Universal Law of Gravitation Honors Physics

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1 Universal Law of Gravitation Honors Physics Newton s Law of Universal Gravitation The greatest moments in science are when two phenomena that were considered completely separate suddenly are seen as just two different versions of the same thing. Newton s explanation that the motion of the moon around the Earth (and the planets around the sun) and of an apple falling to the ground are two expressions of the same phenomenon, gravity, represents a wonderful instance of this sort of synthesis. Newton conjectured that objects near the surface of the Earth fall towards the center of the Earth as a result of a universal attraction of all matter towards all other matter. He then considered that the circular motion of the moon around the Earth could then be explained if that same force of gravity were supplying the force towards the center of the moon s circular orbit, the Earth. That force of gravity would provide the unbalanced force necessary for circular motion. Universal Gravity Newton postulated that all matter in the universe is attracted to all other matter. He named this force of attraction gravity and he formulated the following expression to describe how the force of gravity depends on the mass of each object and the distance between their centers. In this equation, G is a constant that would need to be discovered by experiment, m1 and m2 are the masses of the two objects and r is the distance between their centers. It doesn t matter which mass you call m1 and which one you call m2 : That s a result of Newton s third law which states that the force on one object will be equal in size to the force on the other. Newton had to invent calculus in order to show that the distance between two spherical objects, the r in his equation, should be measured between the centers of the objects. He was able to show that spherical objects act, from the perspective of gravitational force, as if all of their mass were located at a point at their center. As a result, the distance from a spherical mass, such as the Earth or the sun, must always be measured from its center. It took more than a hundred years before Henry Cavendish was able to directly measure the value of G. The modern accepted value of G is: G = 6.67 x N m 2 /kg 2 1 Page

2 Example 1 What is the force between a 5.0kg spherical object and a 10 kg spherical object whose centers are located 2.5m apart? FG=Gm1m2/r 2 FG = (6.67 x N m 2 /kg 2 )(5.0kg)(10kg)/(2.5m) 2 FG = (6.67 x N m 2 /kg 2 )(5.0kg)(10kg)/(2.5m) 2 FG = 5.3 x N Example 2 What is the average gravitational force between the Earth and the sun? The mass of the Earth is 6.0 x kg, the mass of the sun is 2.0 x kg and the average distance between their centers is 1.5 x m. FG=Gm1m2/r 2 FG = (6.67 x N m 2 /kg 2 )( 2.0 x )( 6.0 x kg)(1.5 x m) 2 FG = (6.67 x N m 2 /kg 2 )( 2.0 x )( 6.0 x kg)/( 1.5 x m) 2 FG = 3.6 x N Example 3 What is the distance between two objects whose masses are 28,000 kg and 50,000 kg if the gravitational force between them is 50N? FG=Gm1m2/r 2 To solve for r, first multiply both side by r2 (FG)(r 2 )= Gm1m2 Then divide both sides by FG r 2 = Gm1m2/ FG And take the square root of both sides r= (Gm1m2/FG) 1/2 r= ((6.67 x N m 2 /kg 2 )(2.8 x 10 4 kg)(5.0 x 10 4 kg)/(50n)) 1/2 r = (1.86 x 10-3 ) r = 18.6 x 10-4 r = 4.3 x 10-2 m r = 4.3 cm 2 Page

3 Universal Gravitation:, G = 6.67 x10-11 Nm 2 /kg 2 Examples 1. Two spherical objects have masses of 200 kg and 500 kg. Their centers are separated by a distance of 25 m. Find the gravitational attraction between them. (1.067x10-8 N) 2. Two spherical objects have masses of 1.5 x 10 5 kg and 8.5 x 10 2 kg. Their centers are separated by a distance of 2500 m. Find the gravitational attraction between them. (1.361x10-9 N) 3. Two spherical objects have masses of 3.1 x 10 5 kg and 6.5 x 10 3 kg. The gravitational attraction between them is 65 N. How far apart are their centers? (0.045 m) 4. A 1 kg object is located at a distance of 6.4 x10 6 m from the center of a larger object whose mass is 6.0 x kg. a. What is the size of the force acting on the smaller object? 9.77 N b. What is the size of the force acting on the larger object? 9.77 N c. What is the acceleration of the smaller object when it is released? 9.77 m/s 2 d. What is the acceleration of the larger object when it is released? 1.63 x10-24 m/s 2 Class Work 5. Two spherical objects have masses of 8000 kg and 1500 kg. Their centers are separated by a distance of 1.5 m. Find the gravitational attraction between them N or 3.56 x10-4 N 6. Two spherical objects have masses of 7.5 x 10 5 kg and 9.2 x 10 7 kg. Their centers are separated by a distance of 2.5 x 10 3 m. Find the gravitational attraction between them x10-4 N 7. Two spherical objects have masses of 8.1 x 10 2 kg and 4.5 x 10 8 kg. The gravitational attraction between them is 1.9 x 10-3 N. How far apart are their centers? m 8. Two spherical objects have equal masses and experience a gravitational force of 85 N towards one another. Their centers are 36mm apart. Determine each of their masses kg 9. A 1 kg object is located at a distance of 7.0 x10 8 m from the center of a larger object whose mass is 2.0 x kg. a. What is the size of the force acting on the smaller object? 272 N b. What is the size of the force acting on the larger object? 272 N 3 Page

4 c. What is the acceleration of the smaller object when it is released? 272 m/s 2 d. What is the acceleration of the larger object when it is released? 1.36 x10-28 m/s 2 Homework 1. Two spherical objects have masses of 8000 kg and 5.0 kg. Their centers are separated by a distance of 1.5 m. Find the gravitational attraction between them N 2. Two spherical objects have masses of 9.5 x 10 8 kg and 2.5 kg. Their centers are separated by a distance of 2.5 x 10 8 m. Find the gravitational attraction between them x10-18 N 3. Two spherical objects have masses of 6.3 x 10 3 kg and 3.5 x 10 4 kg. The gravitational attraction between them is 6.5 x 10-3 N. How far apart are their centers? 1.50 m 4. Two spherical objects have equal masses and experience a gravitational force of 25 N towards one another. Their centers are 36 cm apart. Determine each of their masses kg 5. A 1 kg object is located at a distance of 1.7 x10 6 m from the center of a larger object whose mass is 7.4 x kg. a. What is the size of the force acting on the smaller object? 1.71 N b. What is the size of the force acting on the larger object? 1.71 N c. What is the acceleration of the smaller object when it is released? 1.71 m/s 2 d. What is the acceleration of the larger object when it is released? 2.31 x10-23 m/s 2 4 Page

5 Orbital Motion Newton based much of his theory on the work of Johannes Kepler, which was based on the astronomical observations of Tycho Brahe. Newton concluded that in order for the planets to proceed in their orbits around the sun there must be a force attracting them towards the sun. Kepler writings could be summarized into three laws: Kepler s first law states that the orbits of the planets are elliptical, with the sun at one focus. Sun An ellipse has two focal points and no real defined center like a circle. (Actually if Kepler didn t study the data concerning Mars, he probably would never have discovered this fact! Mars has an extremely elliptical orbit, while the other planets are nearly circular!) 5 Page

6 Kepler s second law states that if you extend an imaginary line between the sun and a planet, that that line would sweep out equal areas in equal times. In other words, Kepler found that the planets move faster when they are closer to the Sun and slower when they are farther away from the Sun. This means that the earth is closer to the sun in winter! (strange but true!) The following diagram shows this: 6 Page

7 Kepler s third law states that there is a specific mathematical relationship between the orbital periods of planets (the time to go around the sun) and their mean (average) distance from the sun. Specifically, for all planets orbiting the sun the ratio of their orbital period squared divided by their orbital radius cubed, T 2 /r 3, yields the same number. Thus, if the periods of the planets are TA and TB and the average distances from the Sun are ra and rb, the third law can be expressed as follows: TA 2 = ra 3 TB rb The following data table can be useful for using this equation: Planetary Data planets not shown to scale >> Mercur y Mean Distance to the Sun (AU) Orbital Period (years) Venus Earth Mars Jupite r Satur n Uranu s Neptun e Pluto Period Squared Mean Distance Cubed T 2 /r Equatori al Radius (km) AU=Astronomical Unit 1AU=distance from earth to sun=146,600 km This data can be used for any of Kepler s 3 rd law problems that compare orbits and periods of planets. 7 Page

8 This concept can also be used for orbits and periods of satellites (moons) of planets! The orbits of moons around planets are also ellipses. Example: Galileo discovered four moons of Jupiter. Io, which he measured to be 4.2 units form the center of Jupiter, has a period of 1.8 days. He measured the radius of Ganymede s orbit as 10.7 units. Use Kepler s third law to find the period of Ganymede. Given: For Io Period TB = 1.8 days Radius rb = 4.2 units Unknown: Ganymede s period, TA For Ganymede Radius ra = 10.7 units Basic Equation: TA 2 = ra 3 TB rb Solution: TA 2 = TB 2 ra rb 3 TA 2 = (1.8 days) units units TA 2 = 52.8 days 2 TA = 7.3 days 8 Page

9 Free Response 1. As shown in the diagram below, a 5.0 kg space rock is located 2.5x10 7 m from the center of the earth. The mass of the earth is 6.0x10 24 kg. Earth Space Rock a. Determine the force of gravity acting on the space rock, due to the earth. Calculate the magnitude and state the direction. ( N) b. Compare your answer in a) to the force of gravity acting on the earth, due to the space rock. Indicate that force on the diagram above. ( N) c. On the diagram above, indicate the direction the space rock would accelerate if released. Label that vector a. ( )from rock towards earth d. Calculate the acceleration the rock would experience m/s 2 e. **If instead of falling, the object were in a stable orbit, indicate on the diagram above a possible direction of its velocity. Label that vector v. ( or ) up or down from rock f. **Calculate the velocity the rock needs to be in a stable orbit m/s g. **Calculate the period of the rock orbiting the earth s 9 Page

10 2. As shown in the diagram below, a 2000 kg spacecraft is located 9.2x10 6 m from the center of the earth. The mass of the earth is 6.0x10 24 kg. Earth Spacecraft a. Determine the force of gravity acting on the spacecraft, due to the earth. Calculate the magnitude and state the direction. ( N) b. Compare your answer in a) to the force of gravity acting on the earth, due to the spacecraft. Indicate that force on the diagram above. ( N) c. On the diagram above, indicate the direction the spacecraft would accelerate if released. Label that vector a. ( )from spacecraft towards earth d. Calculate the acceleration the spacecraft would experience m/s 2 e. **If instead of falling, the spacecraft were in a stable orbit, indicate on the diagram above a possible direction of its velocity. Label that vector v. ( or ) up or down from spacecraft f. **Calculate the velocity the spacecraft needs to be in a stable orbit m/s g. **Calculate the period of the spacecraft orbiting the earth s 10 Page

11 3. As shown in the diagram below, a 1000 kg asteroid is located 6.8x10 6 m from the center of the Mars. The mass of the Mars is 6.4x10 23 kg. Mars Asteroid a. Determine the force of gravity acting on the asteroid, due to the Mars. Calculate the magnitude and state the direction. 923 N left b. Compare your answer in a) to the force of gravity acting on the Mars, due to the asteroid. Indicate that force on the diagram above. 923 N right c. On the diagram above, indicate the direction the asteroid would accelerate if released. Label that vector a. ( )from asteroid towards mars d. Calculate the acceleration the asteroid would experience m/s 2 e. **If instead of falling, the asteroid were in a stable orbit, indicate on the diagram above a possible direction of its velocity. Label that vector v. ( or ) up or down from asteroid f. **Calculate the velocity the asteroid needs to be in a stable orbit m/s g. **Calculate the period of the asteroid orbiting the earth s 11 Page

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