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Transcript
Physics 231 Lecture 22 • • Main points of today’s lecture: Newton’s law of universal gravitation: GMm F= • • Kepler’s laws and the relation between the orbital period and orbital radius. 4π 2 2 r 3 T = GM Gravitation potential energy PE grav = − • r2 GM1M 2 r12 Tensile stress and strain ΔF ΔL =Y A L0 • Bulk stress and strain: ΔF ΔV = ΔP = B A V A uniform10 kg beam is support at one end by a hinge and at the other by a cable. The cable from the wall is attached to the beam at an angle of 30°. What is the tension in the cable? a) 9.8 N b) 19.6 N c) 29.4 N d) 39.2 N d) 49 N e) 58.8 N f) 68.6 N e) 98 N Fhinge T 30o L T Wbar 1. Draw the forces. 2. Choose the axis for calculating the torques so as to avoid calculating torques you don’ t want to know. Don’t want to know Fhinge, so pput axis at hinge. g 3. Set the net torque about this axis to zero and solve for the tension. b g L / 2 τ i = 0 = TL sin ( 30°) − m bar i TL sin ( 30°) = m bar g L / 2 T sin ( 30°) = m bar g / 2 1 T = m bar g / 2 T = m bar g T = 98N 2 To get Fhinge , use Fi = 0. i Newton’s law of universal gravitation • All objects (even light photons) feel a gravitational force attracting them to other objects. This force is proportional to the two masses and inversely proportional to the square of the distance between them. m1m2 F12 = G 2 r12 G=6 6.673 673 x 10-11 N m² m /kg /kg² Example • A spaceship hi iis on a journey j to t the th moon. The Th masses off the th earth th andd moon are, respectively, 5.98x1024 kg and 7.36x1022 kg. The distance between the centers of the earth and the moon is 3.85x106m. At what point as measured from the center of the earth, point, earth does the gravitational force exerted on the craft by the earth balance the gravitational force exerted by the moon? This point lies on a line between the centers of GM moon mship the earth and the moon. GM earth m ship Fmoon = 2 = F = earth rmoon _ ship Moon r2 earth _ ship GM moon m ship rmoon _ ship 2 moon _ ship r M moon 2 moon _ ship r rearth _ ship rearth _ ship rmoon _ ship = = = GM earth m ship 2 rearth _ ship M earth 2 earth _ ship r M earth M moon 2 rearth _ ship 2 rmoon _ ship 5.98x1024 kg = = 9.01 22 7.36x10 kg rearth _ ship + rmoon _ ship = 9.01rmoon _ ship + rmoon _ ship = 10.01r Earth rmoon _ ship = 3.85x105 m M earth = M moon rearth _ ship = 3.85x106 m = 3.47x106 m moon _ ship Example • Mp George weighs G i h 100 N on the th Earth. E th What Wh t would ld his hi weight i ht be b on the th surface of another planet that has a planetary mass, which is 3 times the mass of Earth and mass and a planetary radius, which is 2 times the radius of Earth? GM Earth m George WEarth = – a) 75 N R 2Earth – b) 100 N G ( 3M earth ) m George GM planet m George = Wplanet = – c) 125 N R 2planet ( 2R earth )2 – d) 150 N 3GM earth m George 3 GM earth m George = = 2 2 4R earth 4 R earth 3ME Rp 2Re WE 100 N Wp ? Wplanet = Hint: This is a ratio problem. 3 3 Wearth = 100N = 75N 4 4 Conceptual question • An astronaut A t t is i on a spacecraft ft floating fl ti ““weightlessly” i htl l ” in i orbit. bit While her feet are attached securely to the spacecrafte, she shakes a large iron anvil rapidly back and forth. She reports back to Earth that – a) the shaking costs her no effort because the anvil has no inertial mass in space. – b) th the shaking h ki costs t h her some effort ff t b butt considerably id bl lless than on Earth. – c) although weightless, the inertial mass of the anvil is the same as on Earth. E th 3 Kepler’s laws • Johannes Kepler proposed three laws of planetary motion: 1. All planets moved in elliptical orbits with the Sun at one of the focal points. 2 Planetary 2. l orbits bi sweep out equall areas in i equall times. (This is a consequence of angular momentum conservation.) area = 1 1 Δθ 1 Δt ≈ r 2ωΔt r ⋅ rΔθ = r 2 2 2 Δt 2 2a Δθ r r 2ω = const . 3 Th 3. The square off the th orbital bit l period i d off any planet l t is i proportional ti l to t the th cube b of the average distance from the planet to the Sun. For a circular orbit: GM star m / planet m / planet v 2 GM star v 2 2 Fgrav = = m a = = = ω planet c 3 r2 r r2 2r GM star 2π 2 2π 3 4 π = 2 ωT = 2π ω = 3 r T = r T T GM star G 2 2 4π a 3 = T 4. For an elliptical orbit: GM star 4 Conceptual question • Two satellites T t llit A and d B off the th same mass are going i around dE Earth th in concentric circular orbits.The distance of satellite B from Earth’s center is twice that of satellite A.What is the ratio of the centripetal force acting on B to that acting on A? – a) 1/8 GM E ms GM E ms F = F = c,A c.B – b) 1/4 rA 2 rB 2 – c) 1/2 – d) 1 2 GM E ms GM E ms 1 = Fc.B = – e)) 1 cB = F 2 2 RB 2RA Fc,B/Fc,A ? ( 2rA ) Fc,B Fc,A = 4A 4r 4 c,A A 1 4 Hint: This is a ratio problem. 5 Example • One satellite O t llit is i in i an orbit bit about b t Jupiter J it off radius di r1 and d a period i d off 100 days. If a second satellite is placed in an orbit with 4 times the radius (i.e. r2=4r1), what is the period for the orbit of the second satellite? – a)) 200 d 2 3 4 π 2 T1 = r1 – b) 400 d GM Jupiter – c) 800 d – d) 1600 d 2 2 4π 3 4π 3 = 4r T22 = r GM ( 1) 2 GM Jupiter Jupiter r2 4r1 T1 100 d T2 ? 4π 2 3 = 64 r1 = 64T 2 1 GM Jupiter J it d = 800 days d Hint: Hi t this thi can also l T22 = 64T12 T2 = 8T1 = 8 ⋅100 days be solved as a ratio problem 6 Conceptual quiz • Suppose E S Earth th h had d no atmosphere t h and dab ballll were fi fired d ffrom the top of Mt. Everest in a direction tangent to the ground. If the initial speed were high enough to cause the ball to travel in a circular trajectory around Earth Earth, the ball’s ball s acceleration would – a) be much less than g (because the ball doesn’t fall to the ground). Note: someone in a space – b) b be approximately i t l g. ship in the same orbit would – c) depend on the ball’s speed. feel “weightless” and many would call this a “zero g” environment. “weightlessness” weightlessness is a sensation that occurs when one feels no effects of gravity. It happens h when h g=00 or when h someone is falling freely. 7 Example • The Earth Th E th orbits bit the th sun iin an circular i l orbit bit off radius di rE=1.5x10 1 5 1011 m. What is the mass of the sun? 2 2 3 4 π 4 π 2 3 2 T = r M sun T = r GM sun G M sun 4π 2 r 3 = G T 2 ( ) 3 1.50x 10 m 4 ⋅ ( 3.14) 30 2.0x10 kg = = 2 -11 11 6.673 x 10 N m² /kg² 365d 24h 3600s 1y y d d 2 11 8 Example • A uniform steel beam of length 5.00 m has a weight of 4.5x103 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal pposition by y a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25o above the horizontal. A load whose weight is 12.0x103 N is hung from the beam at a point that is 3.5 m from the wall. Find (a) the magnitude of the tension in the supporting cable and (b) the magnitude of the force exerted on the end of the beam by the bolt that attaches to the wall. Choose pivot to be at hinge: τ i = 0 = T(5m) sin ( 25o ) − Wload ( 3.5m ) − Wbeam ( 2.5m) i (3.5m) + W ⋅ ( 2.5m) T=W ⋅ ( ) load R 25o Wbeam load Wload T beam ( ) (5m) sin 25o (5m) sin 25o T = 2.54x10 4 N Hinge forces: Fi = 0; x-comp: R x -Tcos 25o = 0 ( ) i R x = Tcos 25o = 2.29x104 N ( ) y comp: R y − Wload − Wbeam + T sin(25o ) = 0 y-comp: R y = Wload + Wbeam − T sin(25o ) = 12.0x103 N + 4.5x103 N − 10.7x103 N R y = 5.8x103 N R = R 2x + R 2y = 23.6x103 N 5.8x103 N 0 tan(θ R ) = .25 14 above horizontal = θ = R 4 2.3x10 N