14. Gravitation Universal Law of Gravitation (Newton):

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1 14. Gavitation 1 Univesal Law of Gavitation (ewton): The attactive foce between two paticles: F = G m 1m 2 2 whee G = m 2 / kg 2 is the univesal gavitational constant. F m 2 m 1 F Paticle #1 feels a pull towad paticle #2 and paticle #2 feels a pull towads paticle #1 -- action-eaction foces. The law is fo pais of "point-like" paticles. Evey paticle in the univese pulls on evey othe paticle in the univese, e.g. the moon is pulling on you now. The foce does not depend on what is between two objects, i.e. it cannot be shielded by a mateial (e.g. wall) between them.

2 This is one of the fou fundamental foces. 2 Pinciple of Supeposition: The total foce on a point paticle is equal to the sum of all the foces on the paticle. F 1 = F 12 + F 13 + F 14 +L + = n i=2 F 1i F 1n F 1i : foce of the i th paticle on paticle #1. Fo a eal object with a continuous distibution of paticles: df 1 = G m dm 1 2 F 1 = df 1 = G m 1 dm 2 Gavitational foce fom a thin ing:

3 3 dm x θ df df x df m 1 R F x = df x = G m dm 1 2 = G m dm 1 2 cosθ x = G m xdm 1 3 = G m xdm 1 (x 2 + R 2 2 ) 3 = G m 1 mx (x 2 + R 2 ) 3 2 The foce points towad the ing. F x = 0 if x = 0 The foce pependicula to x (df ) cancels by symmety. Use symmety to simplify you poblem.

4 4 Othe ewton's esults: A unifom spheical shell of matte attacts a paticle outside as if all the shell mass was concentated at the cente. Similaly fo a sphee of matte. Like the case of a ing, a paticle inside a spheical shell of matte feels zeo gavitational foce fom the shell. Gavity: Foce on a mass m on Eath's suface: F g = GmM Eath 2 R Eath = m GM Eath 2 R Eath ( )( kg) ( ) 2 GM Eath 2 = m 3 / kg / s 2 R Eath m = 9.83 m / s 2 = g Fo any planet, the acceleation of gavity at its suface is:

5 5 g planet = GM planet 2 R planet Example: What is the acceleation due to gavity fo (a) an aiplane flying at an altitude of 10 km, (b) a shuttle at 300 km, (c) a geosynchonous satellite at km? (a) g plane = (b) g shuttle = GM Eath ( R Eath + h ) 2 plane ( )( kg) ( ) 2 = m 3 / kg / s m m = 9.79 m / s 2 GM Eath ( ) 2 R Eath + h shuttle ( )( kg) ( ) 2 = m 3 / kg / s m m = 9.00 m / s 2

6 Little diffeent fom g on Eath. Astonauts floating in a shuttle is not due to zeo gavity. GM (c) g satelite = Eath ( R Eath + h shuttle ) 2 6 = ( m 3 / kg / s 2 )( kg) ( m m) 2 = 0.22 m / s 2 Much lowe than g on Eath but gavity still pulls on objects in space. Example: Fo a spaceship between the Eath and moon, at what distance fom the Eath will the net gavitational foce be zeo? x M E F E M s F m M m d F x = F Eath F moon = 0

7 GM E M s x 2 GM m M s (x d) 2 = 0 7 M E x 2 M m (x d) 2 = 0 (x d) 2 M m M E x 2 = 0 x 2 2xd + d 2 x 2 = 0 (1 )x 2 2xd + d 2 = 0 x = 2d ± 4d2 4(1 )d 2 2(1 ) = = 2d ± 4d2 2(1 ) 2d ± 2 d 2(1 ) = 1 ± 1 d x = d > d unphysical

8 x = 1 1 d < d OK 8 Poblem 15: Calculate the gavitational foce due to a hollowed sphee, assuming that the mass of the sphee was M befoe hollowing. R d Calculating the foce by integating ove the solid pat of the sphee is difficult Use Pinciple of Supeposition

9 F = F(solid sphee) F(hollowed mateial) = GMm d 2 GM hm m d R 2 = GMm Gm d 2 d R 2 = GMm d 2 GMm d R 2 = GMm d R d 4 3 π 2 M R πr3 3 9 Satellite: Gavitational Foce is the centipetal foce that keeps a satellite moving in a cicula obit.

10 F c = F g 10 m v2 = GM E m 2 v 2 = GM E v = GM E The speed is independent of the mass of the satellite a penny and a semi-tuck would obit at the same speed fo the same adius. A satellite obiting futhe away fom the Eath has lowe speed. The peiod of the obit: T c = 2π v = 2π GM E / = 4π2 3 GM E Communication satellites ae positioned in obit so that they "appea" stationay in the sky by placing them in a cicula obit with a peiod of 1 day.

11 What is the adius of the obit of a geosynchonous satellite? 11 m M E 3 = GM T 2 E c 4π 2 = ( m 3 kg -1 s -2 )( kg)(86400 s) 2 4π 2 = m = 42,000 km Appaent Weightlessness: The weight of an object is defined as the foce of gavity that the object feels. Fo an astonaut in a spacecaft, the weight is just: F g = GMm 2 whee M is the mass of the eath and m is the mass of the astonaut. Howeve, we see pictues of astonauts floating aound in the space shuttle

12 in an appaently weightless envionment. Ae they actually weightless? Since we define weight to be the foce of gavity, the astonauts ae not weightless. We can define an appaent weight as being the nomal foce between an object and the suppot (e.g. floo). 12 Conside a ide in elevato: (+) mg If the elevato is acceleating upwad, mg = ma = mg + ma > mg the eading on a scale will give an appaent weight lage than the actual weight because the scale measues the nomal foce.

13 If the elevato is acceleating downwad, the appaent weight is smalle than the actual weight because nomal foce is less than mg. If the elevato is in a fee fall, mg = mg = 0 the eading on a scale would be zeo, giving a zeo appaent weight. 13 Conside an astonaut in a shuttle: F g s F ga Shuttle: G M Em s v = m s m s = v2 G M E 2 (1)

14 Astonaut: (1) + (2) G M Em a v 2 = m 2 a = G M E m a 2 m s + m a = 0 v2 (2) 14 = 0 astonauts floating in a shuttle. Howeve, astonauts ae not weightless: F g = G M E m a 2 0 Atifical Gavity: can use cicula motion to ceate an appaent weight, simila to the idea of acceleating elevato. v R

15 Conside a spinning space station, the nomal foce exeted by the wall keeps the astonants in cicula motion. They feels the nomal foce like the gavity on Eath. Any value of (atifical gavity) can be selected by adjusting v and. Fo example, if we want a space station spinning at one evolution pe minute to have an appaent gavity equal to that on Eath, what should be the adius of the station? T = 2π v = Tv 2π mg = mv2 v = g = T g 2π = T g 2π = T 2 g 4π 2 = 890 m 15

16 16

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