Problem 3 Solution Page 1. 1A. Assuming as outlined in the text that the orbit is circular, and relating the radial acceleration

Transcription

1 Prolem 3 Solution Page Solution A. Assug as outlined in the text that the orit is circular, and relating the radial acceleration V GM S to the graitational field (where MS is the solar mass we otain Jupiter's orital GM S 4 speed V.36 m/s. The following alternatie solution is also acceptale: B. Since we treat Jupiter s motion as circular and uniform, V π ω, where J is the reolution period of Jupiter, which is gien in the list of the general phsical constants. J. The two graitational forces on the space proe are equal when GMm ρ GM Sm ( ρ (where ρ is the distance from Jupiter and M is Jupiter s mass, whence ( and M ( ρ ρ M (3 M ρ m (4 M + M S S and therefore the two graitational attractions are equal at a distance of aout 3.3 million kilometers from Jupiter (aout 334 Jupiter radii. 3. With a simple Galilean transformation we find that the elocit components of the proe in Jupiter's reference frame are ' x V ' and therefore - in Jupiter's reference frame the proe traels with an angle θ arc tan with V V V respect to the x axis and its speed is ' + V (we also note that cosθ + V '

2 Prolem 3 Solution Page and sin θ. + V ' Using the gien alues we otain θ.653 rad 37.4 and '.65 4 m/s. 4. Since the proe trajector can e descried onl approximatel as the result of a two-od graitational interaction (we should also take into account the interaction with the Sun and other planets we assume a large ut not infinite distance from Jupiter and we approximate the total energ in Jupiter's reference frame as the proe's kinetic energ at that distance: m The corresponding numerical alue is E GJ. E ' (5 5. Equation ( shows that the radial distance ecomes infinite, and its reciprocal equals zero, when E' + + cosθ (7 namel when cosθ (8 E' + We should also note that the radial distance can't e negatie, and therefore its acceptale alues are those satisfing the equation E' + + cosθ (9 or cosθ ( E' + The solutions for the limiting case of eq. ( (i.e. when the equal sign applies are: / E' θ ± ± arccos + ± π arccos ( E' +

3 Prolem 3 Solution Page 3 and therefore the angle θ (shown in figure etween the two hperola asmptotes is gien : θ ( θ θ π + π arccos π arccos E' G M ( In the last line, we used the alue of the total energ as computed in the preious section. 6. The angular deiation is a monotonicall decreasing function of the impact parameter, whence the deiation has a maximum when the impact parameter has a imum. From the discussion in the preious section we easil see that the point of nearest approach is when θ, and in this case the imum distance etween proe and planet center is easil otained from eq. (: 4 ' ' r + + (3 GM G M B inerting equation (3 we otain the impact parameter GM r + r (4 ' We ma note that this result can alternatiel e otained considering that, due to the conseration of angular momentum, we hae L m' m ' r where we introduced the speed corresponding to the nearest approach. In addition, the conseration of energ gies E m' m' GMm r

4 Prolem 3 Solution Page 4 and coing these two equations we otain equation (4 again. The impact parameter is an increasing function of the distance of nearest approach; therefore, if the proe cannot approach Jupiter's surface less than two radii (and thus r 3 B, where B is Jupiter s od radius, the imum acceptale alue of the impact parameter is 6GM 9B + B (5 ' From this equation we finall otain the maximum possile deiation: θ max π arccos π arccos (6 4 4 ' ' + + 6GM + 9 B B G M G M ' and using the numerical alues we computed efore we otain: m 7. B and θ max.56 rad The final direction of motion with respect to the x axis in Jupiter s reference frame is gien the initial angle plus the deiation angle, thus θ + θ if the proe passes ehind the planet. The final elocit components in Jupiter's reference frame are therefore: ' ' x 'cos( 'sin( whereas in the Sun reference frame the are cos( V x 'sin( Therefore the final proe speed in the Sun reference frame is

5 Prolem 3 Solution Page 5 " ( 'cos( V + ( 'sin( + V + V + V ' V cos( ' V V ( cosθ cos θ sinθ sin θ ( V cos θ sin θ ( + V sin θ + V ( cos θ (7 8. Using the alue of the maximum possile angular deiation, the numerical result is.6 4 m/s.

6 Prolem 3 Solution Page 6 Grading guidelines..4 Law of graitation, or law of circular uniform motion.4 Correct approach.4+.3 Correct results for elocit of Jupiter..3 Correct approach.4+.3 Correct results for distance from Jupiter 3. Correct transformation etween reference frames.3+. Correct results for proe speed in Jupiter reference frame.3+. Correct results for proe angle 4..8 Understanding how to handle the potential energ at infinit. Numerical result for kinetic energ 5..6 Correct approach.6 Equation for the orientation of the asmptotes.8 Equation for the proe deflection angle Correct results for imum impact parameter.3+. Correct results for maximum deflection angle 7..5 Equation for elocit components in the Sun reference frame.5 Equation for speed as a function of angular deflection 8..5 Numerical result for final speed For correct results two possile marks are gien: the first one is for the analtical equation and the second one for the numerical alue. For the numerical alues a full score cannot e gien if the numer of digits is incorrect (more than one digit more or less than those gien in the solution or if the units are incorrect or missing.