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I mean sure, there's ambiguity around the word "spacecraft". But Falcon 9's booster isn't orbital, its payload is. So it might still be a first for an orbital s

I mean sure, there's ambiguity around the word "spacecraft". But Falcon 9's booster isn't orbital, its payload is. So it might still be a first for an orbital spacecraft. The gravitational potential energy of a 100kg payload in 100km orbit is 98 MJ. The kinetic energy of a 100kg payload in 100km orbit is 3042 MJ, over 30x more. So surviving reentry from orbit is orders of magnitude more difficult than surviving a straight up/down shot. Now F9 boosters obviously have a horizontal velocity as well that they need to cancel out, but suborbital and orbital spacecrafts are very different beasts and it's absolutely noteworthy. It's the difference between surviving a car crash at 20 mph vs 95 mph. One is trivial, the other requires a heck of a lot of engineering.

davedx 3y ago
F9 landing is not trivial, and despite all the comments on this thread trying to back up the claim this is the “first commercial spacecraft to land on US soil” ludicrous.
- peeters 3y ago
  I wasn't trying to extend the analogy to saying landing F9 is trivial. But F9 surviving re-entry is fairly trivial, which is the important difference between the two. Once you've survived re-entry and slowed down to terminal velocity, there's not a huge difference between landing from orbital and sub-orbital trajectories. I agree that the way it's phrased is fairly meaningless and reeks of baseball commentary cherry-picking, but surviving from orbit is still a very different milestone than surviving from sub-orbit.
aj7 3y ago
Huh? Did you flunk freshman physics? E=K+U=GmM/2r−GmM/r=−GmME/r. We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy. Do the calculation over.
- versteegen 3y ago
  You are correct (aside from the s/ME/M/ typo), but the snark isn't needed.
  - peeters 3y ago
    No, they are not correct either.
- peeters 3y ago
  You're quoting this article, but you're misunderstanding it: https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/13%3A_Gravitation/13.05%3A_Satellite_Orbits_and_Energy#:~:text=E%3DK%2BU%3DG,magnitude%20of%20the%20potential%20energy https://phys.libretexts.org/Bookshelves/University_Physics/U.... This is calculating the gravitational potential between two point masses. In other words, the potential energy if both masses were singular points at a distance from each other. I was quoting the potential energy differential between being 100km over the ground, and on the ground. This is 98 MJ for a 100kg object, and that's what you have to cancel out to land back on Earth. The total gravitational energy of that object to the Earth's centre is more like 6300 MJ, but that's a meaningless number.

davedx 3y ago

F9 landing is not trivial, and despite all the comments on this thread trying to back up the claim this is the “first commercial spacecraft to land on US soil” ludicrous.

peeters 3y ago
I wasn't trying to extend the analogy to saying landing F9 is trivial. But F9 surviving re-entry is fairly trivial, which is the important difference between the two. Once you've survived re-entry and slowed down to terminal velocity, there's not a huge difference between landing from orbital and sub-orbital trajectories. I agree that the way it's phrased is fairly meaningless and reeks of baseball commentary cherry-picking, but surviving from orbit is still a very different milestone than surviving from sub-orbit.

aj7 3y ago

Huh? Did you flunk freshman physics? E=K+U=GmM/2r−GmM/r=−GmME/r. We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy. Do the calculation over.

versteegen 3y ago
You are correct (aside from the s/ME/M/ typo), but the snark isn't needed.
- peeters 3y ago
  No, they are not correct either.
peeters 3y ago
You're quoting this article, but you're misunderstanding it: https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/13%3A_Gravitation/13.05%3A_Satellite_Orbits_and_Energy#:~:text=E%3DK%2BU%3DG,magnitude%20of%20the%20potential%20energy https://phys.libretexts.org/Bookshelves/University_Physics/U.... This is calculating the gravitational potential between two point masses. In other words, the potential energy if both masses were singular points at a distance from each other. I was quoting the potential energy differential between being 100km over the ground, and on the ground. This is 98 MJ for a 100kg object, and that's what you have to cancel out to land back on Earth. The total gravitational energy of that object to the Earth's centre is more like 6300 MJ, but that's a meaningless number.