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I understand the impracticalities of this concept, but humor the 'what-ifs.'

Ignoring physical obstacles and the effects of atmospheric fluctuations affecting the trajectory.

Say it is possible to have a craft capable of orbiting in Earth's atmosphere just above sea level, that in no way generates lift (just powering through that atmosphere).

How would gravity be perceived by the passenger onboard? On one hand I think they'd be weightless since they are technically always falling... But I could be wrong.

Bonus: How fast would a 200 kg spherical (I guess) vessel be traveling?

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  • $\begingroup$ Sea level is not at a constant distance from earth’s center of gravity. $\endgroup$ – Paul Nov 9 '18 at 21:14
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    $\begingroup$ Fine then: equatorial sea level with no moon $\endgroup$ – anon Nov 9 '18 at 21:20
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    $\begingroup$ Related scifi story: The Holes Around Mars by Jerome Bixby scifi.stackexchange.com/questions/143541/… $\endgroup$ – Organic Marble Nov 9 '18 at 22:02
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    $\begingroup$ @OrganicMarble I'll have to find a copy, that looks fun! $\endgroup$ – uhoh Nov 10 '18 at 0:02
  • $\begingroup$ Re equatorial sea level: there are higher order terms ("frequencies") - the deviation even on the equator is still on the order of 100 m. $\endgroup$ – Peter Mortensen Nov 10 '18 at 7:55
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If you're orbiting, and the rocket thrusters are off, you experience weightlessness. This is true pretty much everywhere.

It's a common misconception that earth's gravity doesn't extend beyond the atmosphere. Craft in space are weightless because they are in orbit, not because earth's gravity is really weak out there. In fact, the Hill sphere (the radius at which the earth's gravitation is no longer dominant) is about 4 times the radius of the moon's orbit. That's quite far out.

The velocity of any circular orbit can be found by $ v=\sqrt\frac{GM}{r} $ where G is the gravitational constant, M is the earth's mass, and r is the radius of the orbit.

Plugging in the Eath's mass and its mean radius of 6371 km gives a velocity of 7909 $ms^{-1}$. That's about Mach 23.

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    $\begingroup$ Excellent answer. $\endgroup$ – Russell Borogove Nov 9 '18 at 21:25
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    $\begingroup$ But the boosters would never be off because they would have to be fighting atmospheric drag. Though I still think even then you are still weightless since the passenger is no longer accelerating. The boosters acceleration is countered by the drags deceleration. $\endgroup$ – anon Nov 9 '18 at 21:32
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    $\begingroup$ If you're matching the orbiting velocity at all times, the huge deceleration from plowing through the lower atmosphere at mach 23 much be matched by the huge acceleration from your (presumably nuclear powered) thrusters. Beyond the large amount of vibrations you'd feel, you'd still be effectively weightless. $\endgroup$ – Ingolifs Nov 9 '18 at 21:53
  • $\begingroup$ I think there is a simple but important point here. The first sentence "If you're orbiting, and the rocket thrusters are off, you experience weightlessness." is really written assuming you are in a vacuum. What it means to say is that the only force is that of the central gravitational field of the Earth. 1. thrusters are off in vacuum, or 2. thrusters are on and perfectly compensating for drag both lead to stable orbit and weightlessness. Astronauts on the ISS would drift to the front of the station over time if the air was still, because the station is always decelerating due to drag. $\endgroup$ – uhoh Nov 9 '18 at 23:50
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    $\begingroup$ This "thrust matches drag" is exactly what the pilots of the "Vomit Comet" aircraft do when following a zero-g trajectory... $\endgroup$ – DJohnM Nov 10 '18 at 6:23

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