orbits and applications - university of colorado boulderjcumalat/phys1110/lectures/lec33.pdf ·...

Orbits and Applications

Finish Chapter 13 today,

Will cover part of Chapter 14 on Fluids on Friday

Review on Monday for Exam Next Thursday

Third Midterm will cover material from Chap. 10-13

Newton’s Law of Gravitation can be written as

Newton’s Law of Gravity

Key Fact: A uniform shell of matter exerts no net gravitational force on a particle located inside it.

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.

Newton’s Shell Theorem

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

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

Sun

planet

Johannes Kepler’s Planetary Laws

Kepler’s 3rd Law Kepler's 3rd law: T2/R3=constant for all planets.

The closer a planet is to the Sun, the faster it will move. Any spacecraft going around the Sun in an orbit smaller than Earth's will also soon overtake and move away, and will not keep a fixed station relative to Earth.

However, if a spacecraft is placed between Sun and Earth, Earth's gravity pulls it in the opposite direction and cancels some of the pull of the Sun. With a weaker pull towards the Sun, the spacecraft then needs less speed to maintain its orbit.

This is called a Lagrange Point! There are 5 such points.

Lagrange Points L1 used to monitor the sun.

L2 is planned location for the James Webb telescope.

L3 could be used to monitor sun spots and provide up 7 day advance notice.

L4 and L5 locations very stable and possible locations for future space stations

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 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

Clicker question 1 Set frequency to BA

The Moon does not fall to Earth because

A: It generates a gravitational force equal and opposite the earth's pull. B: The net force on it is zero. C: It is beyond the main pull of Earth’s gravity. D: It is being pulled by the Sun and planets as well as by Earth. E: none of the above


The Moon does not fall to Earth because

A: It generates a gravitational force equal and opposite the earth's pull. B: The net force on it is zero. C: It is beyond the main pull of Earth’s gravity. D: It is being pulled by the Sun and planets as well as by Earth. E: none of the above

The moon falls to earth all the time! There is a large force, F = GM(earth)M(moon)/R(earth to moon)^2 which pulls it towards us. Acceleration towards the center, that's what you mean by "falling"!


A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P). The work done by the force of gravity during this movement is A: 0 B: + C: -


A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P). The work done by the force of gravity during this movement is A: 0 B: + C: -

The force of gravity from the Sun is centripetal (toward the center). The force has a component along the direction of displacement.

€

W = F •Δ r = F Δr cosθ > 0


The planet executes one complete orbit starting from point A and returning to A. The work done by the force of gravity during this orbit is: A: 0 B: + C: -


The planet executes one complete orbit starting from point A and returning to A. The work done by the force of gravity during this orbit is: A: 0 B: + C: -

Zero. By the work energy theorem, or just conservation of energy.


Kepler's 3rd law: T2/R3=constant for all planets. The period T for the earth is 1 year (!) The distance of the earth to the sun is called 1 A.U. Suppose an asteroid is in circular orbit around the sun at a distance of 2 A.U. How long does it take the asteroid to orbit the sun once?

A 2 years B: 3 years C: 2^(3/2) = 2.83 years D: 2^(2/3) = 1.59 years E: Not enough information (An asteroid is not a planet - can't figure it out from Kepler's third law.)


Kepler's 3rd law: T2/R3=constant for all planets. The period T for the earth is 1 year (!) The distance of the earth to the sun is called 1 A.U. Suppose an asteroid is in circular orbit around the sun at a distance of 2 A.U. How long does it take the asteroid to orbit the sun once? A 2 years B: 3 years C: 2^(3/2) = 2.83 years D: 2^(2/3) = 1.59 years E: Not enough information (An asteroid is not a planet - can't figure it out from Kepler's third law.)

T (period) goes like R^(3/2), so if you're 2 times as far, your year is 2^3/2 times longer.

Orbits from an energy perspective

M m

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

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

Remember, potential energy is

So kinetic energy is

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

Gravitational potential energy

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.

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)

Orbital modifications Orbital mechanics is a computationally challenging field.

Shown is the Hohmann transfer orbit, one of the most fuel efficient ways to change orbits.


A rock, initially at rest with respect to Earth and located far away is released and accelerates toward Earth. An observation tower is built 3 Earth-radii high to observe the rock as it plummets to Earth. Neglecting friction, the rock’s speed when it hits the ground is A: twice B: three times C: four times D: eight times E: sixteen times its speed at the top of the tower.


A rock, initially at rest with respect to Earth and located far away is released and accelerates toward Earth. An observation tower is built 3 Earth-radii high to observe the rock as it plummets to Earth. Neglecting friction, the rock’s speed when it hits the ground is A: twice B: three times C: four times D: eight times E: sixteen times its speed at the top of the tower.

Cons. of E with v=0, R=infinity, then 1/2 mv2 - G Mem/R = const. = 0

At “4R” we have 1/2 mv2 = GMm/(4R). At "R" it has1/2 mv2 = GMm/R, i.e. 4x BIGGER KE. So 2x BIGGER v.


An object moves along the x-axis . The potential energy U(x) vs. position x is shown in the graph.

When the object is at x=A, which of the statements must be true? A: The velocity is negative, i.e. along the -x direction. B: The acceleration is negative. C: The total energy is negative. D: The kinetic energy is positive. E: None of these statements is always true.

When the object is at x=A, which of the statements must be true?


An object moves along the x-axis . The potential energy U(x) vs. position x is shown in the graph.

When the object is at x=A, which of the statements must be true? A: The velocity is negative, i.e. along the -x direction. B: The acceleration is negative. C: The total energy is negative. D: The kinetic energy is positive. E: None of these statements is always true.

When the object is at x=A, which of the statements must be true?

F = -dU(x)/dx, so the FORCE at point A is negative. (Therefore the acceleration is negative). We know nothing about the velocity. The total energy could be anything, really. K could be zero.


A planet is in elliptical orbit around the Sun. The zero of potential energy U is chosen at r = ∞, so .

How does the magnitude of U, i.e. |U| compare to the KE? A: |U| > KE B: |U| < KE C: |U| =KE D: depends on the position in the orbit.


A planet is in elliptical orbit around the Sun. The zero of potential energy U is chosen at r = ∞, so .

How does the magnitude of U, i.e. |U| compare to the KE? A: |U| > KE B: |U| < KE C: |U| =KE D: depends on the position in the orbit.

The earth is bound, so K + U < 0. So K < -U; -U= + G M m /r. Thus, |U| > K. A diagram may help:

r

PE

Etot PE KE


A satellite is in circular orbit around a planet that has a very tenuous atmosphere extending up to the altitude of the satellite. Due to atmospheric drag, the speed of the satellite… A: increases B: decreases C: remains constant


A satellite is in circular orbit around a planet that has a very tenuous atmosphere extending up to the altitude of the satellite. Due to atmospheric drag, the speed of the satellite… A: increases B: decreases C: remains constant

Drag makes the satellite's radius get smaller; since v = Sqrt[ G M/r], the speed will INCREASE. K+U = total energy, so K - GM m/R is the total energy.

orbits and applications - university of colorado boulderjcumalat/phys1110/lectures/lec33.pdf ·...

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