gravitation newton’s law of universal gravitation (1687): r sir henry cavendish (1731-1810) (a...
Post on 19-Dec-2015
247 views
TRANSCRIPT
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 !