#### Transcript Document

```Physics 221
Department of Physics
Lecture Notes
S. Yost
November 14, 2008
Universal Gravitation
Announcements
Set 13 – Ch. 13 problems due next
Wednesday.
Next: Chapter 14 – fluids. Read sections 1-3
for Monday, 4-5 for Wednesday, 6-7 for
Friday. I will be away until Thursday.
assign before 5PM Monday for lecture
notes and a set of examples to be worked
on-line.
This will be treated as a homework set.
Universal Gravity
Newton’s Law of Gravity describes a universal
attractive force between all massive objects.
F = G m1 m2 / R 2
where R is the separation, and G is Newton’s
gravitational constant:
G = 6.67 x 10-11 Nm2/kg2
Gravity is very weak between objects on a human
scale. In fact, it is the weakest of all known forces,
by far. It is important only because it is always
attractive, and is universal, so that it can be
significant for large masses.
Mass of the Earth
What is the Earth’s mass?
Weight is the gravitational attraction of an object
to the Earth, but the Earth is big – not at a
single location. The weight is the total
gravitational attraction of an object to all the
earth’s parts: It looks like it should be
complicated: The gravitational force is the sum
of the forces of all the little pieces dM of the
Earth. This gives an integral:
mg = ∫ Gm dM r/r2
dM
r
m
Fortunately, this integral is simple
for spherically symmetric objects.
Gravitating Sphere
Don’t worry, we won’t actually do the integral, but
Consider a hollow uniform spherical shell of mass
M: a cosmic ping-pong ball.
If m is outside a spherical shell of mass M, then
the gravitational force is F = GmM/R2, where R
is the distance from m to the center of the shell.
This is as if the shell
M
m
were collapsed down to
R
a point at its center.
Gravitating Sphere
Interestingly, if you are inside the sphere, the
gravitational force is zero. Intuitively, this is
because, if you look at the mass enclosed by
two back-to-back cones of equal angle, it is
proportional to the square of the distance to the
cone, so that m/r2 is the same for two cones in
opposite directions, and the forces cancel.
m1
r1
m
r2
m2
If you are inside the earth, a
distance r from the center, you
only feel an attraction to the
rock inside radius r, not outside.
Gmm1/r 12= Gmm2/r22
Mass Of Earth
Then the weight of an object on the Earth’s surface
is
mg = GMm/Re2 with Re = 6370 km.
GM = gRe2.
(Useful relation! It saves you from having to remember G
and M in many problems)
M = g Re2/G = (9.8 m/s2)(6.37x106 m)
6.67 x 10-11 Nm2/kg2
= 5.96 x 1024 kg. Actually, 5.98 x 1024 kg.
(The earth isn’t exactly a sphere.)
Fun facts:
You don’t have to look up or remember the
radius of the Earth if you remember that
the meter was defined so the North Pole is
10,000 km from the equator. This implies
that (p/2)Re = 10,000 km, or
Re = 20,000 km/p = 6370 km.
Eratosthenes first measured the radius of the
Earth in 225 BC by noting that the sun’s
location 800 km south of a point where it
was directly overhead. This implies that
800 mi = Re/8, or Re = 6400 km.
p/2
Re
q
q
d
Re
d = Re q
Orbital Velocity Example
We can use the
inverse square law
to find the orbital
velocity of a
spaceship at a
height 2Re above
the Earth’s surface.
(Re = 6370 km)
2Re
Re
Orbital Velocity Example
The spaceship is 3 times as far from the center
of the Earth as the ground is, so at this
height,
a = g/9 = 1.1 m/s2
Setting the gravitational acceleration equal to
the centripetal acceleration gives
a = g/9 = v2/R = v2 / (3Re).
v = (3 x 6370x103 m x 1.1 m/s2) ½
= 4.59 km/s.
Binary Sun Example
Suppose that the earth
were replaced by
another sun. What
would be their orbital
period?
All the other planets are
negligable in mass –
this is a binary system.
Each sun attracts the
other.
M
R = 1AU
M
Binary Sun Example
The centripetal
acceleration of either
sun is a = v2/R.
The force on either sun
is GM2/(2R)2.
So v2 = GM/(4R).
Compare Earth’s
velocity: ve2 = GM/R.
v/ve = ½ . So the period
is 2 earth years.
M
R = 1AU
M
Gravitational Potential Energy
The gravitational attraction is
F(r) = -G Mm/r2.
G = 6.67 x 10-11 Nm2/kg2.
The gravitational potential
energy is
U(r) = - G Mm/r.
The integration constant is
chosen so that U(r) = 0 at
r=∞
m
F
r
M
Gravitational Potential Energy
If an object falls to Earth from rest at a very
great distance, how fast does it enter the
Earth’s atmosphere?
“Very great distance” means infinite, or in
practical terms, far enough away that it
doesn’t affect the result. An object infinitely
far away would, in reality, never get here, so
don’t take this too literally.
Gravitational Potential Energy
The initial energy is zero, since v = 0
implies K = 0 and r = ∞ implies U = 0.
The final energy is then
0 = ½ mv2 + U(Re) = ½ mv2 – gmRe2/Re
Since
0 = ½ mv2 – gmRe
the entry velocity is v = √ 2gRe
= √2(9.8 m/s2)(6.38 x 106 m)
= 11 km/s.
Gravitational Potential Energy
Conversely,
v = √ 2gRe = 11 km/s
is the velocity at which a projectile would
have to be launched to get to r = ∞ with
zero velocity.
This is called the escape velocity.
Energy In Orbit
What is the total energy of an orbiting
satellite?
K = ½ mv2. U = -GMm/R.
But mv2/R = GMm/R2. (Centripetal force)
Then the total energy of an orbiting satellite
is U = - ½ GMm/R: half its potential
energy.
Kepler’s Laws
• Before Newton, Kepler had analyzed a
vast collection of observations by Tycho
Brahe, and created a mathematical model
of the motion of the planets.
• A mathematical model is a precise
description of a phenomena, but one not
necessarily intended to explain the
phenomena.
• This model is known as Kepler’s Laws.
Kepler’s Laws
Law 1: All planets move in
elliptical orbits.
P = perihelion
A = aphelion
Elliptical Orbit
Rp = perigee (perihelion)
Ra = apogee (aphelion)
Rp + Ra = 2R.
R
Rp
R
Ra
Kepler’s Laws
Law 2: A line joining any planet
to the sun sweeps out equal
area in equal times.
Proof of 2nd Law
Angular momentum conservation:
Gravity acts along a line through
the sun: no torque about the sun.
L = r x mv, L/m = |r x v|
vt
A
r
r
Lt/m = r x vt = area of
parallelogram with sides r and vt
= 2 x area of triangle “swept out”
= 2A
Result: A = Lt/2m
dA/dt = L/2m = constant
vt
Proof of 2nd Law
The total area inside
ellipse depends on L:
a
A = LT/2m = pab.
if T = period.
More angular
momentum gives a
“fatter” orbit.
b
A = pab
area of an ellipse
Kepler’s Laws
Kepler’s Third Law:
The square of a planet’s year, divided by the
cube of its semimajor axis, is the same for
all planets:
T2/R3 = TE2/RE3
If T is measured in Earth years, and R in
Astronomical Units (the distance from the
Earth to the sun), then TE = RE = 1, and for
any other planet,
T2 = R3.
Kepler’s Third Law
• 1 AU = 1.50 x 1011 m
• 1 year = 3.16 x 107 s
Example: Saturn’s semimajor axis is
R = 9.58 AU, giving an orbital period
of
T = (9.583) ½ = 29.7 years
It is easy to prove Kepler’s Law for
Circular orbits…
Kepler’s Third Law
Centripetal Acceleration: a = v2/R
Newton’s Law: ma = GmM/R2
Then v2 = GM/R.
v = 2pR/T gives Kepler’s Third Law:
4p2R3 = GMT2.
But don’t bother remembering this –
it’s how we got it that matters.
For an ellipse, replace R by the
semi-major axis to get the
general form of Kepler’s Law.
v
R
Kepler’s Third Law
4p2R3 = GMT2.
still holds for elliptical orbits if R is the
semi-major axis.
We won’t prove it, but it can be done
algebraically by calculating the
energy and angular momentum at
the perigee and apogee, and using
the fact that they are conserved, and
the 2nd Law.
You would also find that the total
energy of the orbit is E = - ½ GM/R,
generalizing the earlier result for the
energy of an orbit.
R
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