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Watching the NASA JPL video NASA Launches InSight to Mars (Part 2) after MECO-2, the orbital parameters shown in the video don't make sense to me; specifically the apogee altitude.

At this point in time I believe the 2nd stage has passed escape velocity from Earth and is on its way to Mars, though there will be further maneuvers to prevent the 2nd stage from hitting the planet.

The perigee altitude is about 62 nautical miles, or about 115 km, but the apogee altitude is -3443.92 nautical miles, which is negative. However, using a factor of 1.852 that converts -6378.14, almost exactly matching the 6378.137 conventional value used for the equatorial radius of the Earth when converting altitudes to distances from center of the Earth.

I'm aware that using a negative semi-major axis is perfectly legitimate in some cases, when done with care, but not sure how to understand an apogee distance from Geocenter of precisely zero. Should the display really be showing NaN or blank because there is no meaningful value to display? Or could there in fact be some real math to understand here?

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  • $\begingroup$ It was not launched on equator. The distance to center is shorter from Cali (if I understand oblateness). Might the center be in the focus of the elliptical pseudo-orbit? But the exact match does not make much sense for such theory.. $\endgroup$
    – jkavalik
    May 6, 2018 at 5:55
  • $\begingroup$ @jkavalik see the answer to which I've linked "converting altitudes to distances from center of the Earth." 6378.137 km is a standard convention. $\endgroup$
    – uhoh
    May 6, 2018 at 6:00

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@Mark Adler did a thorough exposition here (https://www.quora.com/How-can-an-apogee-be-negative) of how, mathematically anyway, a computed apoapsis can have a negative altitude. But I've gone through the algebra of Keplerian orbits and determined that there is only one mathematical situation that can give an apoapsis altitude equal to the negative of the primary's radius: the periapsis velocity is zero and all the primary's mass is concentrated in a point at the barycenter, a singularity. This is a degenerate orbit, a straight line, that is the limit as the periapse velocity approaches zero with the stipulation that there's nothing to run into (like rocks) on the way toward the singularity. In that case the "radius of the primary" is now meaningless, since the primary is now a singularity and doesn't have the radius initially specified. "Apoapsis" is now meaningless as well, since it is at a smaller radius than the periapsis and should be called the periapsis. At the time that negative value was displayed, the periapsis velocity was decidedly not zero!

I conclude that the value on the display is not the result of any meaningful calculation, but is probably the result of some code in the display driver substituting something—anything!—when the usual code that calculates the displayed values returns a parameter that says, "Something is wrong." It should have been coded to display "NAN" or some error code.

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    $\begingroup$ Seems likely that the underlying code works in radii (not altitude), and returned "0" for the invalid apoapsis, from which the display routine subtracted the radius of Earth to get altitude. $\endgroup$
    – Chris
    May 7, 2018 at 1:52
  • $\begingroup$ Good point, @Chris! $\endgroup$
    – Tom Spilker
    May 7, 2018 at 2:21
  • $\begingroup$ Although this answer addresses how the software returns a negative value, it does not address the physics of the situation at hand. $\endgroup$
    – ChrisR
    May 8, 2018 at 0:49
  • $\begingroup$ @ChrisR , take a look at Mark Adler's treatment of it in the Quora discussion linked in my answer above. $\endgroup$
    – Tom Spilker
    May 8, 2018 at 0:52
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There is in fact a very good reason for these values but I admit it doesn't make a whole lot of sense at first.

It comes down to the transition from a closed (elliptical) orbit to an open (hyperbolic) orbit. This occurred for Insight when it escaped the Earths gravity well and it's orbital eccentricity became greater than 1.

Now the meaning of the apogee for a hyperbolic orbit isn't intuitively obvious, but if the spacecraft has 'just' escaped, i.e. it's specific energy is zero then the apogee will be infinity. But if it has a higher specific energy then the apogee will be negative, think of it as going through an asymptote on a graph and becoming negative.

Hope this makes some sense, it's a tricky concept to get your head around. Here is a link that may explain it better than I.

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    $\begingroup$ This is the correct answer. It's a hyperbolic orbit, and as such it has a negative semi-major axis, and positive specific energy around Earth. However, you'll note that the vehicle will have a negative specific energy in a heliocentric orbit, and be in an elliptical orbit as well. (They go hand in hand.) $\endgroup$
    – ChrisR
    May 8, 2018 at 0:48
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    $\begingroup$ @ChrisR , there's no quibble with a negative semi-major axis, uhoh's question states that explicitly. But his question is about the precisely -3443.92 nm altitude which yields a primary-centric radius of zero. There's no non-degenerate solution to the equations that will give that, and that's what I was trying to say in my answer above. The InSight trajectory had better not be a degenerate one! :) I think Chris's comment above is probably right. $\endgroup$
    – Tom Spilker
    May 8, 2018 at 1:31

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