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Lecture 6: Gravity and Motion Review from Last Lecture… Newton’s Universal Law of Gravitation Kepler’s Laws are special cases of Newton’s Laws bound.

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Presentation on theme: "Lecture 6: Gravity and Motion Review from Last Lecture… Newton’s Universal Law of Gravitation Kepler’s Laws are special cases of Newton’s Laws bound."— Presentation transcript:

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2 Lecture 6: Gravity and Motion

3 Review from Last Lecture… Newton’s Universal Law of Gravitation Kepler’s Laws are special cases of Newton’s Laws bound and unbound orbits tides and tidal friction

4 Kepler or Newton? find the mass of the Earth using the fact that the Moon’s orbit has a period of 29 ½ days find the average orbital distance for an asteroid that orbits the Sun with a period of 8 years find the period of a binary star system with a mean orbital distance of 10 pc

5 Tides

6 The Moon’s Tidal Forces on the Earth

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12 Galactic Tidal Forces

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14 Tidal Friction

15 Synchronous Rotation

16 Tidal friction and the Moon Tidal friction from the Moon acting on the Earth causes the Earth’s rotation to slow down. As a result, the Moon also moves further and further away from Earth (due to conservation of angular momentum).

17 Implications… was the Moon’s angular size larger or smaller in the past? was the length of a lunar month longer or shorter in the past? were eclipses (both solar and lunar) more or less frequent in the past?

18 The acceleration of gravity the universal law of gravitation allows us to understand why the acceleration due to gravity is independent of the mass of the object and why our weight is different on other planets

19 Why g is independent of mass Imagine dropping a rock near the surface of the Earth. The force on the rock is: F g = G M Earth M rock / d 2 = G M Earth M rock / (R Earth ) 2 Newton’s Second Law of Motion says that the force is also: F g = M rock a rock = G M Earth M rock / (R Earth ) 2 a rock = g = G M Earth / (R Earth ) 2

20 Finding the value of g g = G M Earth / (R Earth ) 2 g = (6.67 x 10 -11 m 3 /(kg s 2 ) ) x 6.0 x 10 24 kg / (6.4 x 10 6 m) 2 M earth = 6.0 x 10 24 kg R earth = 6.4 x 10 6 m = 9.8 m/s 2

21 What about on the Moon? g = G M Moon / (R Moon ) 2 g = (6.67 x 10 -11 m 3 /(kg s 2 ) ) x 7.4 x 10 22 kg / (1.7 x 10 6 m) 2 M Moon = 7.4 x 10 22 kg R Moon = 1.7 x 10 6 m = 1.7 m/s 2 gravity is weaker on the Moon…therefore things weigh less!  gravity is weaker on the Moon…therefore things weigh less!

22 Matter and Energy

23 Energy is what makes matter move kinetic energy = energy of motion potential energy = stored energy gravitational chemical electrical radiative energy = light

24 Units of Energy calories kilowatt-hours BTU Joules 1 Joule = 0.00024 Calories

25 Quantifying Energy kinetic energy = ½ m v 2 where m = mass (in kg) and v = velocity (in m/s)  answer will be in Joules  (1 J = kg x m 2 /s 2 )

26 Gravitational Potential Energy the amount of gravitational potential energy is proportional to the mass, the force of gravity, and the distance for example, for an object suspended above the earth, the gravitational potential energy is W = G m M Earth /r = m x g x r

27 Conservation of Energy the total amount of energy in the Universe remains the same energy can change forms but cannot be created or destroyed

28 Orbital Energy moving faster  larger kinetic energy moving slower  smaller kinetic energy

29 bound vs. unbound orbits bound orbits  gravitational potential energy balances kinetic energy unbound orbits  kinetic energy greater than gravitational potential

30 gravitational encounters

31 escape velocity We can now derive the escape velocity by setting the kinetic energy equal to the gravitational potential energy: ½ m v 2 = Gm M Earth /R Earth  v escape = (2GM Earth / R Earth ) ½

32 The Escape Velocity from Earth v escape = (2GM Earth / R Earth ) ½ = (2 x 6.67x10 -11 m 3 /(kg s 2 ) x 6.0x10 24 kg/6.4x10 6 m) ½ v escape = 11 km/s

33 The End


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