Chapter 13 Gravitation. Problems: 1, 4, 5, 7, 18, 19, 25, 29, 31, 33, 43


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1 Chapte 13 Gavitation Poblems: 1, 4, 5, 7, 18, 19, 5, 9, 31, 33, 43
2 Evey object in the univese attacts evey othe object. This is called gavitation. We e use to dealing with falling bodies nea the Eath. Fo example a ock dopped off of a cliff. Howeve the same ules apply fo all objects, no matte how lage o how fa apat. (apples falling out of tees, planets, moons, stas, galaxies )
3 4 fundamental types of foces 1) Stong nuclea foce ) Weak nuclea foce 3) Electic foce 4) Gavitational foce Out of the 4 fundamental types of foces, the gavitational foce is the weakest of the 4. The stong nuclea foce is by fa the stongest, howeve it only acts ove vey shot distances (compaable to the size of a nucleus).
4 Thee is a gavitational attaction between evey object in the univese. Howeve, this attaction is only noticeable when eithe the objects ae huge, o when they ae vey close togethe. Newton s Law of Gavitation F G m m 1 is the cente to cente distance between the two objects. G = 6.67x1011 Nm /kg
5 Newton s law of gavitation is a vecto equation. The diection of the foce is in the diection of a line that connects the two objects. ˆ F G m m 1 ˆ 1 The foce is attactive. Object 1 is pulled towads Object Because of Newton s 3 d Law, object is pulled towads object 1.
6 G is the Newton s univesal gavitational constant. It applies eveywhee. We have been using g = 9.8 m/s These two values ae elated but diffeent. g is not constant eveywhee. Find gavitational foce between a peson of mass 100 kg (standing at sea level) and the Eath. F F m1m G 98 N 4 11 Nm (5.98 x10 kg)(100 kg) 6.67 x10 6 kg (6.37 x10 m)
7 F F F at sea level (distance equal to adius of Eath) G M m1m G 6.67x e e N kg g x10 Nm kg m 9.8 s Nm kg (5.98x10 (6.37x10 100kg kg) m) (100kg) (100kg) g At diffeent elevations, g has diffeent values (5.98x10 kg)(100kg) 6 (6.37x10 m)
8 All semeste we have been unde the assumption that g is a constant. This woks as long as objects stay nea the suface of the Eath. Fo noticeable changes in elevation, g vaies. Find the weight of a 100 kg peson standing at: Sea level. = e = 6.38x10 6 m On top of Mt. Eveest = e m In spacecaft 5x10 6 m above Eath, = 11.38x10 6 m
9 Sea level: = e = 6.38x10 6 m w = m p g = m p (Gm E / ) = 100 kg(9.8m/s ) w = 980 N (0. lbs) Mt. Eveest: = e m = 6.39x10 6 m w = m p g = m p (Gm E / ) = 100 kg(9.77m/s ) w = 977 N (19.5 lbs) Spacecaft: = 11.38x10 6 m w = m p g = m p (Gm E / ) = 100 kg(3.08m/s ) w = N (69. lbs)
10 On the Moon To find g on the moon, we use M moon and moon On the suface of the moon, g = g moon =1.6m/s What would a 100 kg peson weigh on the moon? w = (100kg)(1.6m/s ) = 16 N o 36.4 pounds
11 Thee ae othe easons why g is not constant. They have to do with the Eath itself. 1) The Eath s mass is not unifomly distibuted. The thickness and makeup of the Eath s cust vaies. ) The Eath is not a pefect sphee. The Eath is flattened out a bit at the poles and bulges out at the equato. The poles ae close to the cente of the Eath than the equato. 3) The Eath is otating.
12 Take an object sitting on the equato. As the Eath otates, the object moves in a cicle. Applying cicula motion: F N points outwad along the adial diection. F N is the measued weight of the object (mg). a g and v / point inwad towad the cente of the Eath. F N mg ma g m ma g v / mg g a ma g g m Using the values of and at the equato, g vaies fom a g by only about 0.034m/s.
13 Gavitation inside the Eath Newton s Shell Theoem: A unifom shell of matte exets no net gavitational foce on a paticle inside it. Beak the Eath into unifom shells, only the shells that have a adius smalle than the location of the object poduce a net gavitational foce on the object. When you take Physics and see Gauss Law, think of this.
14 Gavitational Potential Enegy Ealie we intoduced the potential enegy of a paticle Eath system and saw that the amount of potential enegy depended on the sepaation distance. Gavitational potential enegy is poduced by the existence of two o moe paticles sepaated by some distance(s). If g was constant U = mgh whee h is the sepaation. We would pick as a manne of convenience the suface of the Eath to be whee U = 0. As paticles wee aised above the suface, the gavitational potential enegy inceased. As paticles wee bought close to the cente of the Eath, the gavitational potential enegy deceased.
15 Since location of the zeo of potential enegy is abitaily picked, only changes in gavitational potential enegy wee meaningful. Let s now allow the sepaation distance to be infinite. Now let s pick the sepaation that poduces zeo gavitational potential enegy to be infinity. Now as paticles get close, they have a finite, but negative amount of gavitational potential enegy. We will now see what happens when vaies noticeably and thus g is not constant.
16 Use the wok done by a vaiable foce: We ae going fom away fom the cente of an object (fo example: Eath) to infinity away. W F( ) d F( ) d cos F is attactive so F will be in the opposite diection of d. = F( ) d d putting this in the integal we get: W W 0 d
17 W W 0 d Using: W =  U U U W setting U U 0 W
18 elation between foce and potential enegy U ( ) F g The minus sign hee tells us the foce is attactive. Fom chapte 8 we found: F(x) = du(x)/dx du d du d d d F( )
19 Gaphs of gavitational foce and gavitational potential enegy F(N) U(J) (m) (m) 1 1 Fo objects that ae infinitely fa away the gavitational potential enegy is zeo.
20 Escape Speed Thow a pojectile upwads. it will slow down because of the gavitational pull of the Eath until it comes to a stop, then will fall back to the gound. The escape speed is the speed necessay fo an object to have at the suface of a planet (o othe object) so that is has enough kinetic enegy to escape the gavitational pull of the planet. Since the gavitational field extends to infinity, the launched paticle will theoetically come to est afte being infinitely fa fom the planet. On the suface of the planet, the pojectileplanet system has gavitational potential enegy: U
21 The paticle is launched so it has some velocity and some kinetic enegy. At the suface of the planet the total enegy is: K U 1 mv When the paticle is infinitely fa away, it comes to a est, so the total enegy will be zeo. 1 mv The necessay velocity at the launch to poduce this is: 0 v GM Escape velocity depends on the mass and adius of the planet you ae tying to escape fom. See table 13
22 Keple s Laws People use to think that the Eath was the cente of the sola system. One poblem with this is the etogade motion of Mas. Keple used data (obsevations of stas and planets) to come up with 3 laws of planetay motion. We will apply the laws to planets obiting the Sun but they wok fo any satellites, natual o atificial that obit a massive cental body. The cental body will be the sun(m) and the planet will be mass (m).
23 1) All planets move in elliptical obits, with the Sun at one focus. Ellipses ae flattened cicles. The eccenticity of an ellipse tells how flattened the ellipse is. A cicle would have an eccenticity of zeo. The two foci would be located at the same point. The eccenticity of the Eath s obit is only nothing at second focus The eccenticity of this pictue is geatly exaggeated.
24 ) The Law of Aeas. A line that connects the planet to the Sun sweeps out equal aeas in the plane of the obit duing equal time intevals If the time inteval to go fom points 1 to is the same as to go fom point 3 to 4, the aeas of the two wedges ae equal. As a esult, the planet moves faste when it is close to the Sun and moves slowe when it is futhe away. Planet is fastest at peihelion (neaest point to the Sun) Planet is slowest at aphelion (fathest point to the Sun)
25 3) Law of Peiods: The squae of the peiod of any planetay obit popotional to the cube of the semimajo axis of its obit. Since the planets have obits of low eccenticity, we can eplace the semimajo axis with the aveage distance fom the Sun. Use Newtons s nd law: F = ma c GM v mv Use: v GM T T T 4 GM 3
26 Poblems: 8, 9, 16, 4, 5
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