Last Time: Centripetal Acceleration, Newtonian Gravitation Today: Two practice multiple choice problems Gravitational Potential Energy (a different look) Kepler’s Laws of Planetary Motion HW #6 due tonight 11:59 p.m.

Exam #2 Details : Thursday, October 21, regular class time (75 minutes) Closed-book, closed-note You CAN AND SHOULD bring a calculator Exam #2 Formula Sheet posted on Blackboard Will be provided (not allowed to bring your own copy) Today’s new material NOT on the exam Multiple Choice: 5  5 points = 25 points Short Answer(s): 10 points Problems: 3 problems, 65 points Bonus Problem: 5 points Possible Total: 105/100

Typo Last Time G = 6.673  10−11 kg−1 m3 s−1 should be … G = 6.673  10−11 kg−1 m3 s−2

Mall of America (Minneapolis, MN) Conceptual Question #1 (practice for exam) A merry-go-round rotates with constant angular speed. As a rider moves from the edge towards the center, what happens to the magnitude of the centripetal force that must be exerted on her ? Mall of America (Minneapolis, MN) (a) It increases. (b) It is not zero, but remains the same. (c) It decreases. (d) It is always zero, because its angular speed is constant. (e) It increases or decreases, depending on the direction of rotation (CCW or CW).

Conceptual Question #2 (practice for exam) The gravitational force exerted on an astronaut on Earth’s surface before launch is 650 N downwards. On the International Space Station, is the gravitational force on the astronaut … (a) Larger (b) Exactly the same (c) Smaller (d) Zero

Gravitational Potential Energy So far, we have calculated the potential energy of an object above the Earth’s surface as : m g h PE = m  g  h force distance But, we learned that the acceleration due to gravity decreases the farther we move away from the surface of the Earth. high above the Earth ! For objects high above the Earth, the gravitational potential energy associated with a mass m at a distance r from the center of the Earth is ME : Earth’s mass [ SI: Joules ]

Negative Potential Energy ? PE = 0 at r   PE is a MAXIMUM at r   For r < , PE < 0, becomes more negative as move towards Earth Consider an object with a mass of 1.0-kg falling towards Earth, which is initially 2  106 m above the surface of Earth at rest. What is its speed when it is 1  106 m above the Earth ? 106 m 106 m RE How much work was done by gravity ? ME

What About PE = mgh ? Is our “new” formula for gravitational PE consistent with PE = mgh on the surface of the Earth ? #2 h #1 Earth’s surface

What Must the Speed of a Rocket be to Escape Earth’s Gravity ? The “escape speed” of a rocket is the speed with which a rocket needs to be launched, in order to escape Earth’s gravity, and reach . At  : At launch: Energy Conservation : 11.2 km/s on Earth “barely” makes it to  [25054 mph]

Example A space capsule with mass 1000 kg is launched from the surface of Earth at the escape speed vesc. What will its speed be when it is 103 km above the surface of the Earth ?

Kepler’s Laws of Planetary Motion German mathematician and astronomer, 1571 – 1630 Work provided the grounds for Newton’s theory of gravitation Before Kepler’s Laws … “Geocentric Model”: Earth is center of universe Greek astronomer Ptolemy, 2nd century A.D. – 1400s “Heliocentric Model”: Earth/Planets orbit Sun in circular orbits Copernicus, 1473 – 1543 Kepler analyzed data for 16 years, and deduced a mathematical model for the motion of planets about the Sun

Kepler’s Laws of Planetary Motion All planets move in elliptical orbits about the Sun (not circular orbits), with the Sun at one of the focal points. #1 Ellipse is a curve drawn so that the sum of the distances from any point on the curve, to two internal points called the focal points, is constant. p + q = constant [for any point on the ellipse] Because orbit is an ellipse, the Planet’s distance from the Sun is constantly changing !

Kepler’s Laws of Planetary Motion A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals. #2 According to this law, the if the areas above are equal, the planet will take the same amount of time to move from AB as it does to move from CD. Planets move faster the closer they are to the Sun ! [Also follows from conservation of angular momentum, later …]

Kepler’s Laws of Planetary Motion For circular orbits, the square of the orbital period of any Planet (or a satellite) is proportional to the cube of the average distance from the Planet to the Sun. #3 MS: mass of the Sun = 1.991  1030 kg OR the object about which it is orbiting r : radius of circular orbit Period T = time required for one revolution = 2πr/v For elliptical orbits, r must be replaced with a, the semimajor axis (half the longest distance).

Example A satellite moves in a circular orbit around the Earth at a speed of 5000 m/s. Determine: (a) The satellite’s altitude above the surface of the Earth. (b) The period of the satellite’s orbit.

“Geosynchronous Orbits” A satellite in a geosynchronous orbit remains at the same location relative to a location on the Earth. These are used in telecommunications. This can only occur if the satellite’s orbital period is the same as the Earth’s period of rotation, 24.0 hours. (a) At what distance from the center of the Earth must the satellite be? (b) What is the satellite’s orbital speed?

Next Lecture Class 8.1 – 8.2 : Torque and Equilibrium