The Sideways Tide

AllPages RecentChanges Links to this page Edit this page Search Entry portal Advice For New Users

Note that there is gravity - you are not in a gravity-free position. You are in free-fall, so you move together, you and the juggling ball.

We need to define "sideways". I'll use the following coordinate system. At any given moment you can picture the line joining you with the centre of gravity of the Earth (or whatever body you're orbiting). Going along that line, towards the centre of gravity, is what I'll call "down". You have a velocity, and that includes a direction of motion, which is parallel to the Earth's surface if you're in a circular orbit. That is what I'll call "forward". The direction at right angles to both of these is what I'll call "sideways". If you orient yourself facing forwards, feet down, then we have a clear "left" and "right". To maintain this orientation you will, relative to the fixed stars, rotate once every orbit to make sure your feet stay "down".

Now give it a gentle push "sideways". As you will probably know, it will then travel away from you in a straight line at a constant speed, because that's what happens in zero gravity.

But this isn't zero gravity. This is free-fall, and that's not what happens.

If you get the direction exactly right (and the tolerances are inhumanly small) then the ball will go away from you, yes, in a straight line (more-or-less), yes, but it will slow down. After about 23.2 minutes it will come to a complete stop, and start to come back.

Another 23.2 minutes sees it return to its original position, but now travelling in the opposite direction. It keeps going, but again will slow down, coming to rest after about 23.2 minutes, before starting on its return journey, arriving back in your hand after about 92.7 minutes.

For most people, the correct reaction is - what?

These assumptions can be revisited later to see how things change in the real situation, but they are pretty good estimates as starters.

So let's see why this is true, how it works, and what the implications might be. For the sake of simplicity, let's assume that the orbit is perfectly circular, has a period of exactly 92 minutes, and that the Earth is a perfect, uniform density sphere.

So when you have the ball "floating" by your helmet, you and the ball are both in circular orbit around the some point, the centre of gravity of the Earth. The two orbits are *really* close, so there's no real difference between them. You travel in close formation, so from your point of view, the ball "halds station" with you.

But the clue is in the phrase "from your point of view".

Looking in from the outside, you are both screaming along in a circle around the Earth at about 7.7 km/s.

So what happens when you give the ball a push? We are assuming it's a gentle push (whatever that means), and in exactly the right direction (whatever that means).

You and the ball

What it means is that we're choosing the direction to result in the ball still travelling in an exactly circular orbit, but we've changed its "tilt" slightly. So now you and the ball are travelling in circles of the same size, with the same centre, and at the same speed. The result is as in this diagram, you in green, the ball in blue, and as you can see, the ball initially travels away from you.

But after 1/4 of an orbit you are travelling in parallel, and then the ball starts to come back to you. After 1/2 an orbit your paths cross, and the ball has returned. Then the dance repeats on the other side, until finally you are reunited after a single, complete orbit.

From you point of view, if the direction you push is exactly right, the ball will bounce back and forth through your position, as if tethered by a piece of elastic. You can look at the displacement, and the ball is literally performing simple harmonic motion through your position. It appears to you that there is a "restoring force" for the ball, but that force is, in the grand scheme of things, fictitious. It's an artifact of your rotating frame of reference.

More accurately, the orbital period of the ISS is about 92.68 minutes.

But we can compute the size of this force. The period is 92 minutes, and for simplicity we'll assume the maximum displacement is 1 metre. Then the formula for the displacement is:

• $D(t)=\sin(\omega t)$

OK. Now acceleration is the double derivative of that, so

• $A(t) = -\omega^{2}\sin(\omega t)$

• $A_{d=1} = -\left( \frac{2\pi}{5520} \right)^2$