Let’s say the prefect the Space Elevator. You get on it and ride it up to the top, 400km above the surface of the Earth. Will you be “weightless” in the sense that astronauts aboard the ISS are?
Neat question.
I have no answer, but it definitely made me wonder, since you would be technically spinning pretty fast, would you experience false gravity and be drawn outwards?
No. You would be weightless only at the height of geosynchronous orbit.
You can’t have a 400 km tall space elevator. A space elevator is basically a ridiculously long satellite in geostationary orbit. One end of the elevator reaches all the way to the ground, but the center of mass of the elevator is at 35,800 km altitude.
P.s. if you are just talking about a 400km tall tower, then no, you wouldn’t be weightless at the top. If you step off the tower at the top, you’ll fall straight down to the ground. Which is why it’s not really a space elevator. A satellite at that altitude is “weightless” only because it’s moving very fast, about 7 km/s. If you had such a tower and wanted to launch a satellite from the top, you’d still have to attach a rocket to it and accelerate it to 7 km/s.
Actually the center of mass would have to be well above the geostationary altitude, and the actual top with the counterweight much higher still, to counteract the weight of the cable and its cargo. Everything above geostationary altitude of course wants to have a longer orbital period than geostationary, so at geostationary velocity it tries to pull away to a higher altitude, producing tension on the cable.
You would be weightless after you stepped off the tower, just like the satellite, until you hit the atmosphere. The weightless satellite is falling as well, only it keeps missing the Earth.
If you count the mass of the cable and its cargo together, the centre of mass needs to be at the geostationary point, or only a tiny bit above to cause some tension in the cable. If you put the centre of gravity too high the cable will fall upwards away from the Earth.
I thought you tied the Earth end to a very large rock or something.
:dubious:
Was Clarke’s tethered in the Fountains of Paradise?
Well, the cable is anchored at the equator, it’s not going anywhere! If you make the center of mass too high you’re going to needless expense and putting needless extra tension on the cable, but it does have to be above the geosynchronous point to a sufficient extent to support the payload, whose maximum weight will be at the surface.
The earth end of Clarke’s space elevator was indeed anchored, with some tension, but just enough to keep the elevator stable. If I remember correctly, in the book, someone asks what happens if the cable were severed, and the engineer replies it’ll just hang there - and someone else says that’s not literally true, there’s some tension to keep it stable.
I remembered that, but I thought it was narration. Thanks!
If we ignore the height you specified, then yes, you would be weightless. The principle of a space elevator is that the top is far enough out, and massive enough that center of mass of the entire structure (from satellite to elevator shaft to anchor point on earth’s surface) is in a stable orbit. So, at the top, you’d also be in a stable orbit and, therefore, weightless.
This is wrong in many ways.
If you somehow built a tower that tall on the Earth, which is rotating, then the top would have to already be moving that fast. Each piece of the tower would be accelerated to the appropriate orbital velocity as it was built, or the tower wouldn’t hold together at all.
Note that they don’t have to put rockets at the top of skyscrapers to accelerate them to the appropriate speed after they’re built, even though the tops are moving faster than the foundations.
Geosynchronous orbit is about 35,800 km above Earth’s surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit. At every point, from the bottom to the middle to the top, your perceived gravity would be the difference between how fast Earth’s gravity is pulling you downward and how fast the floor under your feet is moving away from you centripetally due to rotating once every 24 hours. At the equator, you’re 6,371 km from Earth’s center, rotating 40,030 km in 24 hours, which is 1668 kph or 463.3 meters per second, hence Earth is moving away from your feet at about .03 m/s2. Actual gravity there is 9.81 m/s2. Subtract the two and your apparent gravity is 9.78 m/s2 downward.
Now step in to the elevator and to up to geosynchronous orbit, 35,800 km up. Now you’re 42,164 km from Earth’s center. At that point, you’re moving sideways at 3,066 m/s, so your centripetal acceleration is downward at 0.22 m/s2. You’re nearly seven times farther away from Earth’s center where actual gravity is also 0.22 m/s2. The difference between the two is zero, and you feel weightless. But you’re only halfway up the elevator.
Go all the way to the far end of the elevator and you’re 71,600 km from Earth’s surface (78,000 km from the center). You’re moving sideways at 5,672 m/s, so your centripetal acceleration is 0.41 m/s2 but your actual gravity is down to just 0.05 m/s2. The difference is 0.36 m/s2 NEGATIVE. If you step out of the elevator and let go of the hand rail, two seconds later you’ll find your feet have drifted 36 cm off the floor. In ten seconds, you’ll be 18 meters up and drifting further away at 3.6 m/s. You aren’t really drifting upwards. Both you and the platform are moving downwards; it is moving faster than you are. The other end of the elevator is attached to Earth’s surface and Earth is rotating, pulling the platform away from you, while gravity is pulling you downward at a much smaller magnitude. Sixty seconds after you let go, you’ll be 648 meters away (nearly half a mile).
So, no, you aren’t weightless at the top. You weight is NEGATIVE.
In that post, I was talking about a 400 km tall tower, in case what the OP meant by “space elevator” was just a tall tower, tall enough that the top of it was in space.
A 400 km tall tower is just a tower. It’s just supported by compressive strength of the material. The top of the tower is nowhere near orbital speed for that height. If you dropped something from it, it would drop almost straight down.
Actually you’re right:
Place a massive object at the top; would that not move the center of mass upwards? We move an asteroid to synchronous orbit, and lower a cable from it.
Nope, a rigid tower can’t be at proper orbital velocity at every height. Look at this calculator. At a point 50 km up, orbital period is 1.42 hours. At 100 km, 1.44 hours. At 200 km, 1.47 hours. At 400 km, 1.54 hours. (All figures rounded.) But if it is a rigid tower rising off of Earth, each section would have to have an orbital period of 24 hours.
There is a way to put a 400km high space elevator up that would just hang there; build an orbital ring first. This link explains how that could work.
Note that an orbital ring is even more speculative than a geostationary space elevator. If you stepped off a 400km orbital ring you’d just fall towards the Earth, so I wouldn’t advise it.