Physics 151: Lecture 28 Today’s Agenda

Physics 151: Lecture 28 Today’s Agenda • Today’s Topics • Gravity and planetary motion (Chapter 13)

See text: 13.1 Gravitationaccording to Sir Isaac Newton • Newton found that amoon/ g= .000278 • and noticed that RE2 / R2= .000273 • This inspired him to propose the Universal Law of Gravitation: |FMm|= GMm / R2 amoon g R RE G = 6.67 x 10 -11 m3 kg-1 s-2

See text: 13.1 Gravity... • The magnitude of the gravitational force F12 exerted on an object having mass m1 by another object having mass m2 a distance R12 away is: • The direction of F12 is attractive, and lies along the line connecting the centers of the masses. m1 m2 F12 F21 R12

Example • What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth ? • a. 4.8 m/s2 • b. 1.1 m/s2 • c. 3.3 m/s2 • d. 2.5 m/s2 • e. 6.5 m/s2

See text: 13.4 Kepler’s Laws 1st All planets move in elliptical orbits with the sun at one focal point. 2nd The radius vector drawn from the sun to a planet sweeps out equal areas in equal times. 3rd The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit. • It was later shown that all three of these laws are a result of Newton’s laws of gravity and motion.

Example • Which of the following quantities is conserved for a planet orbiting a star in a circular orbit? Only the planet itself is to be taken as the system; the star is not included. • a. Momentum and energy. • b. Energy and angular momentum. • c. Momentum and angular momentum. • d. Momentum, angular momentum and energy. • e. None of the above.

Example • The figure below shows a planet traveling in a clockwise direction on an elliptical path around a star located at one focus of the ellipse. When the planet is at point A, • a. its speed is constant. • b. its speed is increasing. • c. its speed is decreasing. • d. its speed is a maximum. • e. its speed is a maximum. Animation

Example • A satellite is in a circular orbit about the Earth at an altitude at which air resistance is negligible. Which of the following statements is true? • a. There is only one force acting on the satellite. • b. There are two forces acting on the satellite, and their resultant is zero. • c. There are two forces acting on the satellite, and their resultant is not zero. • d. There are three forces acting on the satellite. • e. None of the preceding statements are correct.

Example • A satellite is placed in a geosynchronous orbit. In this equatorial orbit with a period of 24 hours, the satellite hovers over one point on the equator. Which statement is true for a satellite in such an orbit ? a. There is no gravitational force on the satellite. b. There is no acceleration toward the center of the Earth. c. The satellite is in a state of free fall toward the Earth. d. There is a tangential force that helps the satellite keep up with the rotation of the Earth. e. The force toward the center of the Earth is balanced by a force away from the center of the Earth.

Example • Normally, if I throw a ball up in the air it will eventually come back down and hit the ground. • What if I throw it REALLY hard so that I put it into an orbit ! • How HARD do I have to throw ?

U RE r 0 See text: 13.7 Energy of Planetary Motion A planet, or a satellite, in orbit has some energy associated with that motion. Let’s consider the potential energy due to gravity in general. Define ri as infinity

We can solve for v using Newton’s Laws, Plugging in and solving, See text: 13.7 Energy of a Satellite A planet, or a satellite, also has kinetic energy.

See text: 13.7 Energy of a Satellite So, an orbiting satellite always has negative total energy. A satellite with more energy goes higher, so r gets larger, and E gets larger (less negative). It’s interesting to go back to the solution for v. v is smaller for higher orbits (most of the energy goes into potential energy).

Lecture 28, Act 2Satellite Energies • A satellite is in orbit about the earth a distance of 0.5R above the earth’s surface. To change orbit it fires its booster rockets to double its height above the Earth’s surface. By what factor did its total energy change ? (a)1/2(b)3/4(c)4/3 (d)3/2 (e)2

Lecture 28, Act 2Satellite Energies Note : E2/E1 = 3/4 actually means that the energy is larger (because it is negative) (b)

Escape Velocity • Normally, if I throw a ball up in the air it will eventually come back down and hit the ground. • What if I throw it REALLY hard ? • Two other options • I put it into orbit. • I throw it and it just moves away forever – i.e. moves away to infinity

Orbiting • How fast to make the ball orbit. • I throw the ball horizontal to the ground. • We had an expression for v above,

Escape Velocity • What if I want to make the ball just go away from the earth and never come back ? • (This is something like sending a space ship out into space.) • We want to get to infinity, but don’t need any velocity when we get there. • This means ETOT = 0 Why ??

Example • A projectile is launched from the surface of a planet (mass = M, radius = R). What minimum launch speed is required if the projectile is to rise to a height of 2R above the surface of the planet? Disregard any dissipative effects of the atmosphere.

Example • A satellite circles planet Roton every 2.8 h in an orbit having a radius of 1.2x107 m. If the radius of Roton is 5.0x106 m, what is the magnitude of the free-fall acceleration on the surface of Roton? • a. 31 m/s2 • b. 27 m/s2 • c. 34 m/s2 • d. 40 m/s2 • e. 19 m/s2

Example • A spacecraft (mass = m) orbits a planet (mass = M) in a circular orbit (radius = R). What is the minimum energy required to send this spacecraft to a distant point in space where the gravitational force on the spacecraft by the planet is negligible? a. GmM/(4R) b. GmM/R c. GmM/(2R) d. GmM/(3R) e. 2GmM/(5R)

Recap of today’s lecture • Today’s Topics • Gravity and planetary motion (Chapter 13)

Physics 151: Lecture 28 Today’s Agenda