Presentation on theme: "CH 12: Gravitation. We have used the gravitational acceleration of an object to determine the weight of that object relative to the Earth. Where does."— Presentation transcript:
We have used the gravitational acceleration of an object to determine the weight of that object relative to the Earth. Where does this acceleration come from? The gravitational acceleration is due to the gravitational force, an attractive force that exists between the two masses. M 1 and M 2 – Masses of the two objects [kg] G – Universal gravitational constant G = 6.67x10 -11 N m 2 /kg 2 r – distance separating the center of mass of the two objects [m] F g – Gravitational force between the two objects [N] Example: What is the force a 40 kg boy exerts on his 20 kg dog if they are separated by a distance of 2 m? Boy Dog F BD F DB 2 m This is an example of Newton’s 3 rd Law. The force on each object is equal, but they have opposite directions. This force is often very small unless you are using at least one very large mass!
Now that we know how to determine the magnitude of the gravitational force, how can we determine the magnitude of the gravitational acceleration? If we look specifically at an object on the surface of the Earth. If neither the mass of the Earth nor the radius of the Earth change this acceleration would be constant. This is only valid for an object at the surface of the Earth! Notice that the acceleration would depend on the distance away from the surface of the Earth. When we are looking at an object at some altitude we must modify the expression for the acceleration in the following way: The gravitational acceleration depends on the altitude of the mass above the Earth. We can use g as an approximation of the gravitational acceleration for objects that are close to the surface of the Earth.
Kepler’s Laws of Planetary Motion Kepler’s laws are all derived from the law of universal gravitation and angular motion. 1) All planets move in elliptical orbits. Using the law of gravitation it is possible to prove that the orbital path of plants is elliptical. 2) The radius vector from the sun to the planet sweeps out equal areas in equal time intervals. dA r r v v FgFg Apogee (Aphelion) – Furthest Perigee (Perihelion) – Closest dr L is constant constant FgFg
3) The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit. Semimajor axis Semiminor axis Foci 2b 2a c For an elliptical path we substitute a for r Average velocity of an object in a circular orbit Assuming a circular orbit: