Orbit: impulse making orbit spherical

In summary, the perigee distance decreases while the apogee distance increases when a satellite is given a retarded force.
  • #1
Halleluwah
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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 [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.

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.

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.
 
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  • #2
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.
 
  • #3
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].
 
  • #4
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?
 

1. What is the definition of "orbit"?

Orbit refers to the path of an object as it revolves around a larger object, such as a planet orbiting a star.

2. How is an orbit created?

An orbit is created when an object has enough velocity and the right direction to counteract the pull of gravity from another object, allowing it to continuously fall towards the object without getting any closer.

3. What is "impulse" in relation to orbit?

Impulse refers to the change in momentum of an object. In the context of orbit, it is the force applied to an object to alter its velocity and create a specific orbit around another object.

4. Why is an orbit considered to be spherical?

An orbit is considered to be spherical because it follows a curved path around an object, similar to the shape of a sphere. This is due to the gravitational pull of the larger object, which causes the object in orbit to constantly change direction.

5. How is the path of an orbit calculated?

The path of an orbit is calculated using mathematical equations, specifically the laws of gravity and motion developed by Isaac Newton. These equations take into account the mass and distance of the objects involved to determine the shape and size of the orbit.

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