In the first section of Chapter 12, you're presented with the experimentally-verified law of universal gravitation, Fg = GM1M2/r². While this applies to any two masses, M1 and M2, we see it applied more frequently on a planetary scale. That's because gravitational forces are quite weak. For example, the gravitational force of attraction between two 1-kg masses separated by a meter is only 6.67x10-11 N. If, on the other hand, one of the objects is the Earth (mass ~6x1024 kg) and the distance r is the radius of the Earth (~6x106 m), then the Earth pulls on a 1-kg mass on the surface with a force of (7x10-11 Nm²/kg²)(6x1024 kg)(1 kg)/(6x106 m)² ≈ 10 N. (Note that although we expressed all values to 1 significant figure, the result for the weight of the object is within 2% of the value of 9.8 N expected for a 1-kg object.)
The constant G in the formula above is a universal constant of nature. It has the same value anywhere in the universe as far as we know. This is not like g, which is gravitational acceleration (or field) and depends on mass and distance.
The gravitational force has some interesting characteristics. It acts at great distances across seemingly empty space. In fact, its range is infinite, although gravitational forces are exceedingly weak at very large distances. Moreover, you can't hide or shield yourself from gravitational forces. The Sun attracts you the same whether or not the Moon is between the Earth and the Sun. Gravitational forces exerted by different objects on the same objects add like vectors so that the total gravitational force on the object is the vector sum of the individual forces. This is called the superposition principle.
An equation for gravitational acceleration is derived from the law of universal gravitation together with Newton's 2nd Law. This is a simple derivation. Consider an object of mass m in free fall near the surface of the Earth. The net force on the object is Fnet = mg = GmMe/Re². The mass of the object cancels, leaving g = GMe/Re². Thus, we see that gravitational acceleration isn't constant. For drops of a few or even tens of meters near the surface of the Earth, however, the value of g changes little, since Re is so large in comparison to the distance of fall. Let's see just how big the effect is. Suppose the object falls from height h to the surface of the Earth. The gravitational acceleration at height, h, is gh+Re = GMe/(Re + h)². At the surface of the Earth, the acceleration is gRe = GMe/Re². We compare these two values by forming their ratio,
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Note that in the numerator (Re + h), h is much smaller than Re and will therefore have an extremely small effect on the result. Let's see how small. First, we'll rewrite the ratio as follows.
If h is, say, 6 meters, then h/Re is 6 ÷ 6x106 or 10-6. Again, we've limited ourselves to 1 significant figure, since any greater precision wouldn't gain us anything. Obviously, 1 + 10-6 is close enough to 1 for our purposes. If our job were to measure gravitational acceleration to 6 significant figures, then we'd have to worry about that term of 10-6. But it's not going to be of concern to us in this class for falls over short distances. We will, however, have to deal with the r-dependence of gravitational acceleration when dealing with large distances.
The method of forming ratios in order to compare similar things is a very common method in physics and one that will save you much time in problem solving. Note how, in the problem above, we were able to cancel G and Me. It would have been a waste of time to substitute G = 6.67 x 10-11 Nm²/kg² and Me = 5.97 x 1024 kg and then divide them out.
You've already had the opportunity to use this method in many problems so far in this course. Consider this problem from E.10.1 Physlets: A quarter and a penny rest on a turntable (shown in an animation). Compare the magnitudes of the centripetal accelerations of the coins.
The tedious method of solving the problem would be to calculate the acceleration of each coin and then divide them. Moreover, when you were done, you wouldn't necessarily realize that the answer is the same no matter what the individual values of the acceleration are. The result only depends on the ratio of the radii of the coins' paths. If, on the other hand, you selected the equation ac = rw² and then noted that w was the same for both coins, then you saw that the acceleration was directly proportional to the radius. Thus, you only had to compare the radii. That's smart physics (and math). This kind of reasoning is called proportional reasoning, and there are ample opportunities to use it in the subject of gravitation. Here's another example.
Compare the acceleration of a freely-falling object near the surface of the Earth and the acceleration of a satellite in circular orbit a distance of 0.5Re from the surface of the Earth.
Since we're given the distance in terms of the Earth's radius, that's a clue that we're not going to have to substitute the value for the Earth's radius, because Re will divide out in the end. First, we realize that the orbiting satellite is a falling object, too. While it's in continual free fall, it has an orbital velocity which always keeps it the same distance from Earth. So our method will be the same as above: Compare the gravitational accelerations of the two objects. We see that G and Me are the same for both, so those values cancel out, leaving:
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Substituting, h = 0.5Re, we get gRe / gh+Re = (1.5Re/Re)² = 1.5² = 2.25 (or 2.3 to 2 significant figures). As suspected, Re divided out. Thus, the only arithmetic operation required was to divide 1.5 by 1 and square the result. How much more time would it have taken to substitute in values (twice!) for G, Me, and Re? Be kind to yourself and use ratios in algebraic form.
Here's an example where mass doesn't divide out. (You always have to be careful to examine the problem situation carefully in order to identify what the constants are.)
Compare the acceleration due to gravity at the surface of the Earth to that at the surface of the Moon.
Since the masses and radii of the Earth and the Moon are different, both will figure in the answer. G, however, will divide out.
We'll use the fact given on the inside back cover of the text that the mass of the Earth is about 81 times that of the Moon. Substituting values:
