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I watched the latest SpaceX launch of Intelsat 35e on 05 July 2017 carried by their Falcon 9 rocket into geostationary orbit, and noticed some confusing telemetry data.

Telemetry just before the second burn was as follows:

Velocity: 26503 km/hr Altitude: 248 km

Using the vis-viva equation, I calculated the semi-major axis (a) of this low Earth orbit and found a = 6017.6 km

This means that the semi-major axis of this orbit is SMALLER than the radius of the Earth (R = 6371 km), and therefore, will have a perigee that is BENEATH the Earth's surface....i.e. the orbit is sub-orbital. If the 2nd stage were to continue in this orbit (and not perform the second burn to get it into higher orbit) then doesn't this mean the 2nd stage would crash back into the Earth?

Given that the 2nd stage sat in this sub-orbital low Earth orbit for only about 20 minutes before performing the second burn, it wasn't an issue.

But my question is this: Is parking a 2nd stage in a sub-orbital low Earth orbit safe, even if it is eventually going to get boosted to a higher altitude orbit? Is this routine or common?

Thank you very much! I appreciate it!

-Will

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Excellent question!

That's probably ground velocity or more likely velocity in an ECEF (earth-centered, earth-fixed), also known as ECR (earth-centered rotational) frame that they are displaying, referenced to the rotating frame of the Earth. A hint would be that it is zero before lift-off even though the Earth is rotating.

Wikipedia gives the value of $3.986004418\times 10^{14}$ for $GM_E$ the standard gravitational parameter for Earth. Using $(6378 + 248) \times 1000$ meters for the semi-major axis, the orbital velocity from the vis-viva equation

$$v^2 = GM_E/a$$

would be 7756 m/s. The difference between that an your value of 7362 m/s is only 394 m/s, which is a little less than the Earth's equatorial rotation velocity of

$$\frac{2 \pi \ 6378\times1000 \ m}{36484 \ s} = 464 \ m/s$$

and the ratio is probably close to the cosine of the inclination of the orbit. It's not going to be exact because we're not doing anything exactly, but it should be close:

$$cos^{-1}\left(\frac{394}{464}\right) \approx 32° $$

So I think the answer is that it is not sub-orbital. The velocity in the video is probably referenced to the rotating frame of the Earth. The calm state of the altitude accurately reflects an orbit that will stay outside the atmosphere for at least days or weeks.

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    $\begingroup$ This is probably the right answer, but if you could find a similar velocity figure from a Falcon 9 LEO (not GTO) mission, it would be good additional supporting evidence, showing that they aren't switching from surface-relative to fixed velocity figures during the ascent. $\endgroup$ – Russell Borogove Jul 6 '17 at 18:41
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    $\begingroup$ Oh my gosh! Of course! I totally didn't even remember that the Earth supplies some amount of velocity to the rocket. That makes perfect sense! So I took Russell's advice and found a Falcon 9 launch (on 25 June 2017) that launched 10 Iridium-2 satellites into a polar LEO. Velocity: 27114 km/hr; Altitude: 624 km. If you run the vis-viva equation with this velocity and altitude, you get a semi-major axis (a) of 6964 km (giving an altitude of only 593 km). However, if you use a velocity of 27151 km/hr (factoring in the eastward motion of the Earth at the latitude for Vandenberg AFB)... $\endgroup$ – Will Jul 6 '17 at 19:46
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    $\begingroup$ ...you get a semi-major axis (a) of 6982 km (giving an altitude of 611 km), which is much closer to the telemetry display of 624 km. My only guess that my calculation isn't right on 624 km is that the Earth is oblate in shape? I've been using 6371 km for my Earth radius for all of my calculations, since that is just the mean radius. $\endgroup$ – Will Jul 6 '17 at 19:54
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    $\begingroup$ @RussellBorogove I've done some quantitative spot checks on this in the past and it's the only interpretation that makes sense when the altitude is stable, but I'll look at a few more videos and add confirmation here. It also turns out that I asked 2nd stage speed - with respect to what? (SpaceX webcast of Orbcomm OG2 deployment) back in December 2015 and got a similar answer. I totally forgot about that until I started looking for evidence of spot-checking. $\endgroup$ – uhoh Jul 7 '17 at 3:12
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    $\begingroup$ @Will -- a polar orbit doesn't gain from the surface rotational speed (and 593km is only a little further from 611km than 624 is). Wikipedia gives Earth's polar radius as 6356km, which gets you a 608km altitude... ¯\_(ツ)_/¯ $\endgroup$ – Russell Borogove Jul 7 '17 at 3:19

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