Question

1. (graded] In class we found the solution ro r=1+ ecos for the Kepler orbit problem with U(T) = -7/r, where A12 20, l= omr2 = const., and A is a constant determined by initial conditions (we've set the phase do = 0 for simplicity.) We also saw that we could think of this as a problem with effective potential Ueff = U(T) + 2mr2 i.e. mr? + Ueff = E =const. (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

d'l rral Itt cond: M. toe ste sind (6) i et sindur gun e El Bind m . 1 / 12 /1 + f o 'o + 2 + lost ) + teisino - mkr. - Pm nay Life P& Pai cod -21? - 21?loset amo? l? 16²-17 mr ? relation blo E df is F ile = 1) (>1/unborno
​​​​​​the relation between E and $\epsilon$ tells that if value of $\epsilon$ is less than 1 then energy is negative and it is called bound state.. If $\epsilon$ >1 then energy become positive and orbit become unbounded.

For unbounded orbit energy is greater than zero and orbit is named as hyperbolic orbit.

For bounded orbit energy is less than zero and orbit named as elliptical orbit.

For zero energy orbit is parabolic.

When