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Orbit: impulse making orbit spherical

1. Homework Statement
A satellite moving in a highly elliptical orbit is given a retarded force concentrated at its perigee. This is modeled as an impulse [itex]I[/itex]. By considering changes in energy and angular momentum, find the changes in [itex]a[/itex] (semi major axis) and [itex]l[/itex] (semi latus rectum). Show that [itex] \delta l = \delta a (1-e)^2 [/itex], and hence that the effect is to decrease the period and apogee distance, while leaving the perigee distance unaffected. Show that the velocity at apogee increases, while that at perigee decreases.


2. Homework Equations
[tex]
a= \frac{l}{1-e^2} = \frac{GMm}{2|E|} [/tex]
[tex]
l = \frac{J^2}{GMm^2} [/tex]
where [itex]J[/itex] is the angular momentum.

3. The Attempt at a Solution
So I've tried to take differentials of [itex]a[/itex] and [itex] l[/itex] but haven't really got far. I have that [itex]\delta J = -Ib[/itex] where [itex] b[/itex] is the perigee and [itex] \delta |E| = vI [/itex] where [itex]v[/itex] is the velocity. I'm also a bit confused as to how the perigee remains constant according to the answers. The eccentricity should be changing, so what is [itex]e[/itex] in the answer - the previous one? The perigee is [itex] b =a(1-e)[/itex] so if [itex]a[/itex] changes then [itex]e[/itex] should compensate (and [itex]e[/itex] does depend on the angular momentum and energy). Any hints would be appreciated.
 
Looking at this again, we should have
[tex]
\delta a = \frac{b}{(1-e)^2} \delta e
[/tex]
so that [itex]b \delta e = \delta l [/itex] but I'm not sure how to show this.
 
So I've I think I found the relation which I will post. We have [itex] a (1-e^2) = l [/itex] so [itex] \delta a (1-e^2) -2ea \delta e = \delta l [/itex]. Also since [itex] b = a(1-e) [/itex], we have [itex] 0 = \delta a (1-e) -a \delta e [/itex] or [itex] a \delta e = \delta a (1-e) [/itex]. Writing the first relation as [itex] \delta a (1+e)(1-e) - 2a e \delta a [/itex] and replacing the first term with the second relation we have [itex] (1+e)a \delta e - a 2 e \delta e = \delta l [/itex] or [itex] a \delta e(1-e) = b \delta e = \delta l [/itex].
 
So my solution is cheating really, as it assumes that the perigee does not change. The answer given is that [tex] \delta a = -2(I/m) \sqrt{(1+e)a^3/GM(1-e)}[/tex]

Any hints?
 

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