Download Orbits - High Energy Physics

Orbits
Chapter 13.5-8 Today – Will not cover section 13.9.
Newton’s Shell Theorem
A uniform spherical shell of matter attracts a particle
that is outside the shell as if all of the shell’s mass
were concentrated at its center.
It is also true that a uniform shell of matter exerts
no net gravitational force on a particle located
inside it.
Let’s pretend there is a hole all the way thru the earth
and we could drop an object into it – What would
happen?
1.  Mass outside the object we can ignore
2.  Only mass inside radius of object matters
3.  Via the shell theorem can compute the force
assuming all of the inside mass is at the center
€
€
Mass inside the Earth
Gmobj M ins
F=
r2
3
4 πr
M ins = ρ
3
 4 πGmobj ρ 
F=
r
3


F = −kr
Mins
mobj
The r vector points out from the center, so we have
and forth.
and object would just oscillate back
Clicker question1
Set frequency to BA
A satellite in circular orbit about the
Rock
Earth is directly over Denver, 300 miles
Satellite
above the city, and traveling eastward
at 16,000 mph. At the same time, a
Denver
rock is released from rest 300 miles
above the city, right next to the satellite.
True or false: The acceleration of satellite and rock at this point in
time is the same in both magnitude and direction.
A. True
B. False
Clicker question1
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A satellite in circular orbit about the
Rock
Earth is directly over Denver, 300 miles
Satellite
above the city, and traveling eastward
at 16,000 mph. At the same time, a
Denver
rock is released from rest 300 miles
above the city, right next to the satellite.
True or false: The acceleration of satellite and rock at this point in
time is the same in both magnitude and direction.
A. True If the satellite and rock are a distance r from the center
B. False of the Earth, the acceleration of either one is
independent of whether
it is a satellite or rock
Direction is toward the center of the Earth in each case.
Clicker question 2
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An astronaut is floating around in the space shuttle's cabin 200km up.
Her acceleration, as measured from the earth's surface,
A:
B:
C:
D:
zero - she's floating
very small, in some random direction
quite large, nearly g, directed towards
the center of the earth
quite large, nearly g, directed along the line of
travel of the shuttle
Clicker question 2
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An astronaut is floating around in the space shuttle's cabin 200km up.
Her acceleration, as measured from the earth's surface,
A:
B:
C:
D:
zero - she's floating
very small, in some random direction
quite large, nearly g, directed towards
the center of the earth
quite large, nearly g, directed along the line of
travel of the shuttle
She's in orbit, so she's in uniform circular motion and has a large
acceleration! F(grav) is almost the same up there, so her acceleration is
almost the same as on earth in freefall, namely, nearly g, directed
towards the center of the earth. She's just like a ball that's been tossed up
in the air, accelerating straight down.
Gravitational potential energy
Gravitational force:
& potential energy:
Potential energy due to
Earth’s gravity is
RE = radius of Earth = 6380 km
Total energy is potential plus
kinetic and is constant (since
Distance from center of the Earth (km)
gravity is a conservative force)
For total energy < 0, object is bound by the gravitational field
(and orbits are ellipses). Examples are planets around the sun.
For total energy = 0, object is barely unbound (parabolic orbit).
For total energy > 0, object is unbound with a hyperbolic orbit.
Escape velocity
We define escape velocity as the minimum speed needed to
escape a gravitational field (usually from the surface).
Escaping means being able to reach r = ∞. On the previous
slide we found this is possible if we have a total energy ≥ 0.
Total energy of 0 means
, that is
For object m1 on the surface of a planet
with mass MP and radius RP this translates to
Solving for the speed gives
This equation actually works for any radius outside the planets
radius. Just replace RP with the distance from the planet’s center.
Clicker question 3
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Does escape velocity depend on launch angle?
That is, if a projectile is given an initial speed vo, is it
more or less likely to escape an airless, non-rotating
planet, if fired straight up than if fired at an angle?
A. Yes
B. No
Clicker question 3
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Does escape velocity depend on launch angle?
That is, if a projectile is given an initial speed vo, is it
more or less likely to escape an airless, non-rotating
planet, if fired straight up than if fired at an angle?
A. Yes
We derived the escape velocity using conservation of
B. No
energy. All that is needed is for the projectile to have
enough kinetic energy such that the total energy is 0.
Kinetic energy only depends on the magnitude of the velocity
(squared), not the direction so the angle is irrelevant.
On rotating planets the escape velocity is the same but the initial
velocity is not zero so it does make sense to take off at an angle.
History of solar system understanding
1543: Nicholas Copernicus theorizes that the Earth orbits the sun.
1609 & 1619: Johannes Kepler uses 30 years of data collected
by Tycho Brahe to derive 3 laws for planetary motion
1. Planets move in elliptical
orbits with sun at one focus
planet
Sun
2. A line drawn between sun and planet sweeps
out equal areas during equal intervals of time
3. The square of a planet’s orbital period is
proportional to the cube of the semimajor-axis length
These laws can be derived from Newton’s 1687 gravity theory.
Circular orbits from a force perspective
All closed orbits are ellipses but we will analyze the simpler
case of circular orbits. Very good approximation for planets
and moons. Not so good for comets.
A small object m in orbit around a
large object M is called a satellite.
Newton’s 2nd law is
M
The only force is gravity:
The acceleration is only radial:
Therefore:
Solving for v gives
Note that orbital velocity only depends on M/r.
m
Need for Dark Matter
Solving for v gives
It is observed that disk galaxies have a rotation curve which is flat
from centre to edge (line B in illustration), i.e. stars are observed
to revolve around the centre at constant speed over a large range
of distances. It was expected that these galaxies would have a
rotation curve that slopes down from the centre to the edge
(dotted line A in illustration), in the same way as other systems
with most of their mass in the centre, such as the Solar System of
planets or the Jovian System of moons. Clearly, something else is
happening to these galaxies besides a simple application of the
laws of gravity to the observed matter. Hence Dark Matter!
Law of Areas
A line drawn between sun and planet sweeps
out equal areas during equal intervals of time
or dA/dt is constant
1 2
ΔA ≈ r Δθ
2
r
dA 1 2 dθ 1 2
≈ r
= rω
dt 2 dt 2
rΔθ
L = r × p = rp⊥ = r(mv ⊥ ) = r(mωr)
€ 2
L = mr ω
dA
L
=
dt 2m
€
L must be a constant – Kepler’s 2nd law
is just conservation of angular
momentum
Law of Periods: Orbital period
Orbital period T is the time it takes to complete one revolution.
Orbital speed can be determined by distance covered in one
revolution (circumference) divided by the period.
We already know that
So
Can also write as
proving Kepler’s 3rd law (for circular orbits)
m
M
What is the moon’s orbital period?
The moon is a satellite of the Earth and is
at a distance of 384,000 km.
m
We can directly solve for the period
M
Could also calculate speed:
Geosynchronous Orbits
Consider a satellite whose orbital period is exactly 24 hours. Think about
this - the earth rotates once in 24 hours, and in that time the satellite has
also run around the earth once. From our perspective on the ground, that
satellite is at rest above us! This is convenient for TV or satellite
communication - the satellite is always above the same spot. You aim
your dish at it and don’t have to keep tracking the satellite. We call this a
geosynchronous orbit. How high up is such a satellite?
GM E 2
We have R =
T
2
4π
3
Plugging in all the constants, with T=24 hrs, and taking the cube root , I
get R=42,000 km. This is a very high orbit. Since R_earth = 6,000 km, it
is 36,000 km above the ground. (Compare to the shuttle, which is only
about 250 km up)
Geosychronous Orbits cont.
Geosynchronous satellites have v = GM E /RSAT = 3 km/s ,
slower than the shuttle. The farther out you are in orbit, the slower
you go. (Bigger distance, but still bigger T.)
€
Clicker question 4
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Two communications satellites are in orbit at the same height, but one
weighs twice as much as the other. The speed of the heavier satellite is
A: Less than
B: Equal to
D: (need more information)
the speed of the lighter one..
C: Greater than
Clicker question 4
Set frequency to BA
Two communications satellites are in orbit at the same height, but one
weighs twice as much as the other. The speed of the heavier satellite is
A: Less than
B: Equal to
D: (need more information)
C: Greater than
the speed of the lighter one..
Mass cancels out. Speed depends only on radius, for spherical
orbits.
Clicker question 5
The international space station
altitude gradually decreases
due to drag and is periodically
boosted back up. Assuming
perfectly circular orbits, is the
station velocity higher before
or after the boosts.
A. Before
B. After
C. Same
D. Can’t tell
Set frequency to BA
Clicker question 5
The international space station
altitude gradually decreases
due to drag and is periodically
boosted back up. Assuming
perfectly circular orbits, is the
station velocity higher before
or after the boosts.
A. Before
B. After
C. Same
D. Can’t tell
Set frequency to BA
for circular orbits.
Smaller r results in larger v.