I think you are being a bit hyperbolic there.
It also makes a difference whether you want the 1 g acceleration to be from the stationary perspective, or the onboard perspective. Probably, you’d want 1 g acceleration for those on board, to simulate Earth gravity, but that’s different from the stationary observer seeing 1 g acceleration.
If the stationary observer sees the ship accelerating at 1 g, it’ll see the ship get up to light speed in about a year (hypothetically, since a ship would never get to light speed). Of course, the onboard observers would experience a shorter time, but they’d also experience higher and higher levels of gravity.
If the onboard observer experienced a constant acceleration, the stationary observer would see it gradually reducing its acceleration, never reaching light speed.
Related question:
Is 1g terminal velocity on earth? 120ft/sec^2?
Quick answer: no.
Longer answer: Terminal velocity is defined as the balance between the gravitational and drag forces. Drag of a body is based on many factors, including shape (cross-sectional area). So you falling with a parachute, without a parachute, and a sphere will all have different terminal velocities.
The correct answer: 1g is not a velocity, it is acceleration due to gravity.
Absent an atmosphere, the gravitational acceleration of a falling body on the Earth is 32.2 ft/sec^2.
In the real world, you will accelerate until the force due to aerodynamic drag exactly balances out the gravitational force. Once you’re in equilibrium, you’re at terminal velocity, and there you’ll stay until something changes (your shape, air density), or until there’s a big SPLAT.
Whoops I meant 32 ft/sec^2. IIRC terminal velocity is 120 ft/sec. So then is 1g acceleration 32ft/sec^2?
Yes, 1 g = 32 ft/s[sup]2[/sup] or 9.8 m/s[sup]2[/sup]. As previously stated, V[sub]terminal[/sub] depends on a number of factors, including surface-to-volume ratio, density, shape, etc. It’s about 120 ft/s for the average solo skydiver.
Ok, I thought 120ft/s was in a vacuum.
No, in a vacuum an object can fall at any velocity up to the escape velocity of the body it is falling towards. On Earth that would be 7 mi./s (36,960 ft/s).
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So an airplane can get to 30,000 or so ft. without going anywhere near that velocity because, although the gravitational attraction is greater closer to earth, there is a greater density to the atmosphere to push off from?
Airplanes fly because of the interaction of the wings with the air. The escape velocity of the earth has little to do with it.
How about an object that had zero redshift to the cosmic microwave background radiation in all directions?
I think that was the point of my question. The reason you don’t have to worry about escape velocity with airplanes is because you do have air in the lower atmosphere. Just because the force is lift and not thrust doesn’t change the fact that gravity is the force being contended with, right? I guess my question is, at what point does the lack of air to provide lift become countered by gravity( less whatever thrust the engines provide regardless of lift) so that an airplane can’t go any higher? Or perhaps a simpler wording would be, “How high can a DC10 fly?”
Ok I just read a little about escape velocity and I see that it’s the velocity without further need of propulsion. Got it. But I still want to know how high a DC10 can fly.
The service ceiling of an aircraft is that altitude at which the wings can no longer provide lift while the aircraft is in equilibrium (because of the decreasing air density). This site lists the ceiling of a DC10 at 33,400 feet. More modern airliners can cruise at 37,000 feet or higher, while bizjets can operate at 41,000 feet or so.