If you were riding up on the Great Space Elevator (patent pending), how high would you have to go before you really started feeling lighter? Would you ever start to float?
Thanks,
Rob
Falls as the square of the radius. So two Earth radii out, you’re going to feel a fourth of the pull down.
The ISS is 600 km above teh Earth, so like a tenth of an Earth radii, so if it was fixed in place somehow, it would feel 1.1^-2 = ~80% of the gravity we feel on the Earth surface.
Of course, if you’re in orbit, then you feel weightless because your in free-fall, not because Earth’s gravity is substantially weaker.
On a space elevator that wasn’t accelerating upwards (so it moves at constant velocity up the cable), you’d be transitioning from being fixed in place by the cable at the Earths surface to free fall at the top in GSO. So you’d basically feel yourself go from normal weight to weightlessness as you moved from stationary to free-fall.
What if you could just go straight up, ignoring the rotation of the Earth? Let’s say it wasn’t rotating so that we don’t have to apply a force in the horizontal direction. It will decrease as the square of the distance increases, but never become zero, correct? And were the elevator platform removed, your acceleration would increase as you fell, correct?
Thanks,
Rob
That’s correct to everyone but the most strict pedant, I believe. Of course, you have to ignore other celestial bodies too. Mercury, for instance, sure doesn’t care about Earth’s gravity when it’s got that giant star sitting right next door.
Right, even without being in free-fall though, the gravity would be scanty at GSO, like a thousandth of surface gravity, just because its so far up.
Related question:
I’ve always assumed that “thirty-two feet per second squared” is an approximation that is most true near the surface of the Earth.
With this post I am double-checking on that assumption.
Of course it must be an approximation since if it weren’t, it would be possible to accelerate past the speed of light if I started far enough away from the Earth (with no other objects near us) but my assumption has been that even without relativistic effects taken into account, the value is an approximation. Am I right?
Went and checked for myself.
Fg = G(m1*m2)/r^2 where Fg is the force exerted, G is the gravitational constant, the ms are masses, and r is the distance between their centers.
F=Ma <–Middle school physics
I take it Fg is aptly considered as an instance of F, so I can set them equal, getting
Ma = G(m1*m2)/r^2
I guess M and m1 can be treated as identical, since I’m asking what happens to an object’s acceleration given its mass and distance from the other relevant object. Yielding:
a = G*m2/r^2
Which indicates to me that as the radius increases, acceleration due to gravity decreases, which in turn implies that “32 feet per second squared” is indeed an approximation even given Newtonian gravity.
Correct?
Yea, you’re right. It decreases with the square of your distance from the Earth’s center, but since the Earth’s radius is huge you’re not going to get much of a difference between, say sea-level and Mt Everest. So for everyday use by us surface dwellers, the estimation is pretty good. Even at the height of the Space Station, you’d still feel 80% of the force you felt at sea-level.
That said, you can actually measure the small variance at different latitudes due to the Earth not being a perfect sphere, or even smaller variations due to the fact that the Earth’s mass isn’t evenly distrubuted through its volume.
That’s the most complicated “your mom” joke I’ve ever read.
I think I’ve read once that the maximum velocity for something in “free fall” into earth’s gravity well is the same as the escape velocity out of it, but I’m not so sure.
Altitude of GSO is 22,236 mi, or roughly 5.5 Earth radii. So gravity would be, what, 1/30 of that on the surface? Small, but much larger than 0.001.
This is true, provided the object doesn’t have any initial (or additional) downward velocity imparted by, say, a rocket motor. At escape velocity, the object has enough kinetic energy so that when completely converted to gravitational potential energy (i.e. the object has zero kinetic energy left and is berry berry far away), its velocity is zero. Freefall is the same process in reverse: all that gravitational potential energy is converted to kinetic. Conservation of energy says that the same amount of energy is involved either way: you can be a zillion miles high at zero speed, or at ground level moving at 25,000 MPH, and your total energy will be the same.
Heh, oops, I glanced at wikipedia real fast to get the height of GSO, and apparently read the length of the orbit instead of the altitude. So yea, off by a power of ten, its a couple hundreths of Earth gravity, not thousandths.
There is one point being missed here, the one that makes the space elevator (theoretically) possible. When you fall “straight down”, it is not a strict Euclidean straight line, but rather a trajectory from the point above the Earth’s surface from which you feel to the surface as it rotates, carrying you with it. In other words, to a hypothetical observer in the solar photosphere, you move in the course of the daylight hours from the west edge of the visible disk to its east edge (hence the apparent motion of the sun in the sky from east to west). While your fall is nearly a straight line, it’s actually a very slight curve from that frame of reference.
Now, why this becomes relevant is that at the Clarke (geosynchronous) orbit, the speed of something in free fall orbit around the Earth is 24 hours – meaning that if you are lifted up the space elevator from the surface to the Clarke orbit, you arrive in free fall, and could, suitably garbed for an EVA, go out the airlock and … float! You would be in free fall orbit yourself at that point.
Someone with a much better grasp of physics than I will have to spell out how this affects the apparent effect of the basic inverse-square law for the intermediate steps of going up the elevator. But the result is inescapable – at Clarke orbit, your space-elevator terminus is in fact in orbit, even though tethered to a point on the moving surface of the Earth.
??? If the elevator was flexible and the end was a ‘station’ that weighed twice or more than the total weight of the elevator and it was:
a) 10km short of ‘Clark’;
b) 10km farther out than ‘Clark’;
Would angular momentum be a factor at all?? 
If the counterweight is closer than geosynchronous orbit, then the thing can’t work at all, and would fall back to Earth. It’d fall slower than if it weren’t rotating, but fall it would.
If the counterweight is a mere 10 km past geosynchronous height, then you could in principle make a space elevator, but it’d have to be a heck of a lot more than twice the total cable weight. Much more practical to have your cable extend significantly past geosynch (10 km is not significant, on these scales) so you can use a smaller counterweight.
How far out would it take for the angular momentum to make the person feel ½ “G” assuming we are not talking a mass large enough to have a perceptible gravity well of it’s own. ( still assuming a flexible elevator and tensile strengths good nuff to keep it all together )
Or would a rotating station design be better because of the distances required?
A somewhat remotely related query: Would it be safe to say that if I could find a way to instantly materialize on the surface of Jupiter or Saturn in an atmospheric-pressure and temperature proof spacesuit, I would instantly be crushed to death by my own weight because of the gravity?
If by “surface”, you mean the top of the opaque layers, then no. Jupiter’s gravity is only about two and a half times Earth’s, so if you’ve got good leg muscles, you could even stand up and walk around (though it’d be very tiring).