# Elliptical Orbits

Got a couple of problems when lookin over my notes for Energy and velocity in elliptical orbits when studying for the space dynamics part of my exam next wed and been hit by a few problems:

(i)Wit v1 being the speed at the perihelion (position of closest approach, distance a(1-e) and v2 the speed at the aphelion (distance a(1+e)) and taking into account conservation of angular momentum since there is no torque acting why doe v1*a(1-e)=v2*a(1+e)?

(ii) after accepting the previous equation got one other problem after determining v1 (V1^2=GM[2/(a(1-e)) + D] where D=2E/GMm) and similarly for v2 my notes go on to say we can now form 2 equations for [v1/v2]^2 and set them equal and with a bit of manipulation gives D=-1/a hence showing a=-GMm/2E. How do i go about this?

Dont feel i can move on til other parts of the course till i gain a firm understanding of this. thanx

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Doc Al
Mentor
retupmoc said:
(i)Wit v1 being the speed at the perihelion (position of closest approach, distance a(1-e) and v2 the speed at the aphelion (distance a(1+e)) and taking into account conservation of angular momentum since there is no torque acting why doe v1*a(1-e)=v2*a(1+e)?
This is just a statement of conservation of angular momentum. (Draw yourself a picture showing the velocity of the body at both points.)

any ideas on the second part?

worked out how to do that other part finally just had to do a fair bit of algebraic juggling. Got another problem in understanding my notes (shouldve went to the lectures more really but the recommended text book for the course, university physics is inadequate for this section of the course) Anyway my problem arises in 'Analysing orbits using conservation Laws'

Keeping the discussion general we allow both radial and tangential velocities and again singe angular momentum is conserved between the two interacting bodies (either gravitationally interacting masses or electrostatically interacting charges in this case) i can understand why:

1/2mvr^2=E-U(eff)

U(eff) - effective potential L^2/2mr^2(tangential component of kinetic energy) + C/r (where C=-GMm for gravitational attraction etc))

My prob arises in trying to understand this following part of the lecture notes

For Repulsive forces:
Both terms in the effective potential are positive (i understand this)
Orbit has a minimum balue of r, but no maximum, so orbit is unbound and parabolic ( why is this? im presuming E remains constant here because only conservative forces are acting on the system)

Having similar problems for attractive forces, why is there now a min and max value of r, meaning the orbit is bound (understand that the positive term in Ueff will dominate for small are and the negative term for large r)

Can anyone help me out in understanding this as cant move on to transfer orbits and gravitational slingshots until i understand this

Doc Al
Mentor
To find the max and min values of r, start with the equation for total energy expressed in terms of radial velocity and effective potential. Now set the radial velocity equal to zero and solve (the quadratic) for r. (Remember that only positive values of r make physical sense.)
retupmoc said:
For Repulsive forces:
Both terms in the effective potential are positive (i understand this)
Orbit has a minimum balue of r, but no maximum, so orbit is unbound and parabolic ( why is this? im presuming E remains constant here because only conservative forces are acting on the system)
This should make intuitive sense. (But figure it out as described above.) As the mass approaches the repulsive force center, potential energy increases. The mass can only approach until its radial KE is exhausted. On the other hand, as r increases, the KE increases as PE decreases. PE goes to zero at r = infinity, so the mass never returns. (Yes, E is constant.)
Having similar problems for attractive forces, why is there now a min and max value of r, meaning the orbit is bound (understand that the positive term in Ueff will dominate for small are and the negative term for large r)
In this case PE increases (towards zero) as r increases. Solve the above equation and get two meaningful solutions to the quadratic. Intuitively, as you approach the force center the so-called angular momentum barrier dominates (that's one term in the effective potential) and as r increases the radial KE will expire (assuming the total energy allows it--too much energy and you will escape) as the PE increases (to a max of zero).

I hope this helps a little. 