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Once in orbit, astronauts experience "weightlessness" relative to their capsule or space station because they are moving at the same orbital velocity as the surrounding spacecraft.

I have heard that the SR-71 blackbird traveled above mach 3 which is about 14% of orbital velocity, and that it would therefore only experience 86% of the gravity of craft at rest.

Is this true?

This made me wonder what suborbital but level (non-ballistic) spaceflight would be like, where level means maintaining constant altitude.

Question: How can the weight a aero-spacecraft occupant experiences during sub-orbital but non-ballistic trajectories be calculated?

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    $\begingroup$ Your question does not seem to match the title. It's also unclear. Consider an edit. "Required" for what? $\endgroup$
    – Organic Marble
    Commented Apr 2, 2021 at 18:08
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    $\begingroup$ Your question doesn't seem to make any sense - even to folks who do play Kerbal. What are you actually asking? What do you mean by "only need to accelerate upwards 86%"? $\endgroup$
    – Rory Alsop ♦
    Commented Apr 2, 2021 at 20:05
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    $\begingroup$ Are you asking if the SR-71 flew at 100% of orbital velocity, would it be travelling at orbital velocity? $\endgroup$
    – Starfish Prime
    Commented Apr 2, 2021 at 20:10
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    $\begingroup$ I believe OP means something like this: At 0 horizontal velocity, you need to provide 100% of your weight in upward force to maintain altitude. At orbital velocity, you need to provide 0% of your weight in upward force to maintain altitude. The question: At 14% of orbital velocity, do you only need to provide 100%-14% = 86% of your weight in upward force to maintain altitude, or is the calculation different? $\endgroup$
    – Russell Borogove
    Commented Apr 2, 2021 at 20:31
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    $\begingroup$ [Upward force required to] maintain altitude at zero horizontal velocity. $\endgroup$
    – Russell Borogove
    Commented Apr 2, 2021 at 20:36

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My interpretation of your question is this:

  • At 0 horizontal velocity, an aircraft needs to provide 100% of its weight in upward force to maintain its altitude.
  • At orbital velocity, with the surface of the Earth curving away exactly as fast as the aircraft falls, it needs to provide 0% of is weight in upward force to maintain altitude.
  • The question: For an aircraft traveling at 14% of orbital velocity, does it only need to provide 100%-14% = 86% of its weight in upward force to maintain its altitude, or is the calculation different?

If this interpretation is good, please consider editing your question to clarify.

I heard that the SR-71 traveled at about 14% of orbital velocity,

Earth orbital velocity is ~7700 m/s; 14% would be 1080 m/s, or ~2410 mph. Wikipedia tells me the SR-71's maximum speed is 2200 mph; close enough.

It would only need to accelerate upwards 86% of craft at rest.

At orbital velocity, the downward curvature of the Earth as you go forward exactly matches your falling speed, so that you don't need to accelerate upwards to maintain altitude. However, it's not a linear relationship.

Centripetal acceleration is the inward acceleration needed to maintain a circular path, and the formula for it is $v^2 / r$. When $v$ is $v_{orbit}$ (orbital velocity for any given altitude), then by definition the acceleration is the same as $g$, the acceleration under Earth's gravity. When $v$ is below orbital velocity, $g$ is too large, so you need to cancel some of it out with upward acceleration, i.e. lift some of your own weight. At $v$ = ${0.14} \times {v_{orbit}}$, you'd want to accelerate downward at ${0.0196} \times {g}$ to maintain a circular path, so you need to cancel all but that much of the natural gravitational fall, and still lift 98% of your weight.

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    $\begingroup$ Wouldn't the upward acceleration be v^2/r? So for an SR-71 at 26 K meters (+ Earth radius) and that speed, ~.2 m/s^2 which is ~ 2% of g, pretty close to what you got. $\endgroup$
    – Organic Marble
    Commented Apr 2, 2021 at 21:54
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    $\begingroup$ Yeah, that seems reasonable. $\endgroup$
    – Russell Borogove
    Commented Apr 2, 2021 at 21:56
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    $\begingroup$ I've edited the question and commented there and voted to re-open. I don't think that the edit makes any changes that affect your answer. If it does, please feel free to yell at me and/or edit it further. Also, is centripetal acceleration $v^2/r$ all that's needed here? It's non-linear, zero at zero and $g$ at orbital speed and radius. $\endgroup$
    – uhoh
    Commented Apr 3, 2021 at 2:29
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    $\begingroup$ @OrganicMarble I was focused on writing the "please feel free to yell at me" part. I'd come back a few minutes later to add the centripetal part but failed to scan for prior art. I saw "some sort of trigonometric relationship" here and under the question, got worried, and wanted to offer a simpler alternative. Now, did somebody say something about a deadline (excuses themself and leaves the room) $\endgroup$
    – uhoh
    Commented Apr 3, 2021 at 4:36
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    $\begingroup$ Fixed my cosine nonsense. The square relation seemed right to me at first, but then I outsmarted myself by thinking, "no, the arc of the Earth's surface is a circle, not a quadratic..." Thanks to you sensible people! $\endgroup$
    – Russell Borogove
    Commented Apr 3, 2021 at 17:20

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