Calculating Descent Insertion

[Hlynkacg]

So I am familiar with the equation for determining altitude and velocity at perigee but I could use some help calculating changes in the above.

For instance how would I go about determining the dv required to turn a 200 mile circular orbit into a 200 x 50 mile orbit?

 

[Andy44]

Since the two orbits have the same apogee heights, the maneuver will occur at apogee (200 km), so determine the velocity at apogee for each orbit, than subtract one from the other to get the delta-V. I don't have my notes with me now but that's where you start with this problem.

 

[Keithth G]

Let's call the apogee radius [MATH]r_a[/MATH]; and the perigee radius [MATH]r_p[/MATH]. (Note that this are 'radii' not 'altitudes'. To convert to altitudes you have to subtract the radius of the Earth.

Now, if we assume that orbits are elliptical, we can say that the speed at perigee is:

[MATH]v_p = \sqrt{\frac{r_a (2 \, \mu )}{r_p \left(r_a+r_p\right)}}[/MATH]
and the speed at apogee is:

[MATH]v_a = \sqrt{\frac{r_p (2 \, \mu )}{r_a \left(r_a+r_p\right)}}[/MATH]
where [MATH]\mu[/MATH] is the gravitational constant for Earth (398600.4418 km^3 s^-2)

Now, these equations are always true for elliptical orbits. In particular, they are true if you are in a circular orbit with orbital radius, [MATH]r_a[/MATH]. In this, case

[MATH]v_a = v_p = v_c = \sqrt{\frac{\mu }{r_a}}[/MATH]
So, the delta-V required to convert your circular orbit to an elliptical one is:

[MATH]v_c - v_a = \sqrt{\frac{r_p (2 \, \mu )}{r_a \left(r_a+r_p\right)}} - \sqrt{\frac{\mu }{r_a}} [/MATH]
(N.B.. to be consistent with units, it is easiest to work in 'km' and 'seconds'. So, this expression will calculate a delta-V in km/s. To convert to m/s, multiply by 1000.

Also note that the number should be negative - indicating a retrograde burn - since you are lowering one side of your orbit)

 

[Hlynkacg]

Thank you Kenith, that did the trick. But know I am working a related (and possibly far more difficult) problem.

Given given a sufficient negative/positive dV i know that my current position will become the apogee/perigee and my perigee/apogee will be 180 degrees out from that.

That said how would I go about figuring out the new position of apogee/perigee for a given dv?

To give context, the idea is to calculate the burn required to place my perigee at a specific altitude and lat long, or barring that, ensure that I am within a target altitude/velocity range when I reach a target position.

as an aside I would also like to be able to calculate the change in altitude over a given burn. IE if I burn retrograde for x seconds how much altitude will I loose. I currently have a first order estimate using vs and effective acceleration due to gravity but I'm hoping that there might be a more precise option available .

 

[Keithth G]

as an aside I would also like to be able to calculate the change in altitude over a given burn. IE if I burn retrograde for x seconds how much altitude will I loose. I currently have a first order estimate using vs and effective acceleration due to gravity but I'm hoping that there might be a more precise option available .

This bit is the easy part of your question.

Starting with the equations of motion and expanding to third order, and if Ive got my algebra right, I calculate that (if you start in a circular orbit), the change in altitude during the course of a burn is given by the expression:

[MATH]\delta r = \frac{1}{3}\frac{v_c}{r_c}\delta t^2\,\delta V[/MATH]
where [MATH]v_c[/MATH] is the orbital speed of the circular orbit prior to the burn; [MATH]r_c[/MATH] is the radius of the circular orbit prior to the burn; [MATH]\delta t[/MATH] is the duration of the burn; and [MATH]\delta_V[/MATH] is the change in prograde speed.

For a retrograde burn [MATH]\delta V < 0[/MATH], so there is reduction in height - as expected.

If, then, one executes a 100 m/s retrograde burn over a period of 5 seconds while in a circular parking orbit of around the Earth at, say, an orbital radius of around 6570 km and orbital velocity of approximately 7500 m/s, then the reduction in altitude during the burn is 0.95 m.

It is possible to calculate higher order terms without much difficulty. But unless one is executing a very long burn, higher order corrections will be insignificant.

 

[indy91]

If you are ready for some vector math, the geniuses at MIT thought they could include the calculations for the DOI and lunar landing time in the AGC (they ran out of HDD space though :lol: ), so the guidance equations are all available here: http://www.ibiblio.org/apollo/Documents/Preliminary-GSOP-LunarLandingMission-Section5-LGC.pdf

Page 117 has the equation for the DOI maneuver and on pages 251 and following it has the lunar landing time prediction routine. Maybe the exact equations won't help you much, but there is an illustration of the geometry for the landing maneuvers.