

Lesson 1: Motion Characteristics for Circular MotionThe Centripetal Force Requirement Mathematics of Circular Motion 
Lesson 3: Universal GravitationNewton's Law of Universal GravitationAs discussed earlier in Lesson 3, Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that the force of gravitational attraction between the Earth and other objects is inversely proportional to the distance separating the earth's center from the object's center. But distance is not the only variable affecting the magnitude of a gravitational force. Consider Newton's famous equation


The solution of the problem involves substituting known values of G (6.673 x 10^{11} N m^{2}/kg^{2}), m_{1} (5.98 x 10^{24} kg ), m_{2} (70 kg) and d (6.38 x 10^{6} m) into the universal gravitation equation and solving for F_{grav}. The solution is as follows:

The solution of the problem involves substituting known values of G (6.673 x 10^{11} N m^{2}/kg^{2}), m_{1} (5.98 x 10^{24} kg ), m_{2} (70 kg) and d (6.39 x 10^{6} m) into the universal gravitation equation and solving for F_{grav}. The solution is as follows:
Two general conceptual comments can be made about the results of the two sample calculations above. First, observe that the force of gravity acting upon the student (a.k.a. the student's weight) is less on an airplane at 40 000 feet than at sea level. This illustrates the inverse relationship between separation distance and the force of gravity (or in this case, the weight of the student). The student weighs less at the higher altitude. However, a mere change of 40 000 feet further from the center of the Earth is virtually negligible. This altitude change altered the student's weight changed by 2 N which is much less than 1% of the original weight. A distance of 40 000 feet (from the earth's surface to a high altitude airplane) is not very far when compared to a distance of 6.38 x 10^{6} m (equivalent to nearly 20 000 000 feet from the center of the earth to the surface of the earth). This alteration of distance is like a drop in a bucket when compared to the large radius of the Earth. As shown in the diagram below, distance of separation becomes much more influential when a significant variation is made.
The second conceptual comment to be made about the above sample calculations is that the use of Newton's universal gravitation equation to calculate the force of gravity (or weight) yields the same result as when calculating it using the equation presented in Unit 2:
Both equations accomplish the same result because (as we will study later in Lesson 3) the value of g is equivalent to the ratio of (G•M_{earth})/(R_{earth})^{2}.
Gravitational interactions do not simply exist between the earth and other objects; and not simply between the sun and other planets. Gravitational interactions exist between all objects with an intensity which is directly proportional to the product of their masses. So as you sit in your seat in the physics classroom, you are gravitationally attracted to your lab partner, to the desk you are working at, and even to your physics book. Newton's revolutionary idea was that gravity is universal  ALL objects attract in proportion to the product of their masses. Gravity is universal. Of course, most gravitational forces are so minimal to be noticed. Gravitational forces are only recognizable as the masses of objects become large. To illustrate this, use Newton's universal gravitation equation to calculate the force of gravity between the following familiar objects. Click the buttons to check answers.
(kg) 
Mass of Object 2 (kg) 
Separation Distance (m) 
Force of Gravity (N) 

a. 
100 kg 
Earth 5.98 x10^{24} kg 
6.38 x 10^{6} m (on surface) 

b. 
40 kg 
Earth 5.98 x10^{24} kg 
6.38 x 10^{6} m (on surface) 

c. 
Physics Student 70 kg 
Earth 5.98 x10^{24} kg 
6.60 x 10^{6} m (lowheight orbit) 

d. 
70 kg 
Physics Student 70 kg 
1 m 

e. 
70 kg 
Physics Student 70 kg 
0.2 m 

f. 
70 kg 
Physics Book 1 kg 
1 m 


70 kg 
7.34 x 10^{22} kg 
(on surface) 


70 kg 
1.901 x 10^{27} kg 
(on surface) 

Today, Newton's law of universal gravitation is a widely accepted theory. It guides the efforts of scientists in their study of planetary orbits. Knowing that all objects exert gravitational influences on each other, the small perturbations in a planet's elliptical motion can be easily explained. As the planet Jupiter approaches the planet Saturn in its orbit, it tends to deviate from its otherwise smooth path; this deviation, or perturbation, is easily explained when considering the affect of the gravitational pull between Saturn and Jupiter. Newton's comparison of the acceleration of the apple to that of the moon led to a surprisingly simple conclusion about the nature of gravity which is woven into the entire universe. All objects attract each other with a force which is directly proportional to the product of their masses and inversely proportional to their distance of separation.

1. Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?
2. Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is reduced in half, then what is the new force of attraction between the two objects?
3. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and if the distance between the objects remained the same, then what would be the new force of attraction between the two objects?
4. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?
5. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was tripled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?
6. Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of object 1 was doubled, and if the distance between the objects was tripled, then what would be the new force of attraction between the two objects?
7. As a star ages, it it believed to undergo a variety of changes. One of the last phases of a star's life is to gravitationally collapse into a black hole. What will happen to the orbit of the planets of the solar system if our star (the Sun shrinks into a black hole)? (And of course, this assumes that the planets are unaffected by prior stages of the Sun's evolving stages.)
8. Having recently completed his first Physics course, Noah Formula has devised a new business plan based on his teacher's Physics for Better Living theme. Noah learned that objects weigh different amounts at different distances from Earth's center. His plan involves buying gold by the weight at one altitude and then selling it at another altitude at the same price per weight. Should Noah buy at a high altitude and sell at a low altitude or vice versa?
9. Anita Diet is very concerned about her weight but seldom does anything about it. After learning about Newton's law of universal gravitation in Physics class, she becomes all concerned about the possible affect of a change in Earth's mass upon her weight. During a (rare) free moment at the lunch table, she speaks up "How would my weight change if the mass of the Earth increased by 10%?" How would you answer Anita?
10. When comparing mass and size data for the planets Earth and Jupiter, it is observed that Jupiter is about 300 times more massive than Earth. One might quickly conclude that an object on the surface of Jupiter would weigh 300 times more than on the surface of the Earth. For instance, one might expect a person who weights 500 N on Earth would weigh 150000 N on the surface of Jupiter. Yet this is not the case. In fact, a 500N person on Earth weighs about 150 N on the surface of Jupiter. Explain how this can be.
Lesson 3: Universal Gravitation
 Gravity is More Than a Name
 The Apple, the Moon, and the Inverse Square Law
 Newton's Law of Universal Gravitation
 Cavendish and the Value of G
 The Value of g
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19962007