Question

1. graded] In class we found the solution о T= 1+Ecos for the Kepler orbit problem with U(r)= -7/r, where 12 Al2 lomr2 = const. EE ут ут and A is a constant determined by initial conditions (we've set the phase 0o =0 for simplicity.) We also saw that we could think of this as a problem with effective potential 12 Ueff U(r) 2mr2 i.e = const Ueff = E (a) By plugging the solution into the energy equation, find the energy E (b) Find the relation between E and e, and comment on the relation in terms of bound and unbound orbits.

Answer 1

2mit According to question total aneasy & Imut + V (6) here, V(m) = - K (for Central fore). So that, E= 1 mer le – k. 1 X2 amir as & is Constant, we com eraluate it at any convenient point on the orbit. suppose we take the Froming point at which so is This at so is mim mum, min mun say smin the turning points will be no. fromen & ut se - K for a o to bottom p Hecord at min of Cono-1 again: PA Kein - THE pe e Wfr the value of s put in (3) Ay mini and (2) m mk min - rek(+6) - am rimir for sein, becomes which on E = substituting mik (hy), ale

26. k (ED - (L+ PEF) - mkr / o oln I Care bounding & co, giving conic is ellipse, periodic & - . k-6 р, с, . o And for unbounding Exo giring 671 - comic is hyperbola – von Periodic motion. According to Englion, just replaces the values of then paso, k=8, 024. Calenlab.