Knock, knock, knockin' on heaven's door Bob Dylan, Knocking on Heaven's Door

Assignment: Sign up for a time slot for Observing Lab #2. Study for Monday's Exam! Here's an image of last year's test cover page:

In Class: Question to Ponder In a desperate attempt to elude the evil Zorg Fleet, Spaceman Spiff deftly execuates a tight left hand turn. During his turn, his craft briefly follows a circular path whose radius is 10 m. If his speed is 30 m/s through the turn, what is the acceleration he (and his spacecraft) feels? a) zero b) 30 m/s/s c) 60 m/s/s d) 90 m/s/s e) 120 m/s/s Newton Tests the Strength of Gravity Over Distance He knows that Kepler's Third Law will tell him the semi-major axes of the planets' orbits if he know the periods (He knows the periods). He calculated the speed of each planet in its orbit (We assumed the orbits were circular 'cuz it's easier and pretty close to correct, but Newton did it for the real elliptical orbits, macho guy that he is). He can calculate the acceleration the planets feel as they orbit the Sun, using a = v2/r. So let's do the math he did: Speed of a planet = distance/time = v = 2 pi r/ P So v2 = 4 pi2 r2/ P2 But Kepler's Third Law says: P2 = r3 ("r" here 'cuz we assume the orbits are circles) So v2 = 4 pi2 r2/ r3 = 4 pi2/r The acceleration the planet must feel is a = v2/r, so a = 4 pi2/r2 If the acceleration that the planets feel is due to the Sun's pull, then Newton finds that the gravitational acceleration caused by the Sun decreases with distance from the Sun as one over the distance squared. Called an "inverse-square law." Danger! Danger! -- Note that this representation of the acceleration works only for objects that orbit our Sun, and even than, it only holds for orbital radii expressesed in A.U., and even then produces an acceleration in units of A.U./yr2. Avoid using this expression, and instead make use of the more general expression below. But Will This Work for Earth Gravity, Too? Newton reasons that he can test whether the gravity responsible for keeping you and me stuck to the Earth is the same gravity that's keeping the Moon in orbit. The Earth's surface is 6.38 x 106 m from its center. The Moon is 3.84 x 108 m from the center of the Earth. If the inverse square law holds for Earth gravity, too, then the acceleration felt by the Moon must be weaker than that felt here at the surface of the Earth by a factor of (6.38 x 106 m / 3.84 x 108 m)2 = (1/60)2 = 1/3600. Well 1/3600th of 9.8 m/s2 is 2.7 x 10-3 m/s2 --- just what Newton measured for the acceleration that the Moon feels in its circular orbit! Therefore the inverse square law works for the Sun's influence on the planets and the Earth's influence on the Moon. It's a universal law of gravitation -- one that applies in both the celestial and terrestrial domains. Mass as the Source of Gravity Newton further postulates that mass is the source of gravity. The larger the mass, the stronger the gravity (all else being equal). His universal gravitation law now states: acceleration due to gravity = a = GM/r2 where G is Newton's gravitational constant = 6.67 x 10-11 m3 / s2 kg, and M is the mass of the object creating the gravity. Measuring the Mass of an Object Dynamically Using Newton's gravity law and what we know about circular orbits, we can measure the mass of planets or other objects. First, put something in orbit around the planet (e.g., a moon, or an artificial satellite). Measure the radius of the satellite's orbit and its speed. Use a = v2/r to determine the acceleration it must feel. We know that this acceleration must also be equal to the acceleration caused by the gravity of the planet. Therefore, a = v2/r = GM/r2 and M = v2 r/G.