Central Force Motion Challenge Problems

Central Force Motion Challenge Problems Problem 1: Ellitic Orbit A satellite of mass m s is in an ellitical orbit around a lanet of mass m which is located at one focus of the ellise. The satellite has a velocity v a at the distance r a when it is furthest from the lanet. a) What is the magnitude of the velocity v of the satellite when it is closest to the lanet? Exress your answer in terms of m, m, G, v, and r as needed. s a a b) What is the distance of nearest aroach r? c) If the satellite were in a circular orbit of radius r 0 = r, is it s velocity v 0 greater than, equal to, or less than the velocity v of the original ellitic orbit? Justify your answer. Problem 1 Solution: The angular momentum about the origin is constant therefore L! L = L a. (8.1.1) Because m << m, the reduced mass µ! m and so the angular momentum condition s s becomes Thus L = m r v = m r v (8.1.) s s a a r = r v / v. (8.1.3) a a Note: We can solve for v in terms of the constants G, m, r and v as follows. Choose a a zero for the gravitational otential energy for the case where the satellite and lanet are

searated by an infinite distance, U (r =!)= 0. The gravitational force is conservative, so the energy at closest aroach is equal to the energy at the furthest distance from the lanet, hence Substituting Eq. (8.1.3) into Eq. (8.1.4), yields 1 Gm m s 1 Gms m m! = m v s v a s!. (8.1.4) r a 1 Gm m Gm m v s 1 s msv a! = msv!. (8.1.5) r a ra v a After a little algebraic maniulation, Eq. (8.1.5) becomes r Gm (v! v ) = (v + v )(v! v ). (8.1.6) r v a a a a a We can solve this equation for v v = Gm! v a. (8.1.7) ra v a b) We can use Eq. (8.1.3) to find the distance of nearest aroach, ra v a 1 r = = r a. (8.1.8) v! Gm " $ #1 % & ra v a ' c) The seed of the satellite undergoing uniform circular motion can be found from the force equation, So the seed is Gm m s m v! =! s 0. (8.1.9) r 0 r 0 v 0 = Gm / r 0. (8.1.10)

Note that if we assume our original orbit was circular and set v a = v 0 and r a = r 0 in Eq. (8.1.8) then the distance of closest aroach becomes roviding a second check on our algebra. If we substitute Gm = r0 v 0 into Eq. (8.1.7), we have that 1 r = r = r (8.1.11) 0 0 " Gm % $! 1 ' # r 0 v 0 & r v 0 0 v =! v a. (8.1.1) r v a a Let s comare our ellitic orbit to a circular orbit with r = r 0. How does v comare to v 0? If we substitute r = r 0 into Eq. (8.1.1) we have that v = r v 0 r v a a! v. (8.1.13) a We now use the angular momentum condition that r v = r v to rewrite Eq. (8.1.13) as a a Thus v 0 v =! v. (8.1.14) a v v + v v a = v. (8.1.15) 0 We know that v a < v, so v + v v a < v. Thus v 0 < v or v 0 < v. (8.1.16) So the circular orbit with r = r 0 has seed v 0 less than the seed v of closest aroach for the ellitic orbit. This is not surrising because suose we give the satellite that is in a circular motion a small increase in velocity in the tangential direction, v new = v 0 +!v. This imlies that the energy increases. The satellite will no longer travel in a circular orbit since for the same radius r 0,

m v Gm m s new s >. (8.1.17) r 0 r 0 The satellite will move away from the lanet entering into an ellitical orbit. So any velocity v greater than v 0 will form an ellitic orbit.

Problem : Planetary Orbits Comet Encke was discovered in 1786 by Pierre Mechain and in 18 Johann Encke determined that its eriod was 3.3 years. It was hotograhed in 1913 at the ahelion distance, r a = 6.1! 10 11 m, (furthest distance from the sun) by the telescoe at Mt. Wilson. The distance of closest aroach to the sun, erihelion, is r = 5.1! 10 10 m. The universal!11! 30 gravitation constant G = 6.7 "10 N # m # kg. The mass of the sun is m s =.0!10 kg. a) Exlain why angular momentum is conserved about the focal oint and then write down an equation for the conservation of angular momentum between ahelion and erihelion. b) Exlain why mechanical energy is conserved and then write down an equation for conservation of energy between ahelion and erihelion. c) Use conservation of energy and angular momentum to find the seeds at erihelion and ahelion. Problem Solutions Since the only forces are gravitational and oint toward the focal oint!! "!! focal = r focal,comet " F grav = 0 Therefore, angular momentum is conserved. At erihelion, since v is tangent to the orbit L = µr v At aohelion: L a = µr a! a

L a = L! µr a v a = µr v! r a v a = r v The gravitational force is conservative so mechanical energy is conserved. 1 Gm 1 m 1 Gm 1 m Ea = µv a =, E = µv = r a r thus E a = E! 1 Gm 1 m 1 Gm 1 m! =! µv a µv Since the mass m 1 of the comet is much less than the mass of the sun m. r a r the energy conservation equation becomes: m 1 m m 1 m µ =! = m1 m 1 + m m 1 Gm 1 m 1 Gm 1 m 1 Gm 1 Gm = or = m 1v a! r a m 1v! v a! v! r r a r The condition v = energy equation. r a v a r from conservation of angular momentum can now be used in the 1 1 " % Gm r a v Gm v a a! =! $ ' # & r a r r Solve for v a : 1!!r $ a $! 1 1$ # ' 1 & r & v a = Gm ' " # # & " % % "r r a %

" " 1 1 % % $ Gm $! ' ' $ # r r a & ' v a = $ ' $ " r a %! 1 ' $ $ ' # # r & ' & 1/ " Gm r % 1/ v a = $ ' # r a (r a + r )& ( ) " (6.67 ( 10!11 N ) m ) kg! )(.0 ( 10 30 kg )(5.1 ( 10 m) 10 % 1/ v a = $ ' = 5.8 ( 10 3 m ) s!1 $ (6.1 ( 10 11 m) (6.1 ( 10 11 m + 5.1 ( 10 10 m) ' # ( ) & The velocity at erihelion is then r a v a (6.1! 10 m) 11 v = = r (5.1! 10 10 m ) ( 5.8! 10 3 m " s #1 ) = 6.9! 10 4 m " s #1

Problem 3: Satellite Motion A sherical non-rotating lanet (with no atmoshere) has mass m 1 and radius r. A 1 rojectile of mass m << m 1 is fired from the surface of the lanet at a oint A with a seed v A at an angle! = 30! with resect to the radial direction. In its subsequent trajectory the rojectile reaches a maximum altitude at oint B on the sketch. The distance from the center of the lanet to the oint B is r = (5 / )r 1. In this roblem you will find the initial seed v A in terms ofg, m 1 and r 1. a) Is there a oint about which the angular momentum of the rojectile is constant? If so, use this oint to determine a relation between the seed v B of the rojectile at the oint B in terms of v A and the angle! = 30!. b) Now use conservation of mechanical energy constant to find an exression to find v A in terms ofg, m 1 and r 1. Problem 3 Solutions: a) The only force of interest is the gravitational force, which is always directed toward the center of the lanet; hence angular momentum about the center of the lanet is a constant. At oint A, the comonent! of the satellite s linear momentum erendicular to the radius vector is m v A! = m v A sin" =, (8.3.1)

Using sin30! = 1/. The magnitude of the angular momentum about the center of the lanet is then r 1 m v A LA = r 1! =. (8.3.) At oint B (the aogee), the velocity vector is erendicular to the radius vector and the magnitude of the angular momentum is the roduct of the distance from the center of the lanet and the seed, 5 L B = r v B = r 1 v B. (8.3.3) There is no torque on the satellite, so L B = L A ; so equating the exressions in Equations (8.3.3) and (8.3.) yields 5 r m v 1 A r v = 1 B v A v =. B 5 (8.3.4) b) The mechanical energy E of the satellite as a function of seed v and radius (distance from the center of the lanet) r is m 1 m E = 1 m v! G. (8.3.5) r Equating the energies at oints A and B, and using r B = r = (5 / )r 1, r A = r 1 and v B = v A / 5 from art a) (Equation (8.3.4) above),

1 m 1 m 1 m 1 m m v! G = m v! G A B r A r B 1 m 1 1 " v A % m 1 va! G = ' r 1 # $ 5 &! G 5r 1 / 1 " 1 % m " % v 1 1! ' = G 1! '. (8.3.6) A # $ 5& r 1 # $ 5& va = 5 G m 1 4 r1 v = A 5 G m 1 4 r 1 It s worth noting that v A is less than the escae velocity of the lanet, but not by much; G m1 v = (8.3.7) escae r 1 v A = 5! 0.79. (8.3.8) v 8 escae 1 m 1 m 1 m 1 m m v A! G = m v! G r A B r B 1 m 1 1 " v A % m 1 va! G = ' r 1 # $ 5 &! G 5r 1 / 1 " 1 % m 1 " % ' 1! ' (8.3.9) v A # $1! 5& = G r 1 # $ 5& 5 m G 1 va = 4 r1 v = A 5 G m 1 4 r 1 It s worth noting that v A is less than the escae velocity

G m v = 1 escae (8.3.10) r 1 of the lanet, but not by much; v A 5 =! 0.79. (8.3.11) v 8 escae

Problem 4: Inverse Square Central Force: Lowest Energy Solution The effective otential energy for an inverse-square restoring central force! Gm 1m F1, =! r ˆ is given by r L Gm 1m effective! U = µr r Make a grah of the effective otential energy U effective ( r ) as a function of the relative searation r. Find the radius and the energy for the motion with the lowest energy. What tye of motion does this corresond to? Problem 4 Solution: For lotting uroses, the horizontal scale is the ratio r / r 0 and the vertical scale is in units of U effective /U effective ( r 0 ), as found in the second art of the roblem. The uer (red) curve is roortional to 1/ r and the lower (blue) curve is roortional to!1/ r. The sum is the solid (green) curve in between. To find the minimum energy, differentiate the effective otential U ( ) r with resect effective to the radius r and set the derivative equal to zero at r 0,

The energy at this minimum value is L 1 Gm 1 m! + = 0 µ r 3 r 0 0 L r 0 =. µgm 1 m L! µ Gm 1m " µ Gm 1m effective r 0 = $ # Gm m % 1 µ & L ' L U ( ) µ (Gm m ) 1 = #. L The radius of this orbit does not change; the orbit is a circle.

Problem 5 A article of mass m moves under an attractive central force of magnitude F = br 3. The angular momentum is equal to L. a) Find the effective otential energy and make sketch of effective otential energy as a function of r. b) Indicate on a sketch of the effective otential the total energy for circular motion. c) The radius of the article s orbit varies between r 0 and r 0. Find r 0. Problem 5 Solutions: a) The otential energy is, taking the zero of otential energy to be at r = 0, is and the effective otential is ( ) = " ( br! 3 )dr! = r b r 4 U r # " 0 4 U ( ) L L b r = + U ( r ) = + r 4. mr mr 4 eff A lot is shown below, including the otential (yellow if seen in color), the term L / m (green) and the effective otential (blue). The minimum effective otential energy is the horizontal line (red). The horizontal scale is in units of the radius of the circular orbit and the vertical scale is in units of the minimum effective otential. b) See the solution to art (a) above and the lot to the left below.

c) In the left lot, if we could move the red line u until it intersects the blue curve at two oint whose value of the radius differ by a factor of, those would be the resective values for r 0 and r 0. A grah of this construction (done by comuter, of course), showing the corresonding energy as the horizontal magenta is at the right above, and is not art of this roblem. To do this algebraically, we find the value of r 0 such that U eff ( r 0 ) = U eff ( r 0 ). This is L b 4 L b 4 + r = + (r ). mr 4 0 m(r ) 0 4 0 0 Rearranging and combining terms, and then solving for r 0, 3 L 1 15 4 = b r 0 8 m r 0 4 6 1 L r 0 =. 10 mb Thus, r 0 =(1/ 10 )r the right above. circular (not art of the roblem), consistent with the auxiliary figure on

Problem 6: The effective otential corresonding to a air of articles interacting through a central force is given by the exression L 3 U eff ( r ) = + Cr (30.1) µr where L is the angular momentum, µ is the reduced mass and C is a constant. The total energy of the system is E. The relationshi between U eff (r) and E is shown in the figure, along with an indication of the associated maximum and minimum values of r and the minimum allowed energy E min. In what follows, assume that the center of mass of the two articles is at rest. a) Find an exression for the radial comonent f (r) of the force between the two articles. Is the force attractive or reulsive? b) What is the radius r 0 of the circular orbit allowed in this otential? Exress your answer as some combination of L, C, and µ. c) When E has a value larger than E min, find how raidly the searation between the articles is changing, dr / dt, as the system asses through the oint in the orbit where r = r 0. Give your answer in terms of some combination of E, E min, L, C, µ and r 0. d) Does the relative motion between the articles sto when r = r max? If not, what is the total kinetic energy at that oint in terms of some combination of E, L, C, µ, r max and r min?

Problem 6 Solutions: a) The effective otential is given by and f ( r ) =!du L eff / dr for a central force, and so U ( r ) = + U ( r ), (8.6.) µr d d 3 r =! Cr =!3Cr. (8.6.3) dr dr f ( r ) =! U ( ) From the figure, C > 0, so f ( r ) < 0, a restoring force. b) The circular orbit will corresond to the minimum effective otential; at this radius the kinetic energy will have no contribution from any radial motion. This minimum effective otential, and hence the radius of the circular orbit, is found from basic calculus and algebra,! d " L $ U eff ( r ) % = # + 3Cr 3 0 & dr ' r = r µr0 0 5 L ( L ) 1/ 5 r 0 =, r 0 = * 3µC 3µC +., - (8.6.4) c) Recall that the kinetic energy is L 1! dr " K = + µ # $ µr % dt &. (8.6.5) The difference E! E is then found by evaluating U at r = r, min eff 0 1! dr " E # E min = µ, (8.6.6) $ dt % & r=r 0 ' or dr / dt = ( / µ )( E! E min ). d) No; dr / dt = 0, but the kinetic energy, from Equation (8.6.5), is

K = min L. (8.6.7) µr max

Problem 7: Determining the Mass of a Neutron Star A binary system known as 4U0900-40 consists of a neutron star and a normal otical star. You are given two grahs of actual data obtained from observations of this system. The to grah shows the time delays (in seconds) of X-ray ulses detected from the neutron star as a function of time (in days) throughout its orbit. These delays indicate the time of flight for an X-ray ulse (traveling with the seed of light) to cross the orbit of the binary in its tri to the earth. (Ignore the heavy dots scattered about the x -axis.) The bottom grah dislays the velocity comonent of the otical star toward (or away from) the Earth as a function of its orbital hase. The velocities (in km er second) are determined from Doler shifts in its sectral lines. Time on this lot is given in units of orbital hase, where the time between hases 0.0 and 1.0 corresonds to one orbital eriod. Assume that the orbit of 4U0900-40 is circular and that we are viewing the system edge on, i.e., the Earth lies in the lane of the binary orbit. In each grah, the solid curve is a comuted fit to the individual data oints. You are to find from the data the orbital eriod, orbital radii and the masses of the neutron star and the otical star. You need only give answers to two significant figures. For this roblem, use c = 3.0! 10 8 m " s #1 for the seed of light and G = 6.7! 10 "11 N # m # kg " for the Universal Gravitation Constant. Suggested rocedure (you will need a ruler for the first three arts): a. Determine the orbital eriod, T, in days. b. Estimate the velocity, v 0 of the otical star in its orbit around the center of mass. c. Use T and v 0 to find the radius of the orbit of the otical star, a 0, around the center of mass. d. Estimate the size of the orbit, a N, of the neutron star around the center of mass. First exress your answer in light! seconds, then in meters. e. Find the ratio of the mass of the neutron star, m N, to the mass of the otical star, m O. [Hint: Recall that in any binary m 1 r 1 = m r.] f. Use Keler s law to find the total mass of the binary system. g. Find m N and m O ; exress your answers in units of the mass of the Sun (! 10 30 kg ).

Problem 7 Solutions: a. Between the first and third zero-crossings on the uer grah, the distance is 8.98 cm. The scale of the time axis is 1.00 cm to 1 day, so the orbital eriod, P, is 8.98 days. b. The Doler shift velocities range from a maximum of 16.0 km!s -1 to a minimum of -30.8 km!s -1. (The velocities are not symmetric around zero because the binary system as a whole is moving relative to the solar system.) The otical star s orbital velocity is just half the difference between the maximum and minimum Doler velocities, or r 0 =3.4 km!s -1. c. In one eriod P, moving at velocity v 0, the otical star travels a distance v 0 =!a 0. So the orbital radius, a 0 is just a 0 = 1 #(9.98)(4)(60)(60 $% 3.4km & s '1 ) =.89 ( 10 6 km! " )s ( d. The smallest time delay is -117 seconds, while the largest is 111 seconds. The radius of the neutron star s orbit (in light seconds) is half the difference between these, or a N = 114lt! sec. Since the seed of light is.998 x 10 10 cm!s -1, the distance a N is equal to 3.4 x 10 1 cm. e. The ratio M N /M 0 is just equal to a 0 /a N. That is M N M 0.89! 10 11 cm = = 8.45! 10 " 3.4! 10 1 cm f. The semimajor axis of the absolute orbit is a = a N + a 0 = 3.71! 10 1 cm and the orbital eriod is 7.76 x 10 5 s. Inserting these values into Newton s form of Keler s Third Law. we have, M N + M 0 + 4! a 3 = (! ) G ( M N + M 0 ) (3.71 " 10 1 ) 3 ) = 5.0 " 1034 g (7.76 " 10 5 ) (6.67 " 10 #8! M N $ M g. We use the fa(8.7.1)ct that M N 0 1 + " # M 0 % & = M 0 + M N, and both and M 0 (M 0 + M N ) are known. Then we have 5.0! 10 34 g M 0 = = 4.63! 10 34 g 1 + 8.45! 10 " The remaining mass in the mass of the neutron star, M N = 3.9! 10 33 g. The mass of the sun is 1.99 x 10 33 g, so the masses of the otical star and its neutron star comanion are 3.3 and.0 solar masses, resectively.

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