GravitationNewton’s law of universal gravitation (1687):

221

r

mGmF

r

2211 /1067.6 kgNmG Sir Henry Cavendish (1731-1810) (a torsion balance)

Principle of superposition of forces from different sources: j

jFF

Concept of fieldsfor forces that act at a distance

Weight depends on r.

On earth’s surface w = mg

2r

mGmFw Egrav

2E

E

R

Gmg

Cavendish’s mass of the earth

kgG

gRm E

E24

2

106 (since RE = 6380 km)

Magnitude F~1μN for m1=m2= = 100 kg at r = 1 m is very small.

Gravitational Potential Energy: Wgrav= U1 - U2

1222

2

1

2

1r

GMm

r

GMm

r

drGMmdr

r

GMmW

r

r

r

r

grav

r

GMmU

Let us choose U to be zero at infinity r = ∞:

Gravitational force is conservative !

2r

GMm

r

GMm

dr

d

dr

dUFr

Derivation of the gravitational potential energyU = mgy near the earth’s surface ( r = RE+y )

ygmyR

GMm

yRR

ymGM

R

mGM

yR

mGMRUrUU

E

E

EEE

E

E

E

EEE

2)(

)()(

Journey to the center of the earth

rrR

Gmm

R

rm

r

Gm

r

GmMF

e

E

eEg

33

3

22

For a point mass m insidespherical shell (ME – M)

r

MMGmU E )(

The force exerted by the shell is zero!

Only the “inner” mass Mcontributes to the total gravitational force

Exam Example 27: Motion in the gravitational field of two bodies

(problem 13.62)

0X1

X2

X0

X

ΔX=X-X0M

1M

21F

2Fm

Data: masses M1

, M2

, m; positions x1

, x2

, x0

, x; v(t=0)=0

Find: (a) change of the gravitational potential of the test particle m;

(b) the final speed of the test particle at the final position x;

(c) the acceleration of the test particle at the final position x. Solution: (a)

(b) Energy conservation:

(c) Newton’s 2nd

law:

2202

11010

1111

xxxxmGM

xxxxmGMUUU

mUvUmvUK /22

1 2

2

1

12

2

212 /

xx

M

xx

MGmFFa xxx

The Motion of Satellites

v

Closedorbits

Open orbits

V1(r=RE)

V2(r→∞) Orbits are closed or open. Open orbits are parabolic or hyperbolic.Closed orbits are elliptical or circular.

There is only one speed that a satellitecan have in a given circular orbit r :

r

GMv

r

mv

r

mGMmaF EE

cg 2

2

Period:

“Synchronous satellites” for digital satellite system TV:T=1 sidereal day= 24 h(365-1)/365= 23 h 56 m, r =40000 km

Global Positioning System (GPS) : 24 satellitesSpace speeds: 1st- to stay in orbit r = RE V1=(GME/RE)1/2=(gRE)1/2=8 km/s 2nd- to leave the earth K = - U, mV2

2/2 = GMEm/RE , V2 =(2GME/RE)1/2 =(2)1/2 V1 =11.2 km/s (parabolic or escape speed)

3rd- to leave the sun V3= 16.7 km/s > V2(sun) – Vearth-sun=

=[(2)1/2-1](GMsun/Rearth’s-orbit)1/2 = 0.41 (29.8 km/s)= 12.3 km/s since both the sun’s and earth’s grav. attractions act.

V3

Ve-s=29.8km/s

Re=150 Mkm

Earth

Sun

Remark: Conservation of energy yields another relation between r and v,

, that does not contradict to r×v=const along a trajectory.constrGMmmv /2/2

Pioneer-10: Earth (1972)→ Jupiter→ Pluto (1983)→ Solar system’s border (2002)→ star in Mly

EGM

r

v

rT

2/322

On December 18, 2004, Voyager 1 passed the termination shock. This marks the point where the solar wind slows to subsonic

speeds. This is the unofficial date of departure from the Solar System. While the spacecraft still remains under the sun's influence,

at the termination shock particles from the interstellar medium interact with solar particles, signaling that the hypothetical heliopause

is not far from this point. Six years later in 2010 Voyager 1 entered an area of the heliosheath where the solar wind outward speed

is 0, or flowing sideways relative to the sun. This signals that Voyager 1 is getting very close to entering the interstellar medium.On December 5, 2011, it was announced that Voyager 1 had entered a new region referred to as a "cosmic purgatory" by NASA.

Within this stagnation region, charged particles streaming from the sun slow and turn inward, and the solar system's magnetic field

has doubled in strength as interstellar space appears to be applying pressure. Energetic particles originating in the solar system

have declined by nearly half, while the detection of high-energy electrons from outside has increased by 100 fold. The inner edge

of the stagnation region is located approximately 113 astronomical units from the sun, while the outer edge is unknown.At a distance of 120 astronomical units (1.8×1010

km) as of February 2012, it is the farthest man-made object from Earth.

Voyager 1 is now in the heliosheath, which is the outermost layer of the heliosphere.

It will most likely be the first probe to leave the Solar System.

Exam Example 28: Satellite in a Circular Orbit

RE=6380 km

r

v

cgrav FF

ME

mData: r = 2RE , RE = 6380 kmFind: (a) derive formula for speed v and find its value;(b) derive formula for the period T and find its value; (c) satellite’s acceleration.

Solution: use the value g = GME/RE2 = 9.8 m/s2

ca

(a) The only centripetal force is the gravitational force:

skmmsmgR

r

Rg

r

R

R

GM

r

GMv

r

mv

r

mGMmaFlawndsNewton

EEE

E

E

EEcg

/6.51019.3/8.92

2'

6222

2

2

2

(b) The period T is a time required for one orbital revolution, that is

hsm

m

g

R

R

r

RGM

R

r

GM

r

rGM

r

v

rT E

E

E

EE

EE

4/8.9

1038.6222

22222

62/3

2/3

2

2/32/3

(c) Newton’s second law with the central gravitational force yields

atan = 0 and arad = ac = Fg/m = GME/r2 = (GME/RE2) (RE/r)2 = g/4 = 2.45 m/s2

Kepler’s Laws of Planetary Motion (1609, 1619)

Heliocentric world system (Copernicus, 1543) vs. Geocentric world system (Ptolemei)

1. Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse Eccentricity e = (Center O to Focus S) / (Semi-major axis a) The earth’s orbit has e = 0.017. Pluto (e =0.248) is not 9th planet, it is a dwarf planet!

2. A line from the sun to a given planet sweeps out equal areas in equal times.Proof is based on the angular momentum conservation that follows from the fact that the gravitational force is a central force 0/ gravFrdtLd

constdtdArvconstvmrL /2sin||

3. The periods of the planets are proportional to the 3/2 powers of the major axis lengths of their orbits:

sunGM

aT

2/32

Recent discovery: planets in orbit around thousands of other stars via detection of the apparent “wobble” of a star near the center of mass.

Note: Kepler (1571-1630) discovered laws of planetary motion ~100 years before Newton (1642-1727) formulated laws of mechanics (1687) !

1600-Giordano Bruno burned at the stake by Church for heresy: Copernican system, sun=star, eternal plurality of worlds

1633- Galileo Galilei sentenced by Church to imprisonment

Exam Example 29: Satellite in an Elliptical Orbit (problem 13.77)

(perigee) (apogee)

hp ha2RE

Data: hp , ha , RE= 6380 km, ME= 6·1024 kg

av

pv

rada

gravFFind: (a) eccentricity of the orbit e; (b) period T; (c) arad;

(d) ratio of speed at perigee to speed at apogee vp/va;(e) speed at perigee vp and speed at apogee va;(f) escape speeds at perigee v2p and at apogee v2a.

pv2Solution: (a) rp =hp+RE, ra= ha+RE, a =(rp+ra)/2,

ea = a – rp, e = 1 – rp/a = 1- 2rp/(rp+ra) = = (ra- rp)/(ra+ rp) = (ha-hp)/(ha+hp+2RE)

(d) Conservation of angular momentum (La= Lp) or Kepler’s second law: rava= rpvp, vp/va= ra/rp

(b) Period of the elliptical orbit is the same as the period of the circular orbit with a radius equal to a semi-major axis R = a, i.e.,

EGM

aT

2/32

(e) Conservation of mechanical energy K + U = const :

;)(

2

222 2

2222

app

aEp

a

E

a

pp

a

Ea

p

Ep

rrr

rGMv

r

mGM

r

rmv

r

mGMmv

r

mGMmv

)(

2

apa

pE

a

ppa rrr

rGM

r

rvv

(f) Conservation of mechanical energy for an escape from a distance r (the second space speed) :

a

Ea

p

Ep

EE

r

GMv

r

GMv

r

GMv

r

mGMmv 2,

22

2 222

22

(c) Newton’s 2nd law and law of gravitation: arad= Fgrav/ m = GME/r2.

Apparent Weight and the Earth’s Rotation

radamww

0

At the equator:

2

2

0

/)03.08.9( smg

R

vg

m

wg

E

Small additional variations:(i) imperfect spherical symmetry,(ii) local variations in density,(iii) differences in elevation.

Black HolesEvent horizon at Schwarzschild radius

cR

GMV

c

GMR

SescapeS

222

Supermassive black hole at the center of our Milky Way galaxySagittarius A*: RS = 8·106 km, M = 4·106 Msun = 8·1036 kg

“Black holes have no hair” and can be entirely characterized byenergy, momentum, angular momentum, charge, and location.

Primordial Black Holes and γ-Ray Bursts

Black-hole potential well

Objects fall into the black hole in a finite proper time Δt ~ RS/c,but infinitely long for a distant observer, and are affected by gravitational red shift, time dilation, and tidal forces effects !All information is lost inside black hole for outside observers.

E0=K0+U0<0 (trapped state)

E=E0+Wnc (dissipative fall into black hole)

Energyr

r

GMmU

0E0=K0+U0>0 (free escape)

ncWEr

GMmmv 0

2

2

RSsun= 3 km, RSearth= 1 cm

2

322/3 42

GT

aM

GM

aT

a2M

At r = 1.5 R its velocity is v = c.

Circular motion can occur only at r > 1.5 RS.

4RS 2RS

1.5RS

RS

Cone of gravitational capture ( for v2/2 = GM/r )

Black hole is a bright sourcedue to an accretion disk !