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The story of the apple

The story of the apple. When Newton observed the apple fall, he wondered if the force that caused the apple to fall to the ground was the same force that kept celestial bodies in orbit. Universal Gravitation.

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The story of the apple

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  1. The story of the apple When Newton observed the apple fall, he wondered if the force that caused the apple to fall to the ground was the same force that kept celestial bodies in orbit

  2. Universal Gravitation • Newton’s Law of Universal Gravitation states that all objects with mass exert attractive forces on all other objects with mass. • The force is proportional to the product of the masses and inversely proportional to the square of the separation distance Where: Fg = Force due to gravity (N) m1 & m2 = mass of each body (kg) r = separation distance (m) G = 6.67 x 10-11 Nm2/kg2

  3. HENRY CAVENDISH Thin flat wire In 1797 Henry Cavendish devised a very sensitive torsion balance to study gravitational attraction. He was the first to observe gravity in the lab. He actually measured the force which existed between two known masses, and used this data to calculate a proportionality constant. Twist in wire measures force F G = 6.67 x 10-11 Nm2/kg2 F

  4. Understanding Universal Gravitation All objects experience a force of gravity that points toward the center of the earth, even when they are not in contact. Does the moon experience a force of gravity as well? of course

  5. Understanding Universal Gravitation The force of gravity is proportional to the mass of the object exerting the force. The force of gravity on an object is proportional to the mass of the object The force of gravity is inversely proportional to the square of the separation distance. G = 6.67 x 10-11 Nm2/kg2

  6. Determine the force of attraction between the earth and the moon What do we need to know? m1 = mass of the moon = 7.36 × 1022 kgm2 = mass of the earth = 5.97 × 1024 andr = distance between the earth and the moon = 384,402 km Fg = 1.99 x 1026 N

  7. The gravitational force between two electrons 1.00 m apart is 5.42 x 10-71 N. What is the mass of an electron? F= 5.42 x 10-71 N r = 1.00 mm1 = m2 = ? m1 = m2 (m)(m) = m2 m = 9.0 x 10-31 kg

  8. Satellites • To stay moving in a circle, the object must experience a centripetal force. • Gravity provides the centripetal force.

  9. The Andromeda galaxy, known as M31, is 2.1 x 106 light years away. Measurements show that a star out at its extremities 7.5 x 1020 AU (1.125 x 1032 m) from the center orbits the center at 200,000 m/s. Approximate the mass of M31. All the mass can be taken to act at the center of gravity. r = 1.125 x 1032 m v = 200,000 m/sm= ? m = 6.7 x 1052 kg

  10. KEPLER’S THREE LAWS OF PLANETARY MOTION • The planets orbit the sun in elliptical paths with the sun at one of the foci First Law

  11. Second Law • A planet sweeps out equal areas in equal time intervals as it proceeds around the sun Naturally the real implication of this law is what: Planets move faster when closer to the sun and slower when farther away

  12. Third Law The ratio of the cube of the orbital radius to the square of the orbital period is constant for all planets R13 R23 _____________ ____________ = T12T22 or . . . . . . R3Gm ____________ __________________ = T2 42 Where m is the mass of the central body.

  13. What is the period of Venus’ orbit? Looking up the data for Venus on the chart, we find that Venus is 0.72 AU from the sun. This is its orbital radius R2. Since Venus and the earth both orbit the same central body (the sun), we can apply Kepler’s Third Law. Given: R1 = 1 AU (Earth) T1 = 1 year (Earth) R2 = 0.72 AU (Venus) T2 = ? (Venus) T2 = 0.6 years = 223 days

  14. A 150 kg satellite is launched into orbit 12,800 km above the earth’s surface. What is the weight of the satellite on earth? What is the weight of the satellite while in orbit? What is the speed of the satellite while in orbit? What is its orbital period? Given: m1 = 150 kg m2 = 5.98 x 1024 kg (Earth) radius of Earth = 6370 km altitude of satellite above Earth = 12,800 km radius of satellite's orbit = = 19,170 km = 19,170,000 m Fg = mg Fg = (150)(9.8) = 1470 N on Earth* * Of course you could have used Newton’s Law of Universal Gravitation with the radius of the Earth as r and arrived at the same answer.

  15. Fg = 163 N while in orbit v = 4,564 m/s

  16. v = 4,564 m/s r = 19,170,000 m T = 26,400 seconds = 7.3 hours OR R3 Gm ___________ ____________ = (19,170,000)3 (6.67 x 10-11)(5.98 x 1024) ___________________________ __________________________________________ = T2 42 T2 42

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