CHAPTER GRAVITATION

Transcription

1 Solutions--Ch. 1 (Gavitation) CHAPTER 1 -- GRAVITATION 1.1) Accoding to Newton, the magnitude of the gavitational foce between any two bodies will always be equal to Gm 1 m /. a.) The gavitation foce you exet on you fiend will be F on fiend G m you m f / (6.67x1-11 nt. m /kg )(8 kg)(6 kg)/(5 m) 1.8x1-8 nts. Note If you cut a ing ong ball into a million and a half o so ieces, this foce would aoximately equal the weight of one iece. b.) As fo the acceleation, Newton's Second Law yields F x F f m f a a F f / m f (1.8x1-8 nt)/(6 kg).13x1-1 m/s. Note You can now see why you don't feel a gavitational foce when you bush ast a fiend on the steet. The Univesal Gavitational Constant G is so small that at least one of the masses has to be enomous befoe gavitational effects become noticeable. 1.) Figue I below shows an f.b.d. fo the foces acting on the to mass, comlete with comonents. Figue II below shows how the adius of the obits can be detemined. F cos F g FIGURE I m F cos F g lanet o 3 d/ cente FIGURE II o (d/)/cos 3.577d lanet F sin F sin 557

2 Due to symmety, the hoizontal comonents will add to zeo leaving only the vetical comonents with which to contend. Noting that the masses ae all the same and the adius of the cicula motion is.577d, we can use N.S.L. in the centeseeking diection to wite F cente-seeking -F g cos θ - F g cos θ -ma c [Gm /d ]cos 3 o m(v /) mv /[.577d] v (Gm/d) 1/. 1.3) This is a staight feefall oblem (i.e., although we will be using Newton's gavitational foce, the foce is not acting centietally). a.) The eath to moon distance (cente of mass to cente of mass) is 3.8x1 8 metes lus 1.7x1 6 metes lus 6.37x1 6 metes, o aoximately 3.9x1 8 metes. Summing the foces in the diection of feefall, we can wite F -Gm m / -m m a a G / (6.67x1-11 nt. m /kg )(5.98x1 kg)/(3.9x1 8 m).6x1-3 m/s. b.) The distance between the moon's cente of mass and the eath's cente of mass just befoe they stike one anothe will be the sum of thei adii. Using that obsevation and eeating the calculation done in Pat a, we get F -Gm m / e,m -mm a a G / e,m (6.67x1-11 nt. m /kg )(5.98x1 kg)/(8.11x1 6 m) 6.6 m/s. c.) The acceleation is not a constant (obviously), so kinematics is out. The best aoach is using consevation of enegy. Remembeing that the otential enegy function fo a vaying gavitational field is U -Gm 1 m / (a scala), we can wite 558

3 Solutions--Ch. 1 (Gavitation) KE 1 + U 1 + W ext KE + U () + (-Gm m /) + () (1/)m m v + [-Gm m /( m + e )] v [G [-1/ + 1/( m + e )]] 1/ [(6.67x1-11 nt. m /kg )(5.98x1 kg)[-1/(3.9x1 8 m) + 1/(1.7x x1 6 m)]] 1/ v 981 m/s. d.) If the eath is allowed to move, we have to take into consideation its kinetic enegy at the end of the feefall (its initial kinetic enegy is zeo). This means we now have a second unknown--the eath's velocity--to deal with. To get an exession elating the eath's velocity v e to the moon's velocity v m, note that the only foce acting in the system--gavity--is intenal to the system (that is, the moon exets a gavitational foce on the eath and the eath exets an equal and oosite gavitational foce on the moon). That means momentum is conseved. With eveything stating fom est, the system's initial net momentum (and the system's subsequent net momentum) is zeo. Using that bit of infomation, we can wite o just befoe imact -m m v m + v e v e /v m m m /. Knowing the mass of the eath and moon, we now have the elationshi we need between the eath's velocity and the moon's velocity (i.e., ou second equation). 1.) a.) The magnitude of the foce alied to m 1 a distance fom the lanet's cente is F G(m inside shee )m 1 /, whee m inside shee is the mass inside the shee uon which the body sits. To detemine that mass, define a diffeential shee of adius a and thickness da. Its diffeential volume will be its suface aea times its thickness, o dv ( a )da. Defining the volume density function to be ρ, we can wite 559

4 m Using this, we can wite insidesh ρdv m a m π a πa da 3 ada a m a π m π. [ ] a F G(m inside shee )m 1 / G( m / )m1 / Gm m 1 ( / ). Note that at, we don't find a zeo in the denominato as befoe. b.) The otential-enegy-equals-zeo-oint fo a vaiable foce function is ALWAYS laced whee the foce is zeo. In this case, that is at the lanet's cente. If we assume the body is sitting at some osition y u the vetical axis (i.e., we ae assuming the tunnel is in the vetical), then the foce on the mass m 1 when at an abitay oint y will be -Gm m 1 ( / )y (fom Pat a) Setting U(y ), we can wite U () Uy ( ) F d Gmm π 1 U () y ( dx dy dz ) y + + j i j k Gmm π 1 ydy y 3 Gmm1π y 3 Gmm1π 3. 3 y 56

5 Solutions--Ch. 1 (Gavitation) Again, this makes sense. At, the otential enegy is zeo. 1.5) We know the obital distance fom the eath's cente of mass is 1.3x1 6 m + 6.3x1 6 m 7.6x1 6 metes, and we know that this is additionally the distance between the centes of mass of the two bodies (the two quantities ae the same as the satellite's mass is minuscule in comaison to the eath's mass). a.) One of the things that makes obital oblems ticky is that thee ae a numbe of ways one can detemine a velocity. Consevation of enegy has velocities in it, but so does N.S.L. couled with centietal acceleation. Knowing which aoach to use in a given situation is something that comes only with exeience. In this aticula case, we want to look at the foces acting on the satellite. F c -F g -ma c G / (v /) v (G /) 1/ [(6.67x1-11 nt. m /kg )(5.98x1 kg)/(7.6x1 6 m)] 1/ 7 m/s. b.) If it takes time T fo the satellite to tavel the cicumfeence of the cicle uon which it moves, then the atio of those two quantities (i.e., distance/time) will yield the magnitude of its velocity. As T is defined as the eiod of the motion (i.e., the time fo one comlete obit), we can wite v (cicumfeence)/(eiod) /T T /v (7.6x1 6 m)/(7 m/s) 659 seconds (aox. 1 h, 5 min.). c.) The wok we have to do in getting the satellite u to seed is equal to the body's final kinetic enegy. The wok we have to do in lifting the satellite into the aoiate obit is equal to minus the wok gavity does on the body as the body ises, o U g (emembe, the wok gavity itself does as the body ises is - U). If we add these two wok quantities togethe, we have the amount of enegy we have to ovide to the system to get the satellite into obit.... O, we could simly use the modified consevation of enegy equation. In that exession, the W ext tem denotes the exta enegy needed fom us to ut the satellite in obit. Using that exession, we can wite 561

6 KE 1 + U 1 + W ext KE + U () + (-G / e ) + E you (1/) v + (-G /), whee v is the obital velocity of the satellite and is the obital adius fom the eath's cente. Solving yields E you (1/) v + G [1/ e - 1/].5( kg)(7 m/s) + (6.67x1-11 nt. m /kg )(5.98x1 kg)( kg)[1/(6.3x1 6 m) - 1/(7.6x1 6 m)] E you 1.8x1 1 joules. Note If we had used the obital enegy exession E -G / fom the chate, we would have ended u with -1.5x1 1 joules (this, if you will emembe, is KE + U ). So why doesn't the obital enegy exession equal the wok equied to ut the satellite into obit? The answe is simle. The body stated out with otential enegy. Sitting on the eath's suface with -G / e -.53x1 1 joules of otential enegy (ut in the numbes--that's what you get). Add the enegy needed to get it into obit (i.e., the 1.8x1 1 joules) and you end u with the satellite's net final enegy equaling -1.5x1 1 joules. d.) The eath's angula velocity is adians e hous (give o take a bit). That calculates out to ω 7.7x1-5 adians/second. At the equato, the magnitude of the eath's tanslational velocity is v equ ω (6.3x1 6 m)(7.7x1-5 ad/sec) 58 m/s. A satellite launched in the diection of the eath's otation at the equato will begin its ti with initial kinetic enegy equal to (1/)mv e, whee ve is the velocity of the otating eath. In the case of ou satellite, it would stat out with kinetic enegy equal to (1/)( kg)(58 m/s).x1 7 joules woth of enegy we wouldn't have to suly to the system to get it into its obit. It should now be obvious why launching ads in the U.S. (Cae Canaveal, fo instance) ae as close to the equato as ossible. e.) In 18 evolutions, the enegy lost to the satellite will be 56

7 Solutions--Ch. 1 (Gavitation) E lost (18 ev)(x1 5 joules/ev) 3.6x1 8 joules. The amount of enegy it stats out with as it taveled in its obit is E 1 -G / 1 -(6.6x1-11 nt. m /kg )(5.98x1 kg)( kg)/[(7.6x1 6 m)] -1.9x1 1 joules. Afte losing 3.6x1 8 joules of enegy, the satellite has E E 1 - E lost (-1.9x1 1 joules) - (3.6x1 8 joules) -1.85x1 1 joules. An obit with that amount of enegy must be such that E -G / -G /E -(6.67x1-11 nt. m /kg )(5.98x1 kg)( kg)/[(-1.85x1 1 j)] 7.35x1 6 metes. This is the distance fom the eath's cente. Fom the eath's suface, the height will be 7.35x1 6 metes minus 6.3x1 6 metes equals 1.5x1 6 metes. Although the oblem did not ask fo the velocity of the satellite when in this obit, we will need it late. Using N.S.L., we can wite F F g ma c G / ms (v / ) v (G / ) 1/ [(6.67x1-11 nt. m /kg )(5.98x1 kg)/(7.35x1 6 m)] 1/ 7366 m/s. f.) The etading foce does x1 5 joules e evolution of wok on the satellite as the satellite moves along in its ath. On aveage, the wok done by the etading foce in one obit will equal the aveage foce times the aveage cicumfeence of motion times cos 18 o (the etading foce and the diection of 563

8 motion will be oosite one anothe, hence the 18 o angle). Aoximating the aveage adius to be ( 1 + )/, we can detemine the wok calculation fo one aveage obit. Doing so yields x1 5 F avg ( avg ) cos 18 o -F avg [ ( + 1 )/] -F avg [ (7.35x x1 6 )/] -(.7x1 7 )F avg F avg -.3 newtons g.) The modified consevation of angula momentuxession is L 1 + Γ ext t L, whee the L tems ae angula momentum quantities and the Γ t is a toque diven, imulse-elated quantity (emembe, the modified consevation of linea momentuquation was 1 + F ext t, whee the F ext t tem was the imulse being deliveed by extenal foces ove the time inteval t). The aveage extenal toque about the lanet's obital axis is Γ ext s,avg xf avg. Noting that the fictional foce F avg is eendicula to the adius vecto s,avg, the magnitude of the toque will simly be the oduct of the adius and foce vectos (sin 9 o 1). Assuming the diection of the satellite's angula velocity is ositive and noting that the toque slows the motion, ou fictioninduced toque will be negative. As such, we can wite Γ ext - s,avg F avg. The angula momentum of the satellite will equal L s,avg x, whee again the angle between the vectos is 9 o and the sine of the angle is 1. As the magnitude of the momentum is v s, we can wite in geneal 56

9 Solutions--Ch. 1 (Gavitation) L s,avg v s. Using all this infomation in ou modified consevation of angula momentum equation, we can solve fo t L 1 + Γ ext t L s,1 v s,1 - s,avg F avg t s, v s, (7.6x1 6 m)( kg)(7 m/s) - (7.75x1 6 m)(. nt) t (7.35x1 6 m)( kg)(7366 m/s) t 1.3x1 7 seconds. This ounded-off value is aoximately 11 days, 13.5 hous. 1.6) a.) The conseved quantities fo lanetay motion ae angula momentum (thee ae no extenal toques acting on the lanet), and enegy (thee is actically no fictional effect in sace, and thee ae geneally no aeciable non-consevative foces acting on lanets--asteoid collisions exceted). b.) At min, the sta has U 1-5x1 33 joules (see gah) and kinetic enegy KE 1.5x1 33 joules. That means the total enegy of the system is KE 1 + U 1 -.5x1 33 joules. In addition, the sum of the otential and kinetic enegies must always equal that numbe (enegy is conseved). Because we know the total enegy in x1 33 KE at min Enegy (joules) 33 E -.5x1 tot 33-5x1 ENERGY vs. POSITION at, max 33 KE 1.7x1 joules max KE gah min max min E gah total U gah Position 565

10 the system, and because we have a gah of the otential enegy as a function of osition, we can detemine the kinetic enegy as a function of osition fo any oint in the obit. Doing that fo key oints (examle at max, the gah tells us that the otential enegy is aoximately U max -.x1 33 joules... that means KE max -.x1 33 joules total enegy -.5x1 33 joules, o KE max 1.7x1 33 joules at that oint), and emembeing that the kinetic enegy function mios the otential enegy function when the total enegy is constant, we can daw the kinetic enegy gah as shown below. 566