Chapter 12 Universal Law of Gravity

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Presentation transcript:

Chapter 12 Universal Law of Gravity

Chapter 12: Universal Gravitation The earth exerts a gravitational force mg on a mass m. By the action-reaction law, the mass m exerts a force mg on the earth. By symmetry, since the force mg is proportional to the mass m, the value of g must also be proportional to the mass M of the earth. Isaac Newton realized that the motion of projectiles near the earth, the moon around the earth, the planets around the sun,… could be described by a universal law of gravitation

Gravity If two particles of mass m1 and m2 are separated by a distance r, then the magnitude of the gravitational force is: G is a constant = 6.67  10-11 N·m2/kg2 The force is attractive: The direction of the force on one mass is toward the other mass.

The gravitational force varies like 1/r2 The gravitational force varies like 1/r2. It decreases rapidly as r increases, but it never goes to zero. Example: The gravitational force between two masses is 10-10 N when they are separated by 6 m. If the distance between the two masses is decreased to 3 m, what is the gravitational force between them?

Gravitational Attraction of Spherical Bodies If you have an extended object, it behaves as if all of its mass is at the center of mass. Therefore, to calculate the gravitational force between two objects, use the distance between their centers of mass. Gravitational force between the Earth and the moon.

Gravitation of finite objects Newton invented Differential Calculus to interpret his theory: F = ma Newton invented Integral Calculus to prove that the gravitational force of the earth and motion of the moon is the same as if the earth and moon were each concentrated in a single point.

Example Calculate the gravitational force between a 70 kg man and the Earth. F = m g = (70 kg) (9.8 m/s2) = 686 N, but

Variation of g with height The gravitational force between the Earth and the space shuttle in orbit is almost the same as when the shuttle is on the ground.

Kepler’s Laws of Orbital Motion 1. Objects follow elliptical orbits, with the mass being orbited at one focus of the ellipse. A circle is just a special case of an ellipse.

Kepler’s Laws (cont.) 2. As an object moves in its orbit, it sweeps out an equal amount of area in an equal amount of time. This law is just conservation of angular momentum. Gravity does not exert a torque on the planet Why? perigee apogee

Kepler’s Laws (cont.) 3. The period of an object’s orbit, T, is proportional to the 3/2 power of its average distance from the thing it is orbiting, r: Note: M is the mass that is being orbited. The period does not depend on the mass of the orbiting object.

Example 1. The space shuttle orbits the Earth with a period of about 90 min. Find the average distance of the shuttle above the Earth’s surface. answer: 6.65E6 m

Gravitational Potential Energy The gravitational potential energy of a pair of objects is: When we deal with astronomical objects, we usually choose U = 0 when two objects are infinitely far away from each other. In this case, gravitational potential energy is negative. The formula we have used in the past, U = mgy, is valid only near the surface of the Earth (and has a different location for U=0).

Escape Speed We can use conservation of energy to calculate the speed with which an object must be launched from Earth in order to entirely escape the Earth’s gravitational field. Initially, the object has kinetic (velocity v) and potential energy. In order to escape, the object must have just enough energy to reach infinity with no speed left. In this case, M = mass of Earth and R = radius of Earth.

Example A satellite is orbiting the Earth as shown below. At what part of the orbit, if any, are the following quantities largest? Kinetic energy Potential energy Total energy (d) Velocity (e) Gravitational force (f) Centripetal acceleration (g) Momentum A B