And you’d be right. I’m not sure where that came from.
Stranger
Maybe because 2 ellipses with random orientation
dracoi
August 19, 2010, 5:09pm
23
flight:
I solved a closed-form solution to this. It holds for velocity changes in or opposite the direction of motion when you start with a circular orbit. It gets a bit more complex when you start with an elliptical orbit and depends on where in the orbit you make the change. Anyway, it looks like this:
%r = (%V)^2/(2 - (%V)^2)
where %r is the percent change (divided by 100) in the distance to the Earth at the opposite end of the orbit and %V is the percent change (divided by 100) in velocity. So, if I throw the wrench forward at 1% of my velocity (%V = 1.01), the distance from the Earth at the opposite side of its orbit (the most extreme change) will be 4.1%.
For small changes (which is almost always going to be the case, the percentage change in radius is 4 times the percentage change in velocity. I think this is what you were looking for.
That’s awesome! There are definitely enough near-circular orbits to make this a useful shortcut sometimes.
And thanks to everyone for the additional responses. I’m bookmarking things to come back and read in more detail when I have time to really digest some of the resources posted.
