Universal Gravitation. Paths of Satellites around Earth

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Presentation transcript:

Universal Gravitation

Paths of Satellites around Earth

Kepler’s Laws BBased on careful measurement of motion of planets EEmpirical

1. The path of each planet about the Sun is an ellipse with the Sun as one focus 2. Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal periods of time

Kepler’s Laws 3. The ratio of the squares of the periods T of any two planets revolving around the Sun is equal to the cubes of their mean distances from the sun.

Newton’s Law of Universal Gravitation 1. Force of gravity α 1/r 2  Initially this was arrived at empirically  The surface area of a sphere A=4πr 2  Because of this, force laws are frequently inverse square laws.

Newton’s Law of Universal Gravitation  Assume that there are gravitons traveling away from a large body uniformly in all directions.  Then if another body has a size, Δa, then the fraction of the overall sphere captured by Δa would be:  Δa/a=Δa/4πr 2 α 1/r 2  This fraction would also be the fraction of gravitons captured by the other body

Newton’s Law of Universal Gravitation  Symmetry of masses  Newton postulated that the mass of each object had to play a role in the force of gravitation.  Because of Newton’s 3 rd Law, there has to be a symmetry in the equation between the two masses that are being attracted:

Newton’s Law of Universal Gravitation  As a result of the symmetry, Newton postulated that:

Newton’s Law of Universal Gravitation EEvery particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles. GG=6.67 x N m 2 /kg 2

Geo Synchronous Orbits  Geo Synchronous – Object stays above the same place on earth at all times  Gravitational attraction = Centripetal Force

Geo Synchronous Orbits  If we plug the expression for v into the equation above we get:  r=42,300 km  Subtracting the radius of the earth we get that the distance above the earth is ~ 36000km ~6 r E