The 18-June-2021 Spacex-tweeted below says:

View from Falcon 9's second stage Merlin Vacuum engine

There is information out there about the approximate exit diameter and length of the nozzle and there are images out there of what the 2nd stage looks like that help us understand where the camera is likely to be situation, at least along the z-axis.

Question: What can we learn about the distance and direction of Earth from this cool SpaceX-tweeted image of Earth and a vacuum nozzle? I'm assuming this was taken somewhere along a geostationary transfer trajectory, but roughly where?

"bonus points:" There's both relative size to work with, and the position of the bell relative to Earth. Assuming GTO, is there anything to be said about the radial position of the camera and/or the attitude of the spacecraft at this moment?

SpaceX tweet https://twitter.com/SpaceX/status/1405616713576583168/photo/1

Potentially helpful (click for larger):

Falcon 9 2nd stage during assembly, undated, from https://space.stackexchange.com/q/9166/12102 frame grab from some SpaceX video of the F9 2nd stage nozzle against the Earth, perhaps in LEO?

From left: Does the second stage of the Falcon 9 have RCS thrusters? right: Falcon 9 performs extended mission in test for future U.S. military launches

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    $\begingroup$ Assuming a sufficiently-distorting fish-eye lens, we can determine that the Earth is flat! $\endgroup$ Commented Jun 30, 2021 at 4:13
  • $\begingroup$ We got two diameters and two distances. We know the diameter of Earth, the diameter of the nozzle is unknown. The distances from Earth to nozzle and distance from nozzle to camera are unknown. If we would know the diameter of the nozzle, we could calculate the ratio of both distances. If we would know both the diameter of the nozzle and the distance to camera we could calculate the distance to Earth. So if we find a Space-X webside with nozzle diameter and camera distance, we can solve the puzzle. $\endgroup$
    – Uwe
    Commented Jun 30, 2021 at 16:39
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    $\begingroup$ Might have better luck from examining the trajectory, and the time of the photo. (time can be gotten from the location over Earth, and the phase of light, which is very near but not quite local noon).. But for quick-quotes info: My eyeball put the image at above Kansas city, altitude ~ 17500km (by brute-force matching Google Earth image of Earth, for both location and distortion of landmass relative to globe total) $\endgroup$ Commented Jul 1, 2021 at 7:07

1 Answer 1


I'll take a partial stab at this. But first, I believe this rocket is not on a GTO trajectory (judging by the (relatively) high latitude view of the CONUS + Mexico, indicating inclined orbit). This is probably GPS III-05 (17 June 2021 16:09 UTC launch, image was tweeted ~4 hours later), TLE history for reference:

48859, GPS III-05 TLEs

Not knowing the specs of the camera used, I created this log-log figure of potential altitudes based on combinations of sensor size and focal length, using the orbital data to constrain an upper bound of ~20000 km:

Altitude from focal length and sensor size

The equations to determine this are:

$${FOV}_{radians}=2*{tan}^{-1}(\frac{S}{2f})$$ where $S$ is the sensor dimension in the axis of interest and because the Earth (of known radius, 6378 km) takes up about 5/7 ths (I pixel peeped) of the image vertically:

$$h_{km}=\frac{2*6378}{\tan({FOV}_{radians}*\frac{5}{7})}-6378$$ (arc length = angular size x radius)

Realistically, the sensor size and focal length values are probably in this region:

realistic values

These are indicative of the webcam/dashcam (small sensor, wide FOV) class of camera this likely is.

Unfortunately, this does nothing to pin down a altitude :( The proper camera specs are needed for that (Godspeed, nozzle nuzzlers!)


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