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•  Midterm #2 on Thursday at 7:30pm in same rooms •  Midterm covers Chapters 5-8 (up through Monday’s

lecture). Includes CAPA due Tuesday and the tutorial homework due this week.

•  Old midterm exams are posted in D2L. •  Files with clicker questions are posted in D2L.

Universal gravitation

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Universal Gravitation: Potential energy Gravity is a conservative force and therefore has a potential energy associated with it

Can use conservation of energy to determine things like velocity at a given height or the escape velocity.

UG = −Gm1m2

r

Total energy E = K +UG =12 m1v1

2 + 12 m2v2

2 −Gm1m2

r is conserved

Often one mass is much more massive and is basically motionless and so the total energy is E = K +UG =

12 mv

2 −GMmr

Negative total energy is a bound system (closed orbits or objects eventually colliding). Total energy ≥ 0 is an unbound system.

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The gravitational potential energy of a two mass system, where m << M is plotted. Mass M is stationary at the origin and mass m is at r=r0. Also shown is the total energy Etot. Which arrow represents the kinetic energy of mass m? A.  A B.  B C.  C D.  None of these

Clicker question 1 Set frequency to BA

The total energy E is equal to the sum of the potential and kinetic energies: E = K + U so K = E – U.

r

Etot

r0

A

B C

UG = −GMmr

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Escaping means being able to reach r = ∞. 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 UG +K = 0 K = −UG = +Gm1m2

r

For object m1 on the surface of a planet with mass MP and radius RP this translates to P

P1212

1RMGmvm =

Solving for the speed gives P

Pescape

2RGMv =

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A. Yes B. No

Clicker question 2 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 the angle matters.

On planets with air, take offs are more straight up to reduce the energy loss due to air resistance.

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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

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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: 2rGMmFG =

Newton’s 2nd law is !Fnet =m

!a

The acceleration is only radial: rva2

=

M m

v!

!FG

r

Therefore: rvmr

GMm 2

2 = Solving for v gives vorbit =GMr

Note that orbital velocity only depends on M/r.

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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

v!

!FG

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

r

vorbit =2πrT

vorbit =GMr

We already know that

So T =2πrvorbit

= 2πr rGM

=2πr3/2

GM

Can also write as 322 4 rGMT π=

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Orbits from an energy perspective

M m

v!

!FG

Always before, kinetic and potential energy were independent. For circular orbits, this is no longer true.

r

For a circular orbit we know so we can calculate the kinetic energy

vorbit =GMr

K =12mvorbit

2 =12mGM

r=GMm2r

UG = −GMmrRemember, potential energy is

So kinetic energy is K = − 12UG

Furthermore, the total energy is E = K +UG = −12UG +UG =

12UG

Since Ug < 0, total energy E < 0 which is the requirement for a closed orbit (which of course a circular orbit is).

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Gravitational potential energy

UGE

K

Work is needed to change the energy. Since work is force times displacement, rockets are generally fired in (or against) the direction of motion to change orbits rather than perpendicular.

EΔ

1r 2r

Graph of potential energy, kinetic energy, and total energy for circular orbits.

To get from a low orbit r1 to a higher orbit r2 requires an increase in energy.

While the kinetic energy decreases, potential energy increases (twice as much)

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A.  B.  C.  D.  E. 

Clicker question 3 Set frequency to BA Two satellites are in stable circular orbits. Satellite 2 has a speed of v2 and is twice as far from the center of the Earth as satellite 1, which has a speed of v1. What is the ratio of their speeds?

vorbit =GMr

P

Pescape

2RGMv =

322 4 rGMT π=

UG = −GMmr

Ecircular orbit = Korbit +UG =12UG

Korbit =GMm2r

v1 / v2 =1/ 2v1 / v2 =1/ 2v1 / v2 =1v1 / v2 = 2v1 / v2 = 2

v1 =GMr1

v2 =GMr2

v1v2=

GMr1

r2GM

=r2r1= 2and so