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Lecture #9, June 19 Lecture 8.1 Gravitation 1. Gravitational Force During our discussion of forces we talked about gravitational force acting on any object near the earth's surface. We have already learned that this force provides the same acceleration of magnitude g 9.8m s2 directed to the earth's surface for all the objects near it. The fact that acceleration is the same for all the objects means that this gravitational force is proportional to the mass of the object. This discovery was first made by Isaac Newton during the summer of 1665, when he was watching an apple falling to the ground. Generalizing this idea, one has to accept that if there is force acting between the earth and any other object, it should be a force of the same nature acting between any objects of nonzero mass. Since there is force acting on the apple from the earth, it should also be force acting from the apple on the earth. According to the Newton's third law, these forces have to have the same magnitude. This means that gravitational force has to be proportional not only to the mass of the first object, m1 , but also to the mass of the second object, m2 . In the case of the earth's gravitational force this second mass is the mass of the earth itself. So FG m1m2 . (8.1.1) Since the mass of any other object near the earth's surface is much smaller than the mass of the earth itself, this is the reason why we can not detect gravitational forces between the other bodies. However, this is not always true. Anybody who lives near the ocean knows about the rise and the fall of the tide. This phenomenon is due to gravitational force, but not the earth's gravitational force. Water rises and falls because of gravitational attraction to the Moon. We also know that this effect occurs on periodic basis, depending on the Moon's position. So, gravitational force depends not only on the masses of interacting objects, but also on their positions. To be precise, it depends on the distance between the interacting objects. The next question to ask is how this force depends on that distance. Newton was the first to work on that question. He made calculations similar to the following example. Example 8.1.1. What is the Moon’s acceleration due to gravitational force acting on it from the Earth? Let us notice that gravitational force of the Earth is the only force acting on the Moon (ignoring gravitational forces coming from the other celestial bodies). This force is responsible for the Moon's motion around the Earth, meaning that it provides centripetal acceleration, which is, at the same time, gravitational acceleration, so 2 v 2 F 2R I 1 4 2 R a M M M , G J 2 RM H TM K RM TM 8 where RM 3.86 10 m is the radius of the Moon's orbit and 6 TM 27.3days 655hours 2.36 10 s is the Moon's rotation period, so 4 2 3.86 108 m d i 3 2 1 a 2 2.7310 m s g . d2.36 106 si 3590 This means that gravitational acceleration at the distance from the Earth's center is 3590 times less than that on the Earth's surface. Newton noticed that 3590 is about 602 and the distance from the Earth to the Moon is about 60 times larger than the radius of the Earth, so he made an assumption that the force of gravity must be inverse proportional to the square of the distance between the objects. This is how he came up with his law of gravitation, which states the existence of gravitational force between any two particle-like objects and this force has a magnitude of m m F G 1 2 , (8.1.2) r2 where r is the distance between the objects and proportionality coefficient G 6.67 1011 Nm2 kg2 is known as gravitational constant. This force is the force of attraction. It acts on each of the interacting particles in the direction along the straight line connecting these particles. This gravitational force is not altered by the presence of other bodies. For all we know, all the particles discovered so far have positive mass, so gravitational force is always the force of attraction. The value of gravitational constant, G, was determined in the laboratory, using the experiment known as the Cavendish experiment. As you can see this constant is extremely small, so the experiment to determine it must be very sensitive. It can be calculated, if one could measure gravitational force between the two particles, knowing their masses and distance between them, so Fr 2 G . m1m2 Henry Cavendish in 1797 designed an experiment helping to measure such a small force. He made use of the fact that the force needed to twist a long thin quartz fiber by a few degrees is very small. He had two lead spheres mounted at the ends of the light rod hanging from the fiber. Then he placed two large lead spheres near the smaller spheres and measured the angular deflection of the rod to find the G-constant. Example 8.2.2. Newton did not know about Cavendish’s method, so he estimated gravitational constant based on his estimation of the earth's density 5103 kg m3 . What was Newton's estimation for the G-value? The gravitational force acting on any object near the earth's surface is F=mg. On the other hand this force can be calculated according to the Gravitational law from the equation 8.2.2, so we have mM E mg G 2 , rE M E g G 2 , rE gr 2 G E . M E Here ME is mass of the Earth and rE is radius of the Earth. Note that we have considered the earth as a particle for which all the mass is concentrated at its center. It is all right to do so and we will discuss the reasons for that shortly. M Considering earth to be a perfect sphere, its density can be calculated as E , 4 r 3 3 E so gr 2 gr 2 3g 3 9.8m s2 Nm2 G E E 7.351011 . 4 3 3 6 2 ME r3 4rE 4 510 kg m 6.37 10 m kg 3 E So the Newton's estimation was in the 10% range of the correct value. As it was sated earlier, the Gravitational law is valid for particles or the objects, separated by the large distances, so that they can be considered as particles. For instance the distance between the Earth and the Moon is large enough compared to their sizes, so one can treat them as particles. This is also true for the apple, which definitely looks like a particle from the earth's perspective. But the earth itself is not a particle-like object compared to the apple, so how can we treat it as such? This question bothered Newton too. Even though he guessed that this was acceptable to consider the interaction of the spherical Earth, with a particle-like object as if they both were particles but he could not prove this at first. So he invented the new area of science, now known as integral calculus. It took him almost 20 years to prove the shell theorem: A uniform spherical shell of matter attracts a particle that is outside of the shell as if all the shell's mass were concentrated at its center. The important physical principle behind this theorem is known as the principle of superposition. This is the general principle applied to many phenomena, not just gravitational phenomena. It states that the net effect is the sum of individual effects and they do not interfere with each other. For gravity this means, that gravitational force acting on a selected particle is a vector sum of all gravitational forces due to each of the other particles in the system. If the system consists of n particles, we have the force acting on the first particle to be n F1 F1i , (8.1.3) i2 where F1i are the forces acting on the first particle from each of the others n-1 particles. If one considers not a system of the particles, but a large object, such as earth, he/she can consider it as continues system of particles of elementary masses dm, producing elementary gravitational forces dF . So the summation in the equation 8.1.3 should be replaced with integration over the object's volume F dF . (8.1.4) 1 z This is how one can calculate the force acting on the particle near the large object such as earth. In the case of the uniform spherical shell, equation 8.1.4 results in the shell theorem. Example 8.1.3 Two spheres of mass m and the third sphere of mass M form a equilateral triangle and the forth sphere of mass m4 is at the center of the triangle. The net gravitational force on that central sphere is zero. What is the mass M in terms of m? If we double the value of , what then is the magnitude of the net gravitational force on the central sphere? The picture above shows all four masses and three gravitational forces acting from each of the masses to mass . The choice of the coordinate axes is also shown in this picture. The net force acting on is zero, which means that Fx 0, Fy 0.
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