CAPSTONE Lecture 3 Gravity, Energy and their uses /07/2010CASTONE Lecture 3.Cavendish 1

07/07/2010CASTONE Lecture 3.Cavendish 2 Gravitational Energy, The Cavendish Experiment The Virial Theorem Computation

07/07/2010CASTONE Lecture 3.Cavendish 3 Big G and little g Newton’s force law, F=-GmM/r 2, reduce to F=mg, where g=GM/r 2, M is the mass of the Earth and r is the radius of the Earth (onto which a body is falling). Galileo (~1580 – 1642) derived the last equation for bodies falling on the Earth, and found g=980 cm/sec 2. Newton’s law is a generalization of Galileo’s result, for bodies of any M and r. G and M always appear together. To measure M, one must know G.

07/07/2010CASTONE Lecture 3.Cavendish 4 Newton had to show, mathematically, that for perfect, uniform sphere, one can treat the entire mass of the Earth as if it is a point source at the center of the Earth. (For objects like the Earth, this rule is a very good approximation, even though the Earth is not perfectly homogeneous.) He had to assume that the nature of the material of Earth did not affect the value of G or the (1/r 2 ) form of the force law. (No one has ever been able to show this assumption to be wrong.

07/07/2010CASTONE Lecture 3.Cavendish 5 The modern version of the Cavendish experiment is described here and we will do it in class. You will do the experiment yourself later in the term. a)Suspend, with a torsion ribbon, a dumbbell with two small weights of known mass, separated by a few centimeters in a plastic case (to isolate the balls from air currents. b) On a movable arm, outside the case, place to large balls, at the same separation as the small balls from each other. Place the large balls as close as possible to the small ones and let the system equilibrate (it takes all night for it to reach equilibrium.) THE CAVENDISH EXPERIMENT

07/07/2010CASTONE Lecture 3.Cavendish 6 c) Rotate the movable arm so the big balls are as close as possible to the opposite small balls. d) A force has been removed from one side of each ball and added to the other side of each small ball. The net force change is 2GmM/r 2, where r is the separation of the large and small balls. The small balls will move under this force. A mirror attached to the torsion string reflects a laser beam onto the wall. The acceleration of the laser spot can be measured and the force thus determined. Since m, M and r are known, G can be detected. G=6.67 x cm 3 /g-sec 2.

07/07/2010CASTONE Lecture 3.Cavendish 7 Since the balls move on a plane parallel to the surface of the Earth, the effect of the Earth’s mass is the same on both balls and perpendicular to the force of the large mass balls. The Earth has been “removed” from the experiment.

07/07/2010CASTONE Lecture 3.Cavendish 8 Kinetic and Potential Energy K=(1/2)mv 2, defined as kinetic energy of motion U=-GMm/r, the potential energy of a system of two interacting gravitational forces. Consider two bodies in an otherwise empty Universe, very far apart. Then, K=0 (nothing is making them move fast) and U=0 (r is very large). Energy is conserved, so Total energy at the start of the fall of the two bodies to each other is K+U=0, as for the point when they are closest together.

07/07/2010CASTONE Lecture 3.Cavendish 9 (1/2)m(v end ) 2 -GmM/r end = 0. (1/2)m(v end ) 2 = GmM/r end Solve for v at the end of the fall, v=(2GM/r (closest approach)) 1/2 If two bodies are close to each other to start, and we want to remove them to a large distance (infinity) from each other, we must make them both move just as fast as they were moving when they fell from infinity. Thus, v(free-fall)=- v(escape) Free fall velocity and escape velocity

07/07/2010CASTONE Lecture 3.Cavendish 10 Virial theorem F= - GmM/r 2 = - mv 2 /r (1/2)(-GmM/r 2 )=(1/2) (-mv 2 /r) (1/r)(1/2)(GmM/r)=(1/r)(1/2)mv 2 -(1/2)(-GmM/r)=(1/2)mv 2 -U/2=K K=-U/2 This theorem applies for objects in stable, bound orbits.

07/07/2010CASTONE Lecture 3.Cavendish 11 Units Here are the relations of the above quantities in terms of units. Use [x] to mean “the units of x.” [m] = grams, or g [x] = cm (centimeters) [v] = cm/sec [a] = cm/sec 2 [F] = [ma] = (g-cm)/sec 2 [K] = [mv 2 ] =( g-cm 2 )/sec 2 = ergs [U] = [GmM/r] = [G] g 2 /cm; since U is energy, it has the same units as K. Then, [G] =cm 3 /(g-sec 2 )

07/07/2010CASTONE Lecture 3.Cavendish 12 Dimensional Analysis Two sample problems, to demonstrate the use of dimensional analysis: 1. A ball accelerates at 980 cm/sec 2 and has mass 100 grams. What force is acting on the ball? F=ma=100 g x 980 cm/sec 2 = 9.8 x 10 4 g- cm/sec 2.

07/07/2010CASTONE Lecture 3.Cavendish A planet orbits at 1 AU from the Sun. How fast is it moving? M is the mass of the Sun and r is the separation of the planet and the Sun. v = (GM/r) 1/2 = [{6.67 x cm 3 /(g-sec 2 )} x {2 x g}/(1.5 x cm)] 1/2 v=[ {(6.67 x 2)/1.5} x {(10 -8 x )/10 13 } x {gxcm 3 /(g-sec 2 )}/ cm}] 1/2 v = [9 x cm 2 /sec 2 ] 1/2 v = 3 x 10 6 cm/sec = 30 km/sec (1 km/sec = 10 5 cm)