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

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F =GmobjMins

r2

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Mins = ρ4πr3

3

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F =

4πGmobjρ

3 r

The r vector points out from the center, so we have

and object would just oscillate back and forth.

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F = −k r

Mins

mobj

A. True B. False

Clicker question1 Set frequency to BA A satellite in circular orbit about the Earth is directly over Denver, 300 miles above the city, and traveling eastward at 16,000 mph. At the same time, a rock is released from rest 300 miles above the city, right next to the satellite.

Denver

Satellite Rock

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 Set frequency to BA A satellite in circular orbit about the Earth is directly over Denver, 300 miles above the city, and traveling eastward at 16,000 mph. At the same time, a rock is released from rest 300 miles above the city, right next to the satellite.

Denver

Satellite Rock

True or false: The acceleration of satellite and rock at this point in time is the same in both magnitude and direction.

If the satellite and rock are a distance r from the center 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 Set frequency to BA

An astronaut is floating around in the space shuttle's cabin 200km up. Her acceleration, as measured from the earth's surface,

A: zero - she's floating B: very small, in some random direction C: quite large, nearly g, directed towards

the center of the earth D: quite large, nearly g, directed along the line of

travel of the shuttle

Clicker question 2 Set frequency to BA

An astronaut is floating around in the space shuttle's cabin 200km up. Her acceleration, as measured from the earth's surface, A: zero - she's floating B: very small, in some random direction C: quite large, nearly g, directed towards

the center of the earth D: 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.

For total energy < 0, object is bound by the gravitational field (and orbits are ellipses). Examples are planets around the sun.

Gravitational potential energy

Distance from center of the Earth (km)

RE = radius of Earth = 6380 km

For total energy > 0, object is unbound with a hyperbolic orbit.

For total energy = 0, object is barely unbound (parabolic orbit).

Gravitational force:

& potential energy: Potential energy due to Earth’s gravity is Total energy is potential plus kinetic and is constant (since gravity is a conservative force)

Escaping means being able to reach r = ∞. On the previous slide we found this is possible if we have a total energy ≥ 0.

Escape velocity We define escape velocity as the minimum speed needed to escape a gravitational field (usually from the surface).

This equation actually works for any radius outside the planets radius. Just replace RP with the distance from the planet’s center.

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

A. Yes B. No

Clicker question 3 Set frequency to BA

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 Set frequency to BA

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?

We derived the escape velocity using conservation of 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.

2. A line drawn between sun and planet sweeps out equal areas during equal intervals of time

History of solar system understanding

1609 & 1619: Johannes Kepler uses 30 years of data collected by Tycho Brahe to derive 3 laws for planetary motion

1543: Nicholas Copernicus theorizes that the Earth orbits the sun.

1. Planets move in elliptical orbits with sun at one focus

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.

Sun

planet

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.

The only force is gravity: Newton’s 2nd law is

The acceleration is only radial:

M m

Therefore: Solving for v gives

Note that orbital velocity only depends on M/r.

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

€

ΔA ≈ 12r2Δθ

€

dAdt

≈12r2 dθdt

=12r2ω

€

L = r × p = rp⊥ = r(mv⊥ ) = r(mωr)

€

L = mr2ω

€

dAdt

=L2m

L must be a constant – Kepler’s 2nd law is just conservation of angular momentum

€

rΔθ

r

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.

M m

proving Kepler’s 3rd law (for circular orbits)

We already know that

So

Can also write as

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 m We can directly solve for the period

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?

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) €

We have R3 =GME

4π 2 T2

Geosychronous Orbits cont.

Geosynchronous satellites have , slower than the shuttle. The farther out you are in orbit, the slower you go. (Bigger distance, but still bigger T.)

€

v = GME /RSAT = 3 km/s

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 C: Greater than

D: (need more information)

the speed of the lighter one..

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 C: Greater than

D: (need more information)

the speed of the lighter one..

Mass cancels out. Speed depends only on radius, for spherical orbits.

A. Before B. After C. Same D. Can’t tell

Clicker question 5 Set frequency to BA

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

Clicker question 5 Set frequency to BA

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.

for circular orbits.

Smaller r results in larger v.