Tutorial Exercises: Central Forces

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Hector Wood
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1 Tutoial Execises: Cental Foces. Tuning Points fo the Keple potential (a) Wite down the two fist integals fo cental motion in the Keple potential V () = µm/ using J fo the angula momentum and E fo the total enegy. (b) Eliminate φ and deive an expession fo ṙ in tems of J and E. (c) The tuning points ae the lagest and smallest values of. Show that the tuning points fo the Keple potential ae given by a quadatic equation. (d) Unde what conditions ae the tuning points eal? (e) Unde what conditions ae both tuning points positive?. Solution: (a) The two elevant fist integals ae and (b) Eliminating φ gives and (c) The tuning points ae when ṙ = 0 thus m 2 φ = J 2 m(ṙ2 + 2 φ2 ) µm/ = E 2 m(ṙ2 + J 2 m 2 2 ) µm/ = E 2 m 2 µm/ = E J 2 and in ode to undestand the tuning points it is a good idea to plot the gaph of the left hand side of this equation as a function of. The coesponding quadatic equation is 2Em 2 + 2µm 2 J 2 = 0 (d) The tuning points ae eal if the disciminant is not negative (2µm 2 ) 2 4(2Em)( J 2 ) = 4µ 2 m 4 + 8EmJ 2 0 (e) The tuning points ae both positive if the sum and poduct of the oots ae both positive. This tuns out to be equivalent to E < 0 and combining this with the esult above gives 2. Cicula Obits fo the Keple potential µ2 m 3 2J 2 E < 0 (a) Unde what conditions is the obit fo a Keple potential cicula? [Hint: Use the esults of the pevious question.] (b) Expess the adius fo a cicula obit in tems of E. (c) Expess the adius fo a cicula obit in tems of J. (d) A satellite is to be launched in a cicula obit of adius a, what should be its obital speed? [Hint: What is the speed of a paticle moving in an obit with constant?] 7

2 2. Solution: (a) The obit is cicula if the smallest and lagest values of (i.e. the tuning points ae equal) in othe wods the two solutions to the quadatic equation ae the same, o the disciminant is zeo 4µ 2 m 4 + 8EmJ 2 = 0 o E = µ2 m 3 2J 2 (b) If the disciminant of the quadatic is zeo, then the two oots ae both (c) Combining the two pevious esults gives a = µm 2E a = J 2 µm 2 (d) Conside the fist integal J = m 2 φ and note that fo a cicula obit with = a the angula velocity φ is constant and v = a φ. Togethe this gives v = aj/(ma 2 ) = J/(ma) and using the esult fom c) this simplifies to v = µ/a. 3. Elliptical Obits fo the Keple potential (a) Show that fo a given value of J the smallest enegy occus fo a cicula obit and find this value. [Hint: Remembe that fo bound obits E < 0.] (b) Likewise show that fo a given enegy E the maximum value of J occus fo a cicula obit. (c) Using the esults fom the text wite the eccenticity in tems of E and J. [Hint: Conside the fomula fo an ellipse given in the text and max + min.] 3. Solution: (a) An elliptic obit has two eal tuning points and thus fom Q we have 4µ 2 m 4 + 8EmJ 2 0 If J is fixed, then the smallest allowed value fo E occus when this expession is zeo, which is pecisely the condition fo a cicula obit. Reaanging gives E E min = µ2 m 3 (b) If we e-aange the same fomula and make J the subject we get J J max = µ2 m 4 2Em Remembe to evese the sign of an inequality when dividing o multiplying by negative quantities. 2J 2 8

3 (c) The two tuning points ae the two oots of the quadatic 2Em 2 + 2µm 2 J 2 = 0 thus max + min = µm E Fom the fomula fo an ellipse max = l/( e) and min = l/( + e) so Thus o max + min = 2l e 2 = µm E 2l e 2 e 2 = + 2lE µm = + 2J 2 E µ 2 m 3 Notice that e < fo a bound obit, which coesponds to E < The invese squae potential Conside the potential with a and w constant. V () = 2 m ( a w ) 2 (a) Detemine the conditions fo a cicula obit. (b) Give a pactical intepetation of the paametes a and w in the potential. 4. Solution: (a) Fo a cicula obit = 0 always and fom (3.4) 0 = J 2 m 3 m(aw)2 3 o J = maw. The fist integal with ṙ = 0 on the othe hand gives J 2 m ma2 w 2 = 2ER 2 fo a cicula obit of adius R. So togethe this gives which means E = 0 fo a cicula obit. 0 = 2ER 2 (b) The speed V in an obit o adius R is given by Simplifying gives 2 mv 2 2 m ( a w R RV = aw ) 2 = E = 0 Thus, the poduct of adius and speed is a univesal constant fo all cicula obits in this potential, in paticula if adius is a, then speed is w. 9

4 5. Cicula obits fo abitay potentials Conside an abitay cental potential V (). (a) Show how to combine a fist integal and an Eule-Lagange equation to obtain an equation in the fom (b) Expess A and f() in tems of J and V (). + A + f() = 0. 3 (c) Thus expess the angula momentum J fo a cicula obit in tems of its adius a. 5. Solution: (a) The Lagangian is L = 2 m(ṙ2 + 2 φ2 ) V (). Thee is an immediate fist integal fom L/ φ = 0, The Lagange equation fo is L φ = m2 φ = J. m m φ 2 + dv d = 0 and on dividing by the mass and substituting in J (b) Thus, A = J 2 /m 2 and f() = V ()/m. J 2 m dv m d = 0. (c) Fo a cicula obit = 0, thus A = 3 f() = a 3 f(a), o in tems of angula momentum J 2 = m 2 A = m 2 a 3 f(a) = ma 3 V (a). 6. Hamonic potential in catesian coodinates Conside the cental potential V (x, y) = 2 k(x2 + y 2 ). (a) Woking in catesian coodinates, wite down both Eule-Lagange equations. (b) Solve them. [Hint: Do not use any fist integals, the equations ae easie to solve diectly.] 6. Solution: (a) Let k = mω 2. The Lagangian is The Lagange equations ae ( ) d L dt ẋ d dt ( L ẏ L = 2 m(ẋ2 + ẏ 2 ) 2 mω2 (x 2 + y 2 ). L x = 0 o d dt (mẋ) + mω2 x = 0 ) L y = 0 o d dt (mẏ) + mω2 y = 0. 20

5 These fom two decoupled equations ẍ + ω 2 x = 0 ÿ + ω 2 y = 0, (b) The solutions to these equations descibe simple hamonic motion about the oigin with angula fequency ω, x = A cos ωt + B sin ωt y = A 2 cos ωt + B 2 sin ωt. 7. Hamonic potential in plane pola coodinates Wite the potential as V () = 2 k2. (a) Wite down the Eule-Lagange equation fo and two fist integals. (b) Show that the equation fo the tuning points is a quatic (fouth-ode polynomial). The quatic should be special enough to solve easily. (c) Show that the diffeential equation fo can be sepaated to give a solution of the fom k m (t t d 0) = (A 2 )( 2 B). Give an intepetation of A and B. (d) Evaluate the integal. [Hint: Use the substitution 2 = B + (A B) sin 2 α.] (e) Fom the solution fo (t) go back to the fist integal fo φ and use it to detemine φ(t). ( ) k [Hint: Use the substitution w = tan m (t t 0).] 7. Solution: Let k = mω 2. (a) In pola coodinates T = 2 m(ẋ2 + ẏ 2 ) = 2 m(ṙ2 + 2 φ2 ) V = 2 mω2 (x 2 + y 2 ) = 2 mω2 2, So L = 2 m(ṙ2 + 2 φ2 ) 2 mω2 2. Since L/ φ = 0, thee is a fist integal, L φ = m2 φ = C, a constant. This is the angula momentum mh, so 2 φ = h. Also L/ t = 0, so thee is a second fist integal, ṙ L ṙ + φ L φ L = C 2, anothe constant. This epesents the enegy me, so 2 (ṙ2 + 2 φ2 ) + 2 ω2 2 = E. (H.6) 2

6 (b) The motion in is limited by the points at which ṙ = 0. These ae given by o, using φ = h/ 2, o 2 2 φ2 + 2 ω2 2 = E, h ω2 2 = E, 4 2E ω h2 ω 2. The oots of this quadatic in 2 ae given by = 2E ω 2, = h2 ω 2. (c) Now ewite (H.6) as 2 ṙ 2 = 2E 2 2 φ2 ω 2 4 ( = ω 2 4 2E ) ω 2 + h2 ω 2 = ω 2 ( 2 2 )( 2 2 2) = ω 2 ( 2 2 )( ), whee we assume the odeing 2. Then ṙ = ± ω ( 2 2)(2 2 2 ), which can be sepaated and integated, ( 2 2)(2 2 2 ) d = ±ω k dt = ± m (t t 0). The paametes B = 2 and A = 2 2 ae obviously the squaes of the tuning points, o peihelion and aphelion espectively. (d) If we substitute 2 = 2 + (2 2 2 ) sin2 χ so that 2 d = 2(2 2 2 ) sin χ cos χdχ, the left-hand side is ( 2 2)(2 2 2 ) d = dχ = χ. Hence i.e. (e) Fom and sepaating, we find (t) = φ = h 2 = dφ = χ = ±ωt + C, 2 + (2 2 2 ) sin2 (ωt + C). h 2 + (2 2 2 ) sin2 (ωt + C) h 2 + (2 2 2 ) sin2 (ωt + C) dt. 22

7 Witing C = ωt 0 and τ = tan ω(t t 0 ) the ight-hand side becomes h 2 + (2 2 2 ) sin2 (ωt + C) dt = h ω τ 2 dτ = h ω 2 tan 2τ = tan 2τ, since 2 = h/ω. Hence ( ) φ φ 0 = tan 2 tan ω(t t 0 ). A vey nasty choice of co-odinates indeed! 23

KEPLER S LAWS AND PLANETARY ORBITS

KEPLER S LAWS AND PLANETARY ORBITS KEPE S AWS AND PANETAY OBITS 1. Selected popeties of pola coodinates and ellipses Pola coodinates: I take a some what extended view of pola coodinates in that I allow fo a z diection (cylindical coodinates