Okay, so I’m standing on a space station that is a rotating toroid so I experience One gravity under my feet. I’m holding a coffee cup and let it drop. Isn’t its drop going to follow a V*cos function? Unless the space station is really small it’s going to look like it accelerates at one G, but what is the force it experiences on impact?
It seems that that force won’t be the same as if it were truly accelerating. Now add a little push, in the direction of spin, and anti-spin. Won’t the behavior vary?
I’m writing a science fiction book and would like to get it right.Robert blanton
To answer this question, it is important to emphasize the difference between gravity per se and centripetal acceleration.
|Exists between any two objects with mass|
Stronger between objects of large mass
Stronger as objects are closer together
|Keeps one object in orbit around another object|
Stronger as the radius of orbit increases
Stronger as the time of orbit decreases
For some gory detail on how gravity and centripetal acceleration interplay on the Earth, see Gravity and centripetal acceleration on Earth. TL:DR; gravity is king, centripetal acceleration can mostly be ignored.
On the space station, the situation is exactly reversed. Centripetal acceleration is king, and gravity can mostly be ignored.
Let’s define a “day” as the amount of time it takes for our space station to undergo a single complete rotation. We’ll also define t as the length of the “day”, in seconds.
Let r be the radius of the space station, in meters.
We can manipulate the formula a = v2/r to relate the length of the day and the radius of the space station. Start with a = v2/r and substitute both a = 9.8 m/s2 (which is what humans on Earth are used to) and v = 2πr/t. Then do some manipulation to get:
9.8t2 = 4π2r
The radius is proportional to the square of the orbital time. That is, smaller space stations need to spin faster. The bigger the space station, the longer the day.
For story purposes, if we want a particular size, we can calculate the required length of the day. Solving for t:
t = √(4π2r/9.8) = 2.0√r
For example, a space station of radius 150 meters would need to have a day of 24 seconds; a space station of radius 1,500 meters would need to have a day of 77 seconds.
Or if we want a particular day length, we can calculate the required size. Solving for r:
r = 9.8t2/(4π2) = 0.25t2
So if we want a day to last 60 seconds, we need a space station of radius 900 m; if we want a day to last 120 seconds, we need a space station of radius 3.6 km.
The acceleration varies as you wander throughout the space station. The closer you are to the rotational axis, the weaker the effect. So things near the ceiling will be accelerated slightly less than things on the floor.
The variation will be minimal if the hole in the toroid is large; that is to say, if the ceiling and the floor are both quite far from the rotational axis.
So if the space station is large, the coffee cup will behave almost exactly as it does on the Earth. If you add a little push, the coffee cup will follow a parabolic path to the floor. If the floor is hard, the cup will break.
If the space station is small, and you had a fairly precise scale, you would notice that the cup weighs less on the top shelf than it does on the floor.
Just for fun, suppose that in addition to the inhabited toroidal part of the space station, there was a service area in the center, on the rotational axis. People would be weightless here.
Perhaps the service area is connected to the toroidal part by spokes, so that it can be reached without going outside. You would experience a gradual transition between full Earth gravity and complete weightlessness as you went along the spoke. Perhaps you have a ladder on the spoke which has rungs quite close together near the edge, where there is full gravity; and as you ascend toward the hub, the rungs get farther apart, as you move from climbing, to jumping, to swimming, to floating.