets--the Moon does not orbit the center of Earth, but rather the common CG--and so does the Earth, reacting to the pull of the moon. Of course, since the Earth is much more massive, the CG is not very far from the center of the Earth--in fact, it is closer than the Earth's own surface.)
Suppose next that a lit match is placed against the string, burning it through. As the spring expands, it pushes the balls apart; if it is sufficiently light, its own motion does not matter and we can assume that the balls push each other.
By Mach's formulation of the equations of motion, if the heavy ball receives an acceleration a, then the light one gets 2a, twice as much. For each increment in the velocity of the heavy ball, the light one receives twice as much, and it follows that at any time, its total velocity, as well as the distance covered, are twice those of the heavy ball. Conservation of momentum leads to the same result.
If then the heavy ball is at a distance D from the initial position of the spring, the light one is at distance 2D--as in the earlier figure, reproduced here. No matter how much time passes, the center of gravity stays at the same spot.
That turns out to be a very general principle: in any object or collection of objects, forces which only involve those objects and nothing else ("internal forces") cannot shift the center of gravity.
An astronaut floating in a space suit cannot shift his position without involving something else, e. g. pushing against his spacecraft. The center of gravity--or "center of mass"--is a fixed point, which cannot be moved without outside help (turning around it, however, is possible).
By throwing a heavy tool in one direction, the astronaut could get moving in the opposite direction, though the common center of gravity of the two would always stay the same. Given a bottle of compressed oxygen, the same result follows from squirting out a blast of gas (a scene that appeared in an early science fiction film). A rocket does much the same, except that the cold gas is replaced by the much faster jet of glowing gas produced by the burning of suitable fuel.
| Launch of an
The powerful rockets which lift hundreds or even thousands of tons off the launching pad depend on the same principle. If you ever watched a rocket lift off at Cape Canaveral, it is worth remembering that if you could somehow remove from the scene the launching pad, the atmosphere and the Earth, then the combined center of gravity of the rocket and its exhaust gases would always remain where it started, at the launching point.
It may seem like a round-about way for producing motion. And yet, rockets are (at least for now) the only practical means of leaving Earth and flying into space.--------------------------
As the fuel is burned, it is ejected with some constant velocity w relative to the rocket, creating (we assume) constant thrust. Let us simplify matters by also assuming the launch is from some point in space, so that the thrust of the engine only has to overcome the rocket's inertia. In launches from the ground, part of the thrust is needed to overcome gravity too--see Section #18.
The rocket accelerates gradually. Starting from rest, its moves rather slowly at first. After a while, however, not only is its velocity greater, but its acceleration has grown too: at first, nearly the entire mass 2M must be accelerated, but as fuel is used up, the mass being accelerated is less and less. By the time a mass M of fuel has been burned--half the starting mass of the rocket--its acceleration has doubled, because the same push is applied only to a mass M. Its velocity V at this point may be significant, but to calculate it (given w and the rocket's thrust) requires calculus, so let us just assume we have somehow got that value.
The payload now has gained velocity V.
But we need more! So we build a rocket of mass 4M, of which 2M is fuel, while the payload, also of mass 2M, is the smaller rocket described above, serving as second stage, with half of its mass also given to fuel (to simplify the calculation, we neglect the mass of the bigger rocket itself, although it, too, needs to be accelerated). When the fuel of the big rocket is finished, we reach a velocity V, then the second stage is ignited, adding another V to the velocity of the payload, for a total of 2V.
Still faster! Now the rocket has mass 8M of which 4M is fuel of the first stage, while 4M is the two-stage rocket of the preceding design. The first stage gives velocity V, to which the other two add 2V, for a total of 3V.
By now you can see the trend. If the mass of the final payload is M, then
|Total mass||Gives final velocity|
Each time the velocity increases by one notch, the mass doubles.
Staging of the rocket makes the process more efficient. If the rocket of total mass 8M (say) had just one engine and it burned 7M of its fuel, it seems as if the same effect would be obtained as firing 3 identical rockets simultaneously. (We still need divide the calculation into stages--say, the burning of the first 4M, then the burning of 2M, then of the remaining M. Each time, less mass needs to be accelerated).
However, the initial rocket engine for accelerating 8M would have to be very big, and that creates at least two problems. First, it would be heavy, and towards the end of the burn, it is inefficient to carry along such a massive engine (and a big empty fuel tank, too), in addition to the payload. And second, because it is powerful enough to lift 8M (and rockets cannot very well adjust their thrust), at the end it creates a large acceleration, subjecting the structure of the rocket to a much greater force (see example in Section #18). It is better therefore to drop the big empty tanks and large engines along the way, and continue with smaller ones.
(By the way, the "Atlas" rocket pictured above kept its very lightweight stainless steel tank all the way, but dropped two of its three engines).But the basic pattern remains: the final velocity grows much more slowly than the mass of the rockets required by it (like the logarithm of the mass, if this means something to you). A rigorous derivation (which uses calculus) gives the equivalent of a huge number of little stages, fired one after the other, but leads to the same conclusion.
Next Stop: #26 Robert Goddard and his Rockets
Timeline Glossary Back to the Master List
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 10-23-2004