Transcript No Slide Title

Last Week.
1. Objectives of Telescope design.
- Light Grasp
- resolving power Rayleigh’s criterion
- Angular magnification.
2. Refracting and reflecting telescopes.
R = m = 1.22/b
where m is the angular
separation of the two
objects and b is the
telescope aperture.
3. Aberrations and remedies
4. Effects of the atmosphere
- adaptive optics
5. Telescopes for other parts of the spectrum.
- radiotelescopes and arrays of
radiotelescopes.
Topics related to Gravity
1. Kepler’s Laws
2. Artificial Satellites
2. Escape Velocity
3. Tides
4. Synchronous orbits
5. Planetary ring systems
6.The Roche Limit
7.Masses of stars in a binary system
Starting Point – Kepler’s Laws.
Tycho Brahe (1546 – 1601)
From 1576-1597 made detailed measurements of the positions of
the planets with an accuracy of 1 arcminute.
He did not believe that Earth orbited the Sun because he could not
see “local” stars moving relative to the distant background of stars.
Johannes Kepler (1571-1630)
He used Brahe’s measurements to create a model of planetary motion.
He broke with tradition-it had been believed that heavenly bodies
move in circles. Kepler explained planetary motions in terms of the
planets moving in ellipses.
The Ellipse
y
(0,+b)
(-a,0)
(-f,0)
(+f,0)
• Defn - Locus of all points the sum of
whose distances from two fixed
points (foci) is a constant.
(+a,0) •Position of foci
2 + b2)1/2 +
(a
+
f)
+
(a-f)
=
(f
x
(f2 + b2)1/2
(0,-b)
i.e. f = (a2 - b2)1/2
• Eccentricity e = f = (a2 - b2)1/2 = [1 - b2 / a2]1/2
a
a
• When e = 0, a = b and we have a circle
i.e. x2 + y2 = r2
[ Definition]
Kepler’s Laws (1618)
1.The planets orbit the Sun in ellipses
with the Sun at one focus.
2.The line joining the Sun and a planet
sweeps out equal areas in equal times.
3.The square of the period of a planet
is proportional to the cube of the
semi-major axis of the ellipse.
P2  a3
Convenient Measure of distance
-Astronomical Unit(1 au)
= Average Earth-Sun distance
= 1.496 x 1013 cm
= 1.496 x 1011 m
Note:- Elliptical orbits were an essential innovation but for simple
calculations one can assume that the orbits are circles. In
general it is a good approximation.
Measures of Distance – So Far
1. Convenient Measure of distance
-Astronomical Unit(1 au) = Average Earth-Sun distance
= 1.496 x 1013 cm = 1.496 x 1011 m = 1.496 x 108 km.
2. LIGHT YEAR (ly) is the distance travelled by light in vacuum
in 1 year.
1 ly = 9.461 x 1012 km
Note :- 1ly = 6.324 x 104 au
Kepler’s Third Law
Plot of a3 versus P2 for the planets
in the Solar system
a3
- Here a is in AU and P is in
Earth Years.
P2
Clearly P2  a3
All three of Kepler’s Laws are rigorously obeyed wherever two
objects move under their mutual gravitational attraction.
Planetary Orbits
Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto
Semi-major axis
(106 km)
Sidereal Period
Orbital Eccent.
57.9 108.2 149.6 227.9 778.4 1424
Surface Temp.
5906
0.241 0.615 1.0 1.88
11.9 29.5
84.0
165
248
0.21
0.05 0.06
0.05
0.01
0.25
1.3 2.5
0.8
1.8
17.1
178 23.5 25.2
3.1 26.7
97.9
29.6
122
58.7
243
1.0
1.03
0.41 0.43
0.72
0.67
6.4
1-700
730
300
220
130
58
58
50
86.8
102
0.013
Direction of revn
Angle to plane
7.0
of ecliptic(degs.)
Angle of Plane
0.1
to spin axis(degs)
Rotation Period
(Days)
2871 4499
Mass(1024 kgm) 0.33
0.01 0.02 0.09
All in same direction
3.4
0
1.8
97
4.87 5.98 0.64 1900 569
Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto
All the same
Direction of revolution
Direction of rotation
on axis
Same
Opp.
Same Same
Same Same Sideways Same Sideways
All these characteristics have to be explained by any model of
the formation of the solar system.
Reminder of Newton’s Laws
1.A body remains at rest, or moves in a straight line at constant speed,
unless acted on by an external force.
2.The acceleration of an object is proportional to the force acting on
the object.
F = m = d(mv)
dt
3.Whenever a body exerts a force on a second body, the second body
exerts an equal and opposite force on the first body.
Directon of force to keep
Earth in circular orbit
•At any instant, direction of motion
is tangential to the orbit.
Sun
•Centripetal force(FCP) acts towards
the Sun to keep it in circular orbit.
Direction
of motion
FCP = mv2
r
Newton reasoned that if Sun attracts Earth then Earth attracts Sun
F =G.m1.m2
r2
[Universal Law of Gravitation]
The Effect of Centre-of-Mass
Centre-of-mass
dm
dM
The centre-of-mass is given by
dM = m
.r
(m+M)
m
r
M
dm = M
.r
(m+M)
If m/M  0.01 we cannot assume that we have a fixed, central body.
If so we can no longer simplify our equations and we have to
take account of the fact that the motion is about the centre-of-mass.
The Effect of Centre-of-Mass
1.
Mv2 = G.M.m
dM
r2
Now the period P for an orbit is 2.radius/velocity
i.e. v = 2. dM
P
2. Mv2 = M. (2. d / P)2
= G.M.m Since d = m .r
M
M
2
r
dM
(M + m)
dM
P2 = 42.r3
G.(M + m)
= 42 .(M + m)2 .dM3 = 42 .(M + m)2.dm3
G.m3
G.M3
3. In the case of the Earth and Sun the imbalance in mass is such that the
approximation of a static Sun is a good one. However Kepler’s Laws
apply to all systems moving under gravity and often the masses are
closer together so we cannot assume that one of them is fixed.
[ Mass of Sun = 1.99 x 1030 kg Mass of Earth = 5.97 x 1024kg ]
Artificial Satellites
1.There are many possible orbits for our satellites depending on their
function.One useful orbit is the GEOSTATIONARY orbit.
i.e.orbit keeps the satellite in the same place relative to the Earth.
2.Clear condition is that satellite orbits with same period and direction
as the Earth. Hence
P2 = 42.r3
r3 = G.M.P2
G.M
42
= 6.67.10-11.6.10+24(24.60.60)2
42
r = 42,300 km.
So our satellite orbits at about 5.6 Earth Radii above the surface.
Note that this is independent of the mass of the satellite.
[Earth’s radius is 6370 km.]
Escape Velocity
1.There are many reasons why this is important. It is the velocity
required to escape from the gravitational field of another body.
This is the velocity needed to take it to infinity.
2.To take the object to infinity it is necessary to do work against gravity.
Work done =  F.dr =  G.M.m.dr with the integral from r to 
r2
= G.M.m
where rp is the radius of the planet or
rp
other large body.
3.This is the kinetic energy the projectile must have initially to escape
to infinity.
1.mv2escape =GMm/rp
2
For the Earth vescape = [2x6.67x10-11x 5.98x1024/6378x103]1/2
 11.2 kms-1
Ocean Tides
• Tidal forces are due to differential forces across a body.They act on
oceans and on the Earth itself.
• Tides are closely connected with the position of the Moon.
AccelN due to Moon is greater at A than B,which is greater than at C.
Water is fluid & it drops towards Moon most quickly at A and at C it
is left behind.At A and C we have HIGH TIDE and at B LOW TIDE.
• As the Earth spins on its axis( Moon’s period is 27 times longer ) every
longitude on Earth will experience two high tides and two low tides per
day. The bulge(tide) is always aimed slightly ahead of the Moon.
Ocean Tides
 As the Earth spins on its axis( Moon’s period is 27 times longer )
every longitude on Earth will experience two high tides and two
low tides per day. They occur every 25 hours.
The bulge (tide) is always aimed slightly ahead of the line joining
the Earth and Moon.
There are many local effects and in some places there is only one
high tide a day.
The Moon versus the Sun
• FS
FM =
MS.rM2
2 x 1030 ( 3.84 x 108 )2
MM.rS2 = 7.4 x 1022( 150 x 109 )2
 170
HOWEVER
• F = -2GMm so the comparative effect of the Sun and Moon
r3
r
is given by FS
MS . rM 3
= r 3.M
FM
S
M
 0.465
• So tide-raising ability of Moon is 2.15 times effect of Sun but
effect of Sun is not negligible.
• Effects of Sun and Moon combine linearly and vectorially.
When Sun and Moon are aligned (Full or New Moon) we get a
maximum tide 1.465 times normal.These are Spring tides(every 2
weeks).At Half Moon(at right angles) tides are smaller-Neap Tides.
Synchronous Orbits
a)The gravitational forces
on Earth due to the Moon.
b)The differential forces at
the same points.
1.Normally we think of tides in terms of the Oceans but there are effects
on solid objects such as the Earth or Moon.Thus the Earth feels
different forces on its two sides.The differential forces elongate the
shape of the Earth,which behaves like a fluid.The result is bulges of
approx 10 cm.along the line joining the centres of mass.
2.Again Earth’s tidal bulges are not aligned with the Moon.
Frictional forces on the surface drag the bulge axis ahead of the line
joining the Earth-Moon.
Friction is a dissipative force so rotational K.E.
is constantly being lost and Earth spins more slowly[1.6 cms per
century].The Moon is also drifting away at 3-4 cm per year.
Synchronous Orbits
1.How are these things related?
2.Consider torques on the Earth due to the bulges.
Bulge A leads the Moon and is closer to it.Force on
A is greater than on B.The result is a net torque
slowing down the Earth.It reduces angular
momentum.However A.M. is conserved in this
isolated system.
3.At the same time Bulge A is pulling the Moon forward,speeding the
satellite up and causing it to move further away (mvr = const.).
Note:-We are assuming the influence of planets and Sun are negligible
4.Because of these tidal effects the Earth will eventually slow down so
that the same side of the Earth will always face the Sun in the same
way that the Moon faces the Earth.
In the past the Moon would have been much closer to the Earth with a
period as short as a week. This is a more general phenomenon.
The Roche Limit
1.Edouard Roche(1820-1883).-When a comet or Moon or other large
object is spiralling in on a planet (mass M) it is unlikely to collide
with it. Instead it breaks up at the ROCHE LIMIT.
2.As the object approaches it has a differential force across it
dF = -2G.M - the differential force on unit mass.
dr
r3
3.Tidal forces increase rapidly as distance decreases.the result is that the
shape of the object[Moon,comet,asteroid etc]becomes elongated.
4.If we assume that the object is a fluid, then when r is small it becomes
impossible to define a shape where the force of gravity is always at
right angles to the surface.Then the material will always flow in the
direction of the net gravitational force.
5.If oscillations start then the object will break up under tidal disruption.
The Roche Limit
1.Roche Limit = Max. orbital radius for which tidal disruption occurs.
Assume disruption occurs when diff. grav. force exceeds “self”
grav. force holding it together and that “moon” and planet are
spherical. We neglect centrifugal effects. For disruption,
GM.m < 2GMR ,
r3
R2
where R is the radius of the “moon”, r is the distance between the
“moon” and the planet and M is the mass of the planet.
2.Now M = 4/3..Rp3p and m = 4/3..Rm3.m
3.Solving for r we find
r < 21/3(p / m)1/3Rp
Roche found 2.456
4.Examples:-Comet Shoemaker-Levy
Rings of Saturn
Shoemaker-Levy
Comet captured into an orbit
round Jupiter.Its orbit took it
1/3AU from Jupiter and Sun
modified orbit so that we
knew it would collide with the
planet in July 1994.
The comet broke up because it
exceeded the Roche limit and
the fragments crashed on the
planet between July 16-24
Here we see the effect of the
impacts as observed with the
HST in the UV part of the
spectrum.
Saturn’s Rings
Saturn’s Rings seen by Hubble Space Telescope at different wavelengths.
Physics of Atmospheres
1.The composition of the atmosphere of a planet is related to the simple
idea of escape velocity.[vESCAPE = [2G.M/rP]1/2 =11.2 kms-1 for Earth]
2.The Temperature is a key parameter in the formation and evolution of
a planet. Inter alia it affects the nature of the atmosphere.
3.For a star L = 4.R2..T4
where we have assumed a spherical star of radius R
4.Under equilm conditions the total energy content of a planet is constant.
Assume planet is a black-body of radius RP and temp.TP in a circular
orbit at distance D from the star.Assume the planet reflects a fraction 
of incoming light[the ALBEDO]
5.It can be shown that TP = TS(1-)1/4[RS/2D]1/2,
i.e. TP is proportional to the surface temp. of the star and does not
depend on the size of the planet.
6.Using  = 0.3 the temp. of the Earth is 255K.This is substantially
below the value of 300K.This is because it ignores the Greenhouse
effect.This is a significant warming due largely to water vapour.
Physics of Atmospheres
7.From Wien’s Law MAX.T = 2.9 x 10-3 m.K
For Earth MAX = 1000 nm. Hence it emits in the infrared.
This infrared radiation is absorbed and re-emitted by the atmospheric
greenhouse gases( CO2,CH4,and chlorofluorocarbons and H2O vapour)
This acts as a thermal blanket and raises T by about 34 degrees.
[This is the reason for the concern about increases in CO2 emissions]
8.Evolution of the atmosphere depends on T of solar nebula during
formation together with planet’s temp.,gravity,composition,&chemistry
following formation.
Later outgassing from rocks and volcanoes plays a part.
9.
Figure shows the MaxwellBoltzmann distribution for the
velocities of molecules or atoms in
a gas.
Some will exceed the escape
velocity.
Law of Gravity as a function of Distance/Time
• Cavendish showed in the lab. that F = G.M1.M2
r2
Where G = 6.672.10-11 Nm2kg-2
• Earth ,Sun etc are spherical-What we expect for aggregation of
material.[Note:-it is oblate spheroid.]
• Moons of Jupiter-originally they appeared to vary in orbital period.
They were ahead(behind) time when close(far) from Earth.this is
effect of finite velocity of light.
• Analysis of motions of Jupiter ,Saturn and Uranus in terms of
grav. Force.First two worked but not third.Adams and Leverrier
independently suggested another planet- Pluto
It works on scale of Solar System
Law of Gravity with Distance/Time
• Binary stars(majority)-example SIRIUS-8.6ly away.
• Picture shows measurements of relative positions of Sirius A&B
-Perfect Ellipse but Sirius A is not at focus.
This is because we see the elliptical orbit at
an angle.
• Globular Clusters are spherical
Coma Cluster - 20Mly across containing thousands of galaxies(300 seen
here). It is about 270 Mly away.Two supergiant galaxies seen in centre.