Chapter 7 Law of Gravity & Kepler’s Laws HMH Physics Ch 7 pages 224-269 Section 6 pages 232-245
Objectives 1. Explain how Newton’s law of universal gravitation accounts for various phenomena, including satellite and planetary orbits, falling objects, and the tides. 2. Apply Newton’s law of universal gravitation to solve problems. 3. Describe Kepler’s laws of planetary motion. 4. Relate Newton’s mathematical analysis of gravitational force to the elliptical planetary orbits proposed by Kepler. 5. Solve problems involving orbital speed and period.
Gravitational Force (Fg) Gravitational force is the mutual force of attraction between particles of matter Gravitational force depends on the distance between two objects and their mass Gravitational force is localized to the center of a spherical mass
G is the constant of universal gravitation G = 6.673 x 10-11 N•m2/kg2
The Cavendish Experiment Cavendish found the value for G. He used an apparatus similar to that shown above. He measured the masses of the spheres (m1 and m2), the distance between the spheres (r), and the force of attraction (Fg). He solved Newton’s equation for G and substituted his experimental values. Discuss the following with students. Newton did not know the value for G when he developed the law of universal gravitation. Cavendish found a value experimentally with the procedure illustrated on the slide. In the experiment, the lead spheres were placed in a glass case to eliminate air currents. In picture (a), the hanging spheres were stabilized. In picture (b), two large masses were brought in on opposite sides to attract the hanging spheres. The hanging spheres swing toward the large spheres until the force of the twisted wire balances the force of gravity. By finding the force of the twisted wire, he found the force of gravity between the spheres. The force of the twisted wire was very difficult to measure. Cavendish calibrated his wire so that he knew the amount of force needed to twist the wire one degree. Then, he measured the amount that the wire rotated to determine the force. The mirror allowed him to measure very small angular rotations.
Newton’s Law of Universal Gravitation Section 2 Newton’s Law of Universal Gravitation Chapter 7 Newton’s Law of Universal Gravitation
Ocean Tides Newton’s law of universal gravitation is used to explain the tides. Since the water directly below the moon is closer than Earth as a whole, it accelerates more rapidly toward the moon than Earth, and the water rises. Similarly, Earth accelerates more rapidly toward the moon than the water on the far side. Earth moves away from the water, leaving a bulge there as well. As Earth rotates, each location on Earth passes through the two bulges each day. The actual high tide is a little while after the moon is directly overhead. Many students may believe that the moon is only overhead at night. Since they don’t see it at 11:00 AM, it is a reasonable conclusion, but it is incorrect.
Gravity is a Field Force Earth, or any other mass, creates a force field. Forces are caused by an interaction between the field and the mass of the object in the field. The gravitational field (g) points in the direction of the force, as shown. Ask students what each arrow represents. (The gravitational force acting on a unit mass at that location.) Then ask why the arrows in the outer circle are smallest. (They are smallest because gravitational force is inversely proportional to distance squared, so the force decreases as distance from Earth’s center increases.)
Calculating the value of g Since g is the force acting on a 1 kg object, it has a value of 9.81 N/m (on Earth). The same value as ag (9.81 m/s2) The value for g (on Earth) can be calculated as shown below. Help students understand that g and ag have the same value but are used for different situations. g is the force acting due to gravity (per unit mass). We use this to find the force on objects at rest, such as a book on your desk. ag is used to calculate the acceleration of falling objects. Also emphasize that the equation on the slide is for the value of g on Earth. Students can find the value on the surface of another planet by using the mass and radius of that planet.
Find the distance between a 0. 300 kg billiard ball and a 0 Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force is 8.92 x 10-11 N.
Kepler’s Laws Johannes Kepler built his ideas on planetary motion using the work of others before him. Nicolaus Copernicus and Tycho Brahe Kepler’s laws are covered in more detail on the next two slides.
Kepler’s Laws Kepler’s first law Kepler’s second law Orbits are elliptical, not circular. Some orbits are only slightly elliptical. Kepler’s second law Equal areas are swept out in equal time intervals. Be sure students interpret the diagram correctly. How does t1 relate to t2 ? (They are equal). How does A1 relate to A2 ? (They are equal.) How does the speed of a planet close to the sun compare to the speed of a planet that is far from the sun? (Planets travel faster when they are closer to the sun. In this example, the distance on the left is greater than the distance on the right, and the times are equal, so the speed on the left is greater.)
Kepler’s Laws Kepler’s third law Relates orbital period (T) to distance from the sun (r) Period is the time required for one revolution. As distance increases, the period increases. Not a direct proportion T2/r3 has the same value for any object orbiting the sun At this point, you may wish to have students work through the derivation mentioned in the Quick Lab on page 249 of the Student Edition. This will help them see the connection between Newton’s law of universal gravitation and Kepler’s laws. The web site: http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=9.0 allows students to visualize the first two laws. If you choose to show the first law, you see an elliptical orbit with the lengths measured from each focal point. The sum of these lengths is constant by the definition of an ellipse. Choosing the 2nd law allows you to see the equal areas swept out in equal amounts of time. The simulations also allows the user to change other parameters. These simulations can be downloaded. The home web site is: http://www.phy.ntnu.edu.tw/ntnujava/index.php Kepler’s Laws
Equations for Planetary Motion The dimensional analysis recommended on this slide is a good practice that provides a refresher on the units from previous chapters and reinforces the use of appropriate units. Using SI units, prove that the units are consistent for each equation shown above.
Classroom Practice Problems A large planet orbiting a distant star is discovered. The planet’s orbit is nearly circular and close to the star. The orbital distance is 7.50 1010 m and its period is 105.5 days. Calculate the mass of the star. Answer: 3.00 1030 kg What is the velocity of this planet as it orbits the star? Answer: 5.17 104 m/s This problem will provide good practice for the students. After they work on it for few minutes, guide them along. One way to solve this problem is to take the equation for period and square both sides. Make sure they square both the 2 and the pi (42). Then they can substitute values and solve for m. Another possible error is the use of days for T. For the velocity calculation, most will probably use the equation on the previous slide. After some time, point out that they could have obtained the velocity without the mass by using the distance divided by the time (or circumference divided by period). This will reinforce the validity of the equations shown on the previous slide. Classroom Practice Problems