PHYS-2010: General Physics I Course Lecture Notes Section IX

Transcription

1 PHYS-200: Geneal Physics I Couse Lectue Notes Section IX D. Donald G. Luttemose East Tennessee State Univesity Edition 2.5

2 Abstact These class notes ae designed fo use of the instucto and students of the couse PHYS-200: Geneal Physics I taught by D. Donald Luttemose at East Tennessee State Univesity. These notes make efeence to the College Physics, 9th Edition 202) textbook by Seway and Vuille.

3 IX. Gavitation A. Newton s Law of Gavity.. Evey paticle in the Univese attacts evey othe paticle with a foce that is diectly popotional to the poduct of thei masses and invesely popotional to the squae of the distances between them. The was fist ealized by Si Isaac Newton, and hence, this is efeed to as Newton s Law of Gavity and can be expessed mathematically as F g = G m m 2 2 ˆ. IX-) a) m and m 2 masses of objects # and #2 measued in kg). b) distance between the masses measued in m). c) G constant of univesal gavitation = this is the popotionality constant between the foce and the dependent paametes. N m2 G = kg 2. IX-2) d) Newton s law of gavity is thus an invese-squae law. 2. The gavitational foce exeted by a spheical mass on a paticle outside the sphee is the same as if the entie mass of the sphee wee concentated at its cente. IX

4 IX 2 PHYS-200: Geneal Physics I cente m R p h M p Note that = R p + h adius of sphee, i.e., planet, plus the height above the sphee). Example IX. Two objects attact each othe with a gavitational foce of magnitude N when sepaated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each? Solution: Let the individual masses be epesented by m and m 2 and the total mass of the two objects be epesented as M = m +m 2 = 5.00 kg). Then we can expess the second mass as m 2 = M m. The distance between the two masses is = 20.0 cm = m, the gavitational foce between them is F = N, and the gavitational constant is G = N m 2 /kg 2. Now make use of Newton s Law of Gavitation: F = Gm m 2 = Gm M m ) 2 2 F 2 = Gm M m ) = GMm Gm 2 0 = Gm 2 GMm + F 2 Algeba has a well know solution to the quadatic equation of the fom ax 2 + bx + c = 0, as x = b ± b 2 4ac 2a.

5 Donald G. Luttemose, ETSU IX 3 Hee, x = m, a = G, b = GM, and c = F 2, so m = GM ± G 2 M 2 4GF 2 2G = GM ± G M 2 4F 2 /G 2G = 2 = 2 = 2 = 2 M ± 5.00 kg ± [ 5.00 kg ± M 2 4F2 G 5.00 kg) kg)0.200 m) N m 2 /kg kg kg 2 ] [ 5.00 kg ±.0 kg 2 ]. As such, m eithe equals 4.0 kg)/2 = 2.0 kg o 6.0 kg)/2 = 3.0 kg eithe answe is coect). If we chose m = 3.0 kg, then m 2 = M m = 5.00 kg 3.0 kg = 2.0 kg. Hence the solution is m = 3.0 kg and m 2 = 2.0 kg. 3. Why do objects fall independent of thei mass on the Eath s suface? a) Gavity: F g = G M m. R 2 b) Motion: F = ma = mg. c) Set F g = F, then o G M m R 2 g = GM R 2 = mg. IX-3)

6 IX 4 PHYS-200: Geneal Physics I d) Plugging in values: g = N m 2 /kg 2 ) kg) m) 2 = N m 2 /kg m 2 = N/kg, but, since N = kg m/s 2, and paying attention to ou input significant digits, we get g = 9.80 kg m/s2 kg = 9.80 m/s 2! e) Since g can be measued fom the gound e.g., motion expeiments) and R can be measued astonomically e.g., simple shadow length measuements at two diffeent latitudes Eatosthenes did this in 200 BC!), we can then detemine the Eath s mass with ewiting Eq. IX-3): 4. The gavitational constant G. M = g R2 G. a) Measuing G in the laboatoy accuately is a difficult task! b) Cavendish was the fist to measue it while he was tying to detemine the density of the Eath in 798 see the textbook fo details). c) Moe sophisticated expeiments have been caied out since that time = G s accuacy is only known to 5 significant digits: G = ) 0 N m 2 /kg 2, whee the 4) digits ae uncetain.

7 Donald G. Luttemose, ETSU IX 5 = the speed of light s accuacy is known to 9 significant digits! c = ) 0 8 m/s. d) Futue space-based expeiments in fee-fall and a vacuum) should incease the accuacy in the measuement of G! 5. The fathe we get fom the Eath s cente, the smalle the acceleation due to gavity: whee distance fom Eath s cente. g = GM 2, IX-4) a) When g is measued on the Eath s suface o some othe planetay suface), it is called the suface gavity. b) When g is measued elsewhee i.e., not on a planetay suface), it is called the acceleation due to gavity, and if the object is in fee-fall, it also is often called the fee-fall acceleation. Example IX 2. Mt. Eveest is at a height of 29,003 ft 8840 m) above sea level. The geatest depth in the sea is 34,29 ft 0,430 m). Compae the Eath s suface gavity at these two points. Solution: Let h e be the height of Mount Eveest and h s be the geatest depth of the sea. Using Eq. IX-4) and the fact that the adius of the Eath at sea level at the Eath s equato is R = m, we get the distances fo the highest and lowest points of the suface fom the cente of the Eath as h = R + h e = m m

8 IX 6 PHYS-200: Geneal Physics I = m l = R h s = m m = m Which gives the suface gavities of g h = GM 2 h = N m 2 /kg 2 ) kg) m) 2 = m/s 2, g l = GM 2 l = N m 2 /kg 2 ) kg) m) 2 = m/s 2, and g l g h g 00% = % = 0.605%, the suface gavity of the lowest point on the Eath s suface is a little moe than half of a pecent lage than at the highest point. B. Keple s Laws of Planetay Motion.. Johannes Keple ) was a Geman mathematician and astonome who used Tycho Bahe s obsevations of Mas to deive the 3 laws of planetay motion. The data showed that the Copenican model of heliocentic Sun-centeed) sola system was coect, except that the planets move in elliptical and not cicula paths aound the Sun as Copenicus had assumed. 2. The laws: a) Law : The obit of a planet about the Sun is an ellipse with the Sun at one focus. The so-called elliptical obit. The equation fo the ellipse in Catesian coodinates

9 Donald G. Luttemose, ETSU IX 7 when the foci ae on the x axis) is x 2 a 2 + y2 b 2 =. IX-5) a = semimajo axis b = semimino axis Sun focus i) Semimajo axis a): Half of the longest axis of an ellipse. ii) Semimino axis b): Half of the shotest axis of an ellipse. iii) Astonomical Unit A.U.) is the length of the Eath s semimajo axis, A.U. = m. iv) The elative flatness of an ellipse is measued by the eccenticity e: e = a2 b 2 a. IX-6)

10 IX 8 PHYS-200: Geneal Physics I v) The distance fom the cente to eithe focus is given by a 2 b 2. vi) The ellipse is just one type of conic section. If a = b, Eq. IX-6) gives e = 0 and we have a cicula obit = a second type of conic section i.e., a cicle). vii) If we let a get bigge and bigge such that a b, then Eq. IX-6) gives e a 2 /a = a/a = as a. When this happens, we have a paabolic obit. Such an obit is said to be open both cicula and elliptical obits ae closed) and neve etun. A paabolic obit is achieved when the velocity of a satellite just equals the escape velocity, v esc. viii) Thee also ae obits that ae moe open than paabolic obits = the so-called hypebolic obits. These obits have e > and can be achieved if v > v esc. b) Law 2: A line joining a planet and the Sun sweeps out equal aeas in equal amounts of time law of equal aeas ). i) This means that planets move faste when close to the Sun in its obit than when it is fathe away. ii) Objects in vey elliptical obits don t stay nea the Sun fo a vey long time = comets. iii) Peihelion: Point on an obit when a planet is closest to the Sun p = peihelion distance).

11 Donald G. Luttemose, ETSU IX 9 iv) Aphelion: Point on an obit when a planet is fathest fom the Sun a = aphelion distance). v) The peihelion and aphelion of a sola obit can be detemined fom the semimajo axis and the eccenticity with p = a e) IX-7) a = a + e), IX-8) also note that p + a = 2a. IX-9) if A = A 2 aea), then t = t 2 time) t A Sun A 2 t 2 p a c) Law 3: The squae of the obital peiod T) of any planet is popotional to the cube of the semimajo axis a) of a planet s obit about the Sun hamonic law): T 2 a 3. IX-0)

12 IX 0 PHYS-200: Geneal Physics I 3. Keple s laws wee empiically deduced. Nealy 00 yeas afte these planetay laws wee developed, Newton came along and showed theoetically why they ae valid. a) Gavity fo a planet in obit about the Sun [ ]): F g = GM M p 2. b) Centipetal foce: F c = M p v 2 c) Set these two foces equal: GM M p = M p v 2 2 v 2. = GM d) Note, howeve, that the obital velocity of a planet can be expessed as the length of the cicumfeence C of the obit divided by its obital path. Fo demonstation puposes, we will assume that the obit is cicula, but the same esult is obtained fo elliptical obits, then v = C T = 2π T. e) Plugging this equation into the foce equation above gives ) 2π 2 v 2 = = GM T 4π 2 2 T 2 = GM 4π 2 3 = T 2. GM Since we have assumed a cicula obit hee, = a the semimajo axis), and T 2 = 4π2 a 3 = K a 3. GM,. IX-)

13 Donald G. Luttemose, ETSU IX i) K = 4π2 GM = s 2 /m 3. ii) K is independent of the planet s mass = K only depends upon the lage, cental object s mass. iii) T 2 a 3 is valid whethe the obit is cicula whee a =, the obital adius) o elliptical whee a is used intact), though the elliptical obit solution has a slightly diffeent fom fo the K constant. The poof fo elliptical obits equies advanced calculus hence we will not show it hee). 4. We can base othe planets in sola system with espect to the Eath: Tp 2 T 2 = K a 3 p K a 3 = a3 p a 3. IX-2) Since T = y and a = A.U., Eq. IX-2) becomes o T 2 p y 2 = a3 p A.U. 3, T 2 y = a3 AU. IX-3) Example IX 3. The Voyage spacecaft has just passed the 20 A.U. mak in its distance fom the Sun. At this time, the Death Sta fom an evil galactic empie intecepts Voyage and figues out which planet sent it based on the gold ecod that was included on the spacecaft. The Death Sta altes its couse and stats to obit the Sun fom that point and sets its heading such that it will each peihelion at the Eath s location. Calculate the following about its obit: a) the semimajo axis in A.U.); b) the eccenticity, and c)

14 IX 2 PHYS-200: Geneal Physics I the length of time in yeas) it will take to get to Eath. Solution a): We ae given that p =.00 A.U. and since it stats its fee fall obit fom the position of Voyage, that position maks it aphelion position, a = 20 A.U. note that we do not have to convet to SI units fo this poblem). Using Eq. IX-9) we can easily calculate the semimajo axis of the Death Sta s obit: a = p + a 2 Solution b): =.00 A.U A.U. 2 Using Eq. IX-7) and solving fo e, we get, = 60.5 A.U. e = p.00 A.U. = = = 0.983, a 60.5 A.U. nealy a paabolic obit! Solution c): Fist we need to calculate the peiod of a complete obit fom Keple s 3d law, then the amount of time it will take to go fom aphelion to peihelion is one-half that peiod. As such, T T T.00 y ) 2 = a a ) 3 = = T = 47 y ) 60.5 A.U. 3 = 60.5) 3 = A.U. Since this is the full peiod of the Death Sta s obit, the time to get to Eath will be t = T = 235 y. 2

15 Donald G. Luttemose, ETSU IX 3 C. Consevation of Enegy in a Gavitational Field.. Up until this point, we handled consevation of mechanical enegy fo gavity fo a constant suface gavity i.e., a constant acceleation due to gavity). Hee, we will elax that equiement and let g vay with as shown in Eq. IX-4). a) Fo a constant acceleation due to gavity, we have seen fom Eq. VI-) that PE = mgh, IX-4) whee m is the mass of the object, g is the suface gavity i.e., acceleation), and h is the height above the gound. b) Howeve, this is not the most geneal fom of the potential enegy of a gavitational field it is an appoximation to the geneal fom. In highe-level physics, potential enegy is elated to a foce field by the equation F = PE) = dpe) d ˆ, IX-5) whee F = F g as descibed by Eq. IX-) fo a gavitational field and the del symbol is the spatial deivative in thee dimensions fo Eq. IX-5) to be valid, we only use the component of in the ˆ diection i.e., d/d) since F g only points in the -diection. i) Since this is an algeba-based couse, we won t solve that diffeential equation with calculus. Howeve, by ealizing that the deivative symbol d just means infinitesimally small delta change of) and we can appoximate Eq. IX-5) with ) GM m PE) = 2 GM m PE PE = whee the tems ae the initial values. 2 ) ), IX-6)

16 IX 4 PHYS-200: Geneal Physics I ii) Now if we define the initial PE as 0 at the cente of the gavitating body so = 0), Eq. IX-6) becomes ) GM m PE = 2 = GM m iii) The calculus solution to the diffeential equation above Eq. IX-5) would have given exactly the same answe.. iv) As such, the most geneal fom of the potential enegy of a mass in a gavitating field is PE = GM m. IX-7) c) With this geneal fom of the gavitational potential enegy, we can see how the potential enegy equation nea the Eath s suface Eq. IX-4) aises. i) Let s say we have a pojectile that we launch fom the gound. While on the gound, = R the adius of the Eath), which gives a potential enegy of PE = GM m R. IX-8) ii) Now, when the pojectile eaches its highest point above the gound, h, it is a distance of = R + h fom the cente of the Eath. At this point, it has a potential enegy of PE = GM m R + h. IX-9)

17 Donald G. Luttemose, ETSU IX 5 iii) The change in potential enegy between these two points is PE = GM m R + h GM m = GM m = GM m R ) R R + h R + h R R + h) R R R + h) = GM m R + h R R R + h) h = GM m R R + h). iv) If R is much geate than h which it will be fo expeiments nea the Eath s suface), h R. As such, R +h R and the equation above becomes PE GM m = GM mh R 2 h = GM m R R ). h R 2 v) Now, emembeing ou defining equation fo suface gavity e.g., Eq. IX-3): g = GM, R 2 we use this in the potential equation we just wote: PE = m GM h = mg h, R 2 hence we have poven Eq. IX-4) fom fist pinciples. 2. We can now use this geneal potential enegy equation to figue out high tajectoy, obital, and space tajectoy poblems. Fist,

18 IX 6 PHYS-200: Geneal Physics I let s develop a elationship between the initial velocity, v, of a pojectile and the maximum height, h, it will each. a) If we wee limiting ouselves to vetically diected tajectoies nea the Eath s suface, we would have KE i + PE i = KE f + PE f 2 mv2 + mgy = 2 mv2 + mgy. y = 0, since it epesents the gound and the pojectile will each its maximum height y = h) when the velocity goes to zeo v = 0). Fom this, we can solve fo the initial velocity and get v = 2gh. b) Using the moe geneal fom of the potential, ou initial position will be on the suface of the Eath = R ) and we will each a distance fom the cente of the Eath of = R + h, which we will simply wite as. The consevation of mechanical enegy then gives KER ) + PER ) = KE) + PE) 2 mv2 GM m = R 2 mv2 GM m. c) Once again, v = 0 at the top of the tajectoy, and as such, the initial velocity is 2 mv2 GM m R = 0 GM m 2 mv2 = GM m GM m R 2 mv2 = GM m ) R v 2 = 2GM m m R )

19 Donald G. Luttemose, ETSU IX 7 ) v 2 = 2GM R v = 2GM R ). IX-20) d) Now if we once again define g to be the acceleation due to gavity at the Eath s suface i.e., suface gavity), we can ewite Eq. IX-3) to ead GM = g R 2. e) Plugging this into Eq. IX-20) gives v = 2gR 2 and eplacing by R + h, we get v = = = = = v = 2gR 2 2gR 2 2gR 2 2gR 2 R ) R ) R + h, IX-2) R + h R R + h) R R R + h) R + h R R R + h) 2gR 2ghR R + h. h R R + h) h R + h ) IX-22) f) Finally, one can immediately see that if h R, then R + h R and Eq. IX-22) becomes v 2ghR R = 2gh.

20 IX 8 PHYS-200: Geneal Physics I As can be seen, this equation as witten above) is just an appoximation to a moe geneal equation i.e., Eq. IX-22). Example IX 4. A satellite of mass 200 kg is launched fom a site on the Equato into an obit at 200 km above the Eath s suface. a) If the obit is cicula, what is the obital peiod of this satellite? b) What is the satellite s speed in obit? c) What is the minimum enegy necessay to place this satellite in obit, assuming no ai fiction? Solution a): The adius of the satellite s obit is = R + h = m m = m, m = 200 kg is the satellite s mass, and M = kg is the Eath s mass. The obital velocity will just be the tangential velocity of the cicula obit. Since the gavitational foce povides the centipetal acceleation, we have m v2 ob F c v 2 ob v ob = = F g = GM m 2 = GM GM = N m 2 /kg 2 ) kg) m = m/s. Fo a cicula obit, the obital peiod is T = 2π = 2π m) v ob m/s

21 Donald G. Luttemose, ETSU IX 9 = s =.48 h. Solution b): The obital speed was computed in pat a): v ob = m/s. Solution c): The minimum enegy to each obit can be detemined by calculating the diffeence between the total mechanical enegy of the satellite in obit and total mechanical enegy of the satellite pio to launch. The satellite itself pio to launch is not moving, but the suface of the Eath is moving due to the Eath s otation we will ignoe the Eath s obital velocity hee since we ae not leaving the Eath s gavitational influence). Hence, minimum launch enegy = E min = KE + PE) ob KE + PE) ot. The Eath s otational peiod is the definition of a day, so T ot =.000 day = 86,400 s. Fom this, we get the Eath s otation velocity of v ot = 2πR = 2π m) T ot s Hence, the minimum launch enegy is E min = 2 mv2 GmM ob [ = m 2 v2 GM ob = m v2 ob v2 ot 2 ) ) = 464 m/s. 2 mv2 ot GmM 2 v2 ot GM + GM R ) R R )] )

22 IX 20 PHYS-200: Geneal Physics I E min = 200 kg) m/s) m/s) N m kg kg) )] m m = 200 kg) m 2 /s m 2 /s 2) = J. 3. The absolute magnitude of the potential enegy due to a lage gavitating body is often times lage than the kinetic enegy of the object in motion. a) We can define the total mechanical enegy E tot of body in motion as the sum of the kinetic and potential enegies: E tot = KE + PE. IX-23) b) Since PE is negative in a gavitational field, E tot < 0 i.e., negative) fo pojectile i.e., bound) tajectoies. 4. Fo an object to just ovecome any gavitating body s like the Eath s) potential field, an object has to be launched with zeo total enegy, that is, KE = PE to escape the pimay body s e.g., Eath s) gavitational field = the escape velocity. a) Hence, to calculate the escape velocity fom the suface of a lage body of mass M and adius R, we just have to set the initial kinetic and potential enegy sum to zeo and solve fo the velocity: E tot = 2 mv2 esc GMm R = 0. IX-24)

23 Donald G. Luttemose, ETSU IX 2 b) This gives the equation fo the escape velocity o escape speed): v esc = 2GM R. IX-25) c) As can be seen fom Eq. IX-25), the escape velocity does not depend upon the mass of the ocket. Using the values fo Eath in Eq. IX-25), we get v esc = 2GM R, IX-26) o v esc =.2 km/s to leave the Eath s gavitational field. d) Fo an object to escape the gavitational field of a pimay body, it must achieve a velocity geate than o equal to the escape velocity: v v esc. Example IX 5. The Sun has a mass of kg and a adius of m. Calculate the escape velocity fom the sola suface in km/s) and compae it to the escape velocity fom the suface of the Eath. Solution: Fom Eq. IX-26), the escape velocity fom the sola suface is: v esc ) = 2GM R = N m 2 /kg 2 ) kg) m ) km = m/s = 67.6 km/s. 000 m Since v esc ) =.2 km/s, v esc ) = 67.6 km/s /.2 km/s) v esc ) = 55. v esc ), o 55. times bigge than the escape velocity fom the Eath s suface!