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Wednesday, April 6, 2011

For those who are wondering ...

... Right now the plan is to do a mailbag next week in advance of the spring game, and one the following week after G-Day.

I have a bunch of content to fill the blog the next few days, and will be off this weekend. (Headed back to D.C. to fulfill a personal obligation.)

In the meantime, faithful reader "Albert" sent in this mailbag question earlier in the week, and because I'm too smart to answer it, I'll post it in case anyone wants to take a swing at it:

Problem: A satellite is injected into an orbit at radius r from the center of attraction. The velocity is exactly the desired speed to go into circular orbit but the flight path angle γ≠0. The flight path angle is the angle between the velocity vector and the plane normal to the position vector.

a. Develop the relationship between the resulting eccentricity and sinγ.

b. Develop a relationship between the resulting periapsis distance and sinγ that is valid even for the rectilinear orbit case. Likewise for apoapsis distance.


As always, please show your work.

2 comments:

trumely said...

Let m be the mass of the satellite, even though we don't need it. Let a be the semi-major axis. Let e be the eccentricity.

For a circular orbit: v_0^2 = GM/r_0 (*1)

For our setup: v_0^2 = (v_0 sin γ)^2 + (v_0 cos γ)^2

For the extreme points in the orbit, there is no inward or outward motion.
So by the conservation of angular momentum, we must have

v_± = (r_0 v_0 cos γ) / r_± = (√(G M r_0) cos γ) / r_± (*2)

But from conversation of energy we have

m v_0²/ 2 - G M m/r_0 = m v_±² / 2 - G M m/r_±
or
v_±² - v_0² = 2GM (1/r_± - 1/r_0) (*3)

Combining *1, *2 and *3 we get

(G M r_0 cos² γ) / r_±² - GM/r_0 = 2GM (1/r_± - 1/r_0)
or
r_± = r_0 ± r_0 √(sin² γ) (*4)

From r_± = a(1 ± e) we see that we have:

e = | sin γ |

trumely said...
This comment has been removed by the author.