Chapter 12 Gravitation. Theories of Gravity Newton’s Einstein’s.

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Chapter 12 Gravitation

Theories of Gravity Newton’s Einstein’s

Newton’s Law of Gravitation any two particles m 1, m 2 attract each other F g = magnitude of their mutual gravitational force

Newton’s Law of Gravitation F g = magnitude of the force that: m 1 exerts on m 2 m 2 exerts on m 1 direction: along the line joining m 1, m 2

Newton’s Law of Gravitation Also holds if m 1, m 2 are two bodies with spherically symmetric mass distributions r = distance between their centers

Newton’s Law of Gravitation ‘universal’ law: G = fundamental constant of nature careful measurements: G=6.67×10 -11 Nm 2 /kg 2

How Cavendish Measured G (in 1798, without a laser)

1798: small masses (blue) on a rod gravitate towards larger masses (red), so the fiber twists

2002: we can measure the twist by reflecting laser light off a mirror attached to the fiber

If a scale calibrates twist with known forces, we can measure gravitational forces, hence G

What about g = 9.8 m/s 2 ? How is G related to g? Answer: Show that g = Gm E /R E 2

Other planets, moons, etc? g p =acceleration due to gravity at planet’s surface density assumed spherically symmetric, but not necessarily uniform

Example: Earth’s density is not uniform

Yet to an observer (m) outside the Earth, its mass (m E ) acts as if concentrated at the center

Newton’s Law of Gravitation This is the magnitude of the force that: m 1 exerts on m 2 m 2 exerts on m 1 What if other particles are present?

Superposition Principle the gravitational force is a vector so the gravitational force on a body m due to other bodies m 1, m 2,... is the vector sum: Do Exercise 12-8 Do Exercise 12-6 Do Exercise 12-8 Do Exercise 12-6

Superposition Principle Example 12-3 The total gravitational force on the mass at O is the vector sum: Do some of Example 12-3 and introduce Extra Credit Problem 12-42

Gravitational Potential Energy, U This follows from: Derive U = - G m 1 m 2 /r

Gravitational Potential Energy, U Alternatively: a radial conservative force has a potential energy U given by F = – dU/dr

Gravitational Potential Energy, U U is shared between both m 1 and m 2 We can’t divide up U between them Example: Find U for the Earth-moon system

Superposition Principle for U For many particles, U = total shared potential energy of the system U = sum of potential energies of all pairs Write out U for this example

Total Energy, E If gravity is force is the only force acting, the total energy E is conserved For two particles,

Application: Escape Speed projectile: m Earth: m E Find the speed that m needs to escape from the Earth’s surface Derive the escape speed: Example 12-5

Orbits of Satellites

We treat the Earth as a point mass m E Launch satellite m at A with speed v toward B Different initial speeds v give different orbits, for example (1) – (7)

Orbits of Satellites Two of Newton’s Laws predict the shapes of orbits: 2nd Law Law of Gravitation

Orbits of Satellites Actually: Both the satellite and the point C orbit about their common CM We neglect the motion of point C since it very nearly is their CM

Orbits of Satellites If you solve the differential equations, you find the possible orbit shapes are: (1) – (5): ellipses (4): circle (6): parabola (7): hyperbola

Orbits of Satellites (1) – (5): closed orbits (6), (7): open orbits What determines whether an orbit is open or closed? Answer: escape speed

Escape Speed Last time we launched m from Earth’s surface (r = R E ) We set E = 0 to find

Escape Speed We could also launch m from point A (any r > R E ) so use r instead of R E :

Orbits of Satellites (1) – (5): ellipses launch speed v < v esc (6): parabola launch speed v = v esc (7): hyperbola launch speed v > v esc

Orbits of Satellites (1) – (5): ellipses energy E < 0 (6): parabola energy E = 0 (7): hyperbola energy E > 0

Circular Orbits

Circular Orbit: Speed v uniform motion independent of m determined by radius r large r means slow v Derive speed v

Compare to Escape Speed If you increase your speed by factor of 2 1/2 you can escape!

Circular Orbit: Period T independent of m determined by radius r large r means long T Derive period T Do Problem 12-45

Circular Orbit: Energy E depends on m depends on radius r large r means large E Derive energy E

Orbits of Planets

Same Math as for Satellites Same possible orbits, we just replace the Earth m E with sun m s (1) – (5): ellipses (4): circle (6): parabola (7): hyperbola

Orbits of Planets Two of Newton’s Laws predict the shapes of orbits: 2nd Law Law of Gravitation This derives Kepler’s Three Empirical Laws

Kepler’s Three Laws planet orbit = ellipse (with sun at one focus) Each planet-sun line sweeps out ‘equal areas in equal times’ For all planet orbits, a 3 /T 2 = constant

Kepler’s First Law planet orbit = ellipse P = planet S = focus (sun) S’ = focus (math) a = semi-major axis e = eccentricity 0 < e < 1 e = 0 for a circle Do Problem 12-64

Kepler’s Second Law Each planet-sun line sweeps out ‘equal areas in equal times’

Kepler’s Second Law Present some notes on Kepler’s Second Law

Kepler’s Third Law We proved this for a circular orbit (e = 0) T depends on a, not e

Kepler’s Third Law Actually: Since both the sun and the planet orbit about their common CM

Theories of Gravity Newton’s Einstein’s

Einstein’s Special Relativity all inertial observers measure the same value c = 3.0×10 8 m/s 2 for the speed of light nothing can travel faster than light ‘special’ means ‘not general’: spacetime (= space + time) is flat

Einstein’s General Relativity nothing can travel faster than light but spacetime is curved, not flat matter curves spacetime if the matter is dense enough, then a ‘black hole’ forms

Black Holes

If mass M is compressed enough, it falls inside its Schwarzschild radius, R s This curves spacetime so much that a black hole forms

Black Holes Outside a black hole, v and r for circular orbits still obey the Newtonian relationship: Also: from far away, a black hole obeys Newtonian gravity for a mass M

Black Holes Spacetime is so curved, anything that falls into the hole cannot escape, not even light Light emitted from outside the hole loses energy (‘redshifts’) since it must do work against the extremely strong gravity So how could we detect a black hole?

Black Holes

Answer: An Accretion Disk (emits X-rays)

Matter falling towards the black hole enters orbit, forming a hot disk and emitting X-rays