MSL Cruise Phase |
MSL Cruise Phase |
Nov 26 2011, 03:50 PM
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#1
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Merciless Robot Group: Admin Posts: 8789 Joined: 8-December 05 From: Los Angeles Member No.: 602 |
Okay, we're off and running! Please post all comments relating to MSL's transit to Mars here.
-------------------- A few will take this knowledge and use this power of a dream realized as a force for change, an impetus for further discovery to make less ancient dreams real.
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Nov 27 2011, 03:34 AM
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#2
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Member Group: Members Posts: 214 Joined: 30-December 05 Member No.: 628 |
I'm having some trouble with "data dropouts" myself while trying to view the videos - I wanted to review the telemetry data on the evolution of perigee and apogee during the second Centaur burn, because the first time through I did not understand what I was seeing. My recollection is that after a steady increase the apogee figures dropped abruptly somewhere over Madagascar. This may have simply indicated a move to a higher power of 10 on the display but it was too blurry to be sure. The perigee seemed to be stuck somewhere in the 80's or -80's (couldn't tell if it was a negative sign or a "star" in the simulation). This I really did not understand because it persisted even after the spacecraft was well on its way to Mars. Is it just that after a certain point the perigee ceased to update? Maybe some rocket scientist here can explain how the perigee figure would be expected to evolve if we actually continued to track it as the spacecraft approaches escape velocity. Seems to me both apogee and perigee would eventually have to go to infinity at the point where the vehicle transitions to a solar orbit but when it becomes possible to view the video without a "please try again later" message I am sure it will confirm that this is not what we actually saw.
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Nov 27 2011, 02:23 PM
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#3
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Junior Member Group: Members Posts: 89 Joined: 27-August 05 From: Eccentric Mars orbit Member No.: 477 |
(Full inline quote removed- Mod)
The apogee for a perfect parabola is infinite, but if you run the formulas to find the perigee and apogee of an ellipse, on a hyperbola, you will get the correct perigee but a finite, negative apogee. Obviously a distance can never be negative (you can never be closer to me than at my same position, with zero distance) but you can run all the formulas in reverse with this negative apogee and get the correct position and velocity of the spacecraft. Which brings me to my second point: There is in theory enough information in the elements to get the position and velocity of the spacecraft during the burns, if they are all consistent. One thing I don't know is how they handle "altitude". A really common way to do it is to take the radius distance from the center and subtract the equatorial radius of the Earth, but since the Earth is not a perfect sphere, this would result in a negative altitude at launch. So I don't know what you have to add to get back the radius vector, and it may be two different things for different altitudes. I remember seeing one of these simulations where the altitude wasn't in between periapse and apoapse. Back to the original point: Since the apogee took one value and stuck with it after escape velocity was achieved, maybe they just put in some fill value, like -9999999 meters, and translated it to nautical miles. In which case, after escape, the orbital elements become insufficient to reconstruct position and velocity. |
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Nov 27 2011, 04:08 PM
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#4
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Senior Member Group: Members Posts: 1670 Joined: 5-March 05 From: Boulder, CO Member No.: 184 |
Back to the original point: Since the apogee took one value and stuck with it after escape velocity was achieved, maybe they just put in some fill value, like -9999999 meters, and translated it to nautical miles. In which case, after escape, the orbital elements become insufficient to reconstruct position and velocity. Yes those numbers were fascinating. I assumed we just had one look at the numbers after the orbit went hyperbolic (eccentricity > 1). It seemed a reasonable value of negative apogee for a hyperbolic orbit. Osculating orbital elements of course can always be converted to an instantaneous position and velocity. I wrote a FORTRAN subroutine a long time ago that does this conversion - at least for heliocentric orbits. I wonder what the earth-relative velocity and eccentricity values were when the engines cut off? It takes about 3.2 km/sec delta-V to go from low-Earth orbit to reach escape velocity (11.3 km/sec). Another 0.6 km/sec or so is needed to get to a Mars transfer orbit, though it looks from this press-kit excerpt that the actual excess velocity is more like 3.3 km/sec. Orbit at SC Separation Perigee: 104.0 km Inclination: 35.5 deg Hyperbolic Departure Hyperbolic Excess Velocity Squared (C3): 10.78 km2/sec2 Declination of the Launch Asymtote (DLA): -1.10 deg Right Ascention of the Launch Asymtote (RLA): 126.6 deg Approximate Values Orbit parameters shown for launch on 25 Nov 2011 at 10:25 a.m. EST. And the following velocity equation from Wikipedia can help get back the semimajor axis, and then the eccentricity: http://en.wikipedia.org/wiki/Elliptic_orbit Steve |
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Nov 28 2011, 08:58 PM
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#5
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Junior Member Group: Members Posts: 89 Joined: 27-August 05 From: Eccentric Mars orbit Member No.: 477 |
(Full inline quote removed- Mod)
The guys at JPL SSD (that do the Horizons ephemeris program) got the spice kernel for a projected launch at what happened to be the actual launch time, 26 Nov at start of window. Since the launch was accurate (<0.1 sigma) this is probably pretty good. You can't get the kernel from them, but you can run Horizons and get any form of vectors or elements you want, which may be even better than a kernel. Earth departure according to the kernel: Kernel starts at 2011-NOV-26 15:52:12.3830 CT (not UTC, about a minute difference. UTC is 2011-11-26T15:51:06.200 at kernel start) Periapse was 798.736 seconds before this, 13m18.736 seconds, so periapse was at 2011-Nov-26 15:38:53.647 CT (15:37:47.464 UTC) Periapse distance: 6572.438km from the center of the Earth, or about 194km altitude Eccentricity: 1.17677 From this, velocity at periapse was 11.490km/s. This was 476m/s above escape speed at this altitude. Hyperbolic excess speed (v_inf, eventual speed of departure from Earth) is 3.274km/s, for a C3 of 10.721 Spaceflightnow reported centaur main engine start 2 at 32:40 MET (15:34:40 UTC) and cutoff at 40:30 MET(15:42:30 UTC) so theoretical periapse is during the centaur burn, which is kind of as expected. The second burn also was used to increase the inclination, so it was not purely in plane. The parking orbit was something like 28deg inclination, while departure was at 34.5deg. This is weird, since you should be able to launch at an azimuth such that no plane change is needed in the second burn. |
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