Orbit: impulse making orbit spherical

AI Thread Summary
The discussion focuses on analyzing the effects of an impulse applied to a satellite in a highly elliptical orbit at its perigee. The impulse leads to changes in the semi-major axis (a) and the semi-latus rectum (l), with the relationship δl = δa(1-e)² being established. It is noted that the impulse decreases the orbital period and apogee distance while keeping the perigee distance constant. The confusion arises regarding how the eccentricity (e) is affected, as changes in a should theoretically alter e. The final solution indicates that the perigee remains unchanged despite the adjustments in a and l, which prompts further clarification on the calculations involved.
Halleluwah
Messages
5
Reaction score
0

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 I. By considering changes in energy and angular momentum, find the changes in a (semi major axis) and l (semi latus rectum). Show that \delta l = \delta a (1-e)^2, 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.

Homework Equations


<br /> a= \frac{l}{1-e^2} = \frac{GMm}{2|E|}
<br /> l = \frac{J^2}{GMm^2}
where J is the angular momentum.

The Attempt at a Solution


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

Any hints?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top