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Guide 12-1. Solving Gravitation Problems using Proportional Reasoning

 The Gravitational Force Law

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.) 

Don't interchange the use of the symbols G and g.

The constant G in Fg = GM1M2/r² 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.


Here are some important facts about gravitational forces and the gravitational force law.

  • Gravitational forces act at great distances across seemingly empty space. In fact, the range of the gravitational force is infinite, although gravitational forces are exceedingly weak at very large distances.

  • You can't hide or shield yourself from gravitational forces. For example, 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 object 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.


 Acceleration due to Gravity

An equation for acceleration due to gravity is derived from the law of universal gravitation together with Newton's 2nd Law. Consider an object of mass m in free fall near the surface of the Earth. The force diagram is shown to the right. (The same diagram applies whether the object is rising or falling.) Fg represents the gravitational force of the Earth on the object. Of course, this is also the weight of the object. The direction of the acceleration is down, so we pick the direction of +y down as well. The net force equation is:

,

where Me represents the mass of the Earth. The distance r is the distance from the center of the Earth to the object, where we assume that the distance between the object and the surface of the Earth is negligible in comparison to the radius of the Earth. Substituting r = Re for the radius of the Earth and applying Newton's 2nd Law we obtain the following.

.

The mass on the left-hand side of the equation is the mass of the object. Dividing out this mass from both sides gives the acceleration. We denote this by ag to emphasize the fact that this acceleration is due to gravity. This plays a role similar to what we have denoted g in the past. Note, however, that we cannot take g to be a constant if the object in question is very far from the surface of the Earth. 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 acceleration at height h is ah+Re = GMe/(Re + h)².  At the surface of the Earth, the acceleration is aRe = GMe/Re².  We compare these two values by forming their ratio,

.

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 the 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 the acceleration when dealing with large distances.

 The Proportional Reasoning Method

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.  onsider the following problem from earlier in the course. The animation shows two coins moving in circular paths on a turntable with constant angular velocity. The problem is to compare the magnitudes of the centripetal accelerations of the points. The tedious method of solving the problem would be to calculate the acceleration for each point and then divide them. Moreover, at the completion of the problem one 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, the equation ac = rω² were used, one sees that sinceω is the same for both coins, the acceleration is directly proportional to the radius. Thus, one only has 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 an example: Compare the acceleration of a freely-falling object near the surface of the Earth to 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 accelerations of the two objects. We see that G and Me are the same for both, so those values cancel out, leaving:

.

Substituting, h = 0.5Re, we get aRe / ah+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

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:

 


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