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

Chapter 12

Gravitation

Theories of Gravity

• Newton’s

• Einstein’s

Newton’s Law of Gravitation

• any two particles m1, m2 attract each other

• Fg= magnitude of their mutual gravitational force

221

g r

mGmF


• Fg = magnitude of the force that:

• m1 exerts on m2

• m2 exerts on m1

• direction: along the line joining m1, m2

221

g r

mGmF


• Also holds if m1, m2 are two bodies with spherically symmetric mass distributions

• r = distance between their centers

221

g r

mGmF


• ‘universal’ law:• G = fundamental

constant of nature

• careful measurements: G=6.67×10-11Nm2/kg2

221

g r

mGmF

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/s2 ?

• How is G related to g?

• Answer:

radiusEarth

massEarth

2

E

E

E

E

R

m

R

Gmg

Show that g = GmE/RE2Show that g = GmE/RE2

Other planets, moons, etc?

• gp=acceleration due to gravity at planet’s surface

• density assumed spherically symmetric, but not necessarily uniform

radius splanet'

mass splanet'

2

P

P

P

PP

R

m

R

Gmg

Example:Earth’s density is not uniform

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

2g r

mGmFw E


• This is the magnitude of the force that:

• m1 exerts on m2

• m2 exerts on m1

• What if other particles are present?

221

g r

mGmF

Superposition Principle

• the gravitational force is a vector

• so the gravitational force on a body m due to other bodies m1 , m2 , ... is the vector sum:

...on on 21 mmmmg FFF

Do Exercise 12-8

Do Exercise 12-6

Do Exercise 12-8

Do Exercise 12-6

Superposition Principle

• Example 12-3

• The total gravitationalforce on the mass at Ois the vector sum:

21 FFFg

Do some of Example 12-3 and introduce Extra Credit Problem 12-42

Do some of Example 12-3 and introduce Extra Credit Problem 12-42

Gravitational Potential Energy, U

• This follows from:

U

rdFW g

grav

r

mGmU 21

Derive U = - G m1m2/rDerive U = - G m1m2/r


• Alternatively: a radial conservative force has a potential energy U given by F = – dU/dr

r

mGmU 21

dr

dUF

r

mGm g2

21


• U is shared between both m1 and m2

• We can’t divide up U between them

r

mGmU 21

Example: Find U for the Earth-moon systemExample: Find U for the Earth-moon system

Superposition Principle for U

• For many particles,U = total sharedpotential energy of the system

• U = sum of potentialenergies of all pairs

231312 UUUU

Write out U for this exampleWrite 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,

r

mGmvmvm

UKE

21222

211 2

1

2

1

Application: Escape Speed

• projectile: m

• Earth: mE

• Find the speed that m needs to escape from the Earth’s surface

r

mGmU E

Derive the escape speed: Example 12-5Derive the escape speed: Example 12-5

Orbits of Satellites


• We treat the Earth as a point mass mE

• Launch satellite m at A with speed v toward B

• Different initial speeds v give different orbits, for example (1) – (7)


• Two of Newton’s Laws predict the shapes of orbits:

• 2nd Law• Law of Gravitation


• 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


• If you solve the differential equations, you find the possible orbit shapes are:

• (1) – (5): ellipses • (4): circle• (6): parabola• (7): hyperbola


• (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 = RE)

• We set E = 0 to find

E

Eesc R

Gmv

2

Escape Speed

• We could also launch m from point A (any r > RE)

• so use r instead of RE :

r

Gmv E

esc

2


• (1) – (5): ellipses

• launch speed v < vesc

• (6): parabola

• launch speed v = vesc

• (7): hyperbola

• launch speed v > vesc


• (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

r

Gmv E

Derive speed vDerive speed v

Compare to Escape Speed

• If you increase your speed by factor of 21/2 you can escape!

r

Gmv

r

Gmv

Eesc

E

2

Circular Orbit: Period T

• independent of m• determined by radius r• large r means long T

EGm

rT

2/3 2

Derive period TDerive period T

Do Problem 12-45Do Problem 12-45

Circular Orbit: Energy E

• depends on m• depends on radius r• large r means large E

02

1

2

E

Ur

mGmE E

Derive energy EDerive energy E

Orbits of Planets

Same Math as for Satellites

• Same possible orbits, we just replace the Earth mE with sun ms

• (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, a3/T2 = constant

Kepler’s First Law

• planet orbit = ellipse

• P = planet• S = focus (sun)• S’ = focus (math)• a = semi-major axis• e = eccentricity

0 < e < 1e = 0 for a circle

Do Problem 12-64Do 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

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

22

3

4SGm

T

a

Kepler’s Third Law

• Actually:

• Since both the sun and the planet orbit about their common CM

22

3

4

)(

ES mmG

T

a

Theories of Gravity

• Newton’s

• Einstein’s

Einstein’s Special Relativity

• all inertial observers measure the same value c = 3.0×108 m/s2 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

Black Holes

• If mass M is compressed enough, it falls inside its Schwarzschild radius, Rs

• This curves spacetime so much that a black hole forms

2

2

c

GMRS

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

r

GMv

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

Documents

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