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
Newton’s Law of Gravitation
• 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
Newton’s Law of Gravitation
• Also holds if m1, m2 are two bodies with spherically symmetric mass distributions
• r = distance between their centers
221
g r
mGmF
Newton’s Law of Gravitation
• ‘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
Newton’s Law of Gravitation
• 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
Gravitational Potential Energy, U
• Alternatively: a radial conservative force has a potential energy U given by F = – dU/dr
r
mGmU 21
dr
dUF
r
mGm g2
21
Gravitational Potential Energy, U
• 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
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)
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 = 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
Orbits of Satellites
• (1) – (5): ellipses
• launch speed v < vesc
• (6): parabola
• launch speed v = vesc
• (7): hyperbola
• launch speed v > vesc
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
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