Physics blooper on the last Stargazing live - throw a ball towards the Earth from the ISS and it will return to you

[robertinventor]

I could hardly believe my ears when the clever physicists and astronomers on the show made this blooper.

Shows how even really clever physicists can sometimes get things wrong.

If you throw a ball from the ISS towards the Earth, ti will not keep going until it hits the Earth's atmosphere. 

It is in orbit just like the ISS. And a moment's reflection shows that it is in an orbit that has to intersect the orbit of the ISS at the point where you threw it.

So - if you throw a ball slowly towards the Earth, then eventually, 90 minutes later, you will see the ball return from the direction of the Earth and come back and hit the ISS at the same speed you threw it.

It is a great show and I know how easy it is to make a mistake when you are talking live on a show also. Expect they realised it soon after the show was over. Anyway just thought might be worth mentioning.

[Tiny Small]

I'll bite...

I don't see how you figured this out as an orbit is a controlled fall (control being over direction, in this case, around) meaning any given mass in a stable orbit has a constant velocity. An object will always react to a force but as the velocity is constant, there is no force (F=ma). With a tennis ball in orbit, you wouldn't even need to throw it. Given that acceleration is defined as a change in velocity, and velocity being speed and vector coordinate, you would simply need to change its direction, therefore changing its momentum into a force, meaning that the ball would move in the direction of the applied force (towards earth in this case) until an equal or greater force stopped it. Although there would be minor differences in the math due to the conservation of energy and conservation of momentum, the greatest part of the original momentum would be transferred to the new vector and with gravity now exerting a stronger pull due to a more efficient trajectory, the ball would also have an increase of speed.

[cantab]

The ball's orbit will intersect that of the ISS, but it need not have the same period. I'm not sure in practice, with a good throw, how far away the ball will be when it next comes near the ISS, but I doubt it will hit.

Certainly, though, you can't get a quick re-entry by throwing an object of the ISS.

[ronin]

If you do as you state throw the ball towards the earth then it has a partial trajectory that is towards the earth so it will carry on towards the earth - Newtons third law ?

If you ignore burn up then it will bounce back, but the restitution and the air resistance mean it will not bounce 100% back.

The ISS is maintained in a stable orbit, so throwing anything in another direction must mean it is will be unable to maintain that stable orbit, When the tool bag drifted off it eventually reentered, and that wasn't thrown.

[Cath]

If you do as you state throw the ball towards the earth then it has a partial trajectory that is towards the earth so it will carry on towards the earth - Newtons third law ?

But that same trajectory will be to take the ball away from the Earth by the time it's orbit is 180 degrees away from where they threw it.

To get the ball to fall back to Earth you'd have to reduce it's orbital speed, ie throw the ball not towards the Earth, but throw it backwards along your orbital path. Although saying that, you could get the ball to renter the atmosphere as they suggest if they threw the ball hard enough, but it would certainly have to be a big and very hard throw!

As Ronin mentions, it would eventually fall back to Earth anyway at only 250 miles above the surface (where the ISS lives), their is too much drag from the Earths exosphere to maintain orbit. They are continually having to push the ISS back into a the desired orbit anyway using various thrusters due to that very thin Earth atmosphere (exosphere) they are in.

[Chris Rowland]

The OP is right that the ball is in a different orbit, one that intercepts the ISS orbit, but his description is wrong, It will return in about 45 minutes and if it misses the ISS will go the other side and return 45 minutes after that, travelling in the same direction and speed as it was when originally released.  As the ball descends it's orbital speed increases and this is why it returns.

This all assumes a 90 minute orbital period, two body solutions for everything and no air resistance.  In reality it's much more complex.

I noticed this but didn't feel that that it was reasonable to expect cosmologists and particle physicists to know the minutiae of orbital motion. After all no one else here does.

If you want to try this there's a program called Orbiter that will simulate all this sort of thing.  Good graphics as well.

Chris

[Tiny Small]

At 250 miles up, acceleration towards earth due to gravity is still very high, around 92% what it is on the surface so anything travelling towards the planet will accelerate at 9.016ms^2. It will also travel sideways at the constant velocity of the iss. The later is momentum, the former is an acceleration and other than air resistance, there is no other force acting on the ball. It WILL fall Ito earth in the same amount of time as if there were no lateral movement, just not straight in front of you. The lateral momentum will alter its location but does not affect the acceleration due to gravity. This is very basic physics and being in orbit doesn't change it. Force is force everywhere in the universe. A force is mass x acceleration. An acceleration is a change in velocity meaning an object can be accelerated with a change of direction but no increase in speed. When you change the direction of the ball from lateral momentum to directly towards the earth by throwing it, you are accelerating it briefly into a state where acceleration due to gravity will pull it towards earth rather than around. All of the gravity acting on the ball is now acting on it straight towards earth, only the momentum is sideways. Everything else is towards the planet. Gravity is an exponential force meaning that an object travelling at x meters per second due to gravity will increase the magnitude of its momentum in the direction of gravity bathe force of gravity every second. At the surface an object will accelerate at 9.8 meters per second every second and the only thing that stops it falling through the floor is the earth exerting a force equal to the objects weight in the opposite direction.

[Tiki]

If you throw a ball from the ISS towards the Earth, ti will not keep going until it hits the Earth's atmosphere. 

It is in orbit just like the ISS. And a moment's reflection shows that it is in an orbit that has to intersect the orbit of the ISS at the point where you threw it.

 This bit is OK.

Consider that until the ball is gently thrown, the ISS and the ball are sharing the same orbit . However, when the ball is thrown (towards the centre of the earth) there is a slight perturbation of it's orbit (owing to the disparate masses of the two objects it can be asssumed that the ISS's orbit remains undisturbed). This new orbit will be very similar to the first except that it will be slightly more eccentric.  The new orbit will intersect the old orbit in four places and have a slightly shorter period (by Kepler III). BTW, note that the speed of orbit >> speed of throw. 

Unless i'm missing something, I doubt that the ball will return within any sensible time scale though.

[Tiki]

Edit: The ball's new orbit should have a slightly longer period.

[robertinventor]

Chris, yes I've just realised that myself. The ball would of course return to the ISS orbit half way around its orbit. So 45 minutes later you would see it return back from the Earth towards the ISS, not ninety minutes later.

If it managed to miss the ISS, then another 45 minutes later it would come back towards the ISS again, this time it would hit the back of your head as facing the Earth.

Cantab - yes the throw would of course change the orbital period. That's why I said, to throw it very slowly. If you throw it slowly enough the period will be almost the same.

I'm not sure how the orbital period will change if you throw it directly towards the Earth - will it increase or decrease, or stay the same? Anyone else know on the top of their head? I'm a mathematician by training (now a programmer) but I specialized in pure maths and logic, not celestial mechanics .

Anyway my intuition would be that if you threw it in the right direction you could surely make its orbital period the same as the ISS. But intuition is easily lead astray as we see.

Now I wondered also - what about that lost tool box - did it return to the ISS. Did they just need to wait 45 minutes and reach out and catch it again?

One complication is that the ISS sometimes adjusts its height, via docked rockets.

Another one is that the gravitational field of the Earth is uneven so since it enters a slightly different orbit, this might get distorted in different ways from the ISS in its orbit around the Earth in a non spherical gravitational field - first is field of an oblate spheroid - but also - with mass concentrations too.

So is just a first approximation probably and would be interesting to see if you can throw a ball away from the ISS gently, and with such accuracy that you can catch it half an orbit later, without leaving the ISS.

[robertinventor]

Tiny, what Cat says is right.

The thing is, that by the time the ball is on the opposite side of the Earth in its orbit, the extra motion you gave to it, though still in the same absolute direction, is now away from the Earth not towards it. So it spends half of each orbit travelling towards the Earth and half traveling away from the Earth, relative to the orbit of the ISS. The combined result is that it stays at the same distance from Earth more or less , as the ISS (for small perturbations, ideal 2 body situation etc).

To get it to fall to the Earth you would need to throw it backwards along the orbit of the ISS at exactly the orbital speed of the ISS and in the opposite direction to the direction the ISS is traveling. That would reduce its speed to 0 around the Earth and it would then fall as you describe.

[robertinventor]

So paradoxically to get it to fall towards the Earth you have to throw it, or rather, fire it, at an extremely fast 17,100 mph, far faster than a bullet, and you would need to aim your gun backwards along the orbit of the ISS roughly in the direction of the horizon of the Earth but up a bit, tangential to the orbit (right angles to the line directly towards the Earth)
 

It would be hard to do, but get the speed exactly right and it would indeed fall towards the Earth in a straight line following Newton's law. Oh, you'd need to make sure you used the exact speed relative to a stationary rather than a rotating Earth, not sure if that 17,100 mph figure is right for that.

[Tiny Small]

That makes sense but I wanted to see if the (very rough) math actually backed it up given the distances and speeds we are talking about.

To start, lets simplify and make a few assumptions.

1. Lets say that gravity is the same the entire distance to earth and is 9.016 ms^2 (ignoring the fact that it increases to about 9.81 ms^2 at the surface).

2. Ignore the effect of air resistance (purely for simplicity).

3. The ISS is exactly 250 miles up.

4. The ISS travels at exactly 17,100 miles per hour.

I know this massively over simplifies it but it serves the purpose for the rough working.

Assuming that you don't throw the ball but just drop it, using s=ut+1/2at^2 and the sequence xn=9.016+(n-1)9.016 (n=1,2,3...), withing 5 minutes, the ball will be traveling at 2,704.8 ms-1 and will have covered a distance of 405,720 m, which is greater than the 402,250m from the ISS to the surface.

The ISS traveling at 7,642.75 ms-1 will cover in that time 2,292,825 m. Now, taking a measurement from the planet core to the ISS we have an orbit radius of 6,773,250 m and using 2pi*r, that gives us an orbit distance of 4.255758488 x 10^7 m. Dividing this by the distance covered in the 5 minutes gives us 18.56 (to 2 dp) as a fraction of the orbit (or 18 and 14/25 in ld money). If we then take the 360 degrees of the orbit and divide it by this, we end up with 19.40 degress (to 2 dp).

So, to summarize, dropping the ball would result in it traveling towards the planet slightly more than the 250 miles in 5 minutes. In the same 5 minutes it would travel laterally 1425 miles covering 19.4 degrees (or 1/(18 14/25ths) of the ISS orbit.

Would that be enough distance to create an orbit for the tennis ball? In reality, with air resistance, the increase in gravity and therefore magnitude of the acceleration and a million and one other factors, these numbers would be substantially different, but still, would covering 19.4 deg of an orbit be enough to create a new orbit?

[Tiny Small]

Edit: Old money.

I expect working out the resultant angle of travel would let you know if it would create and sustain an orbit or not... but it's 3 in the morning and my tired brain has forgotten how to do that.

[Tiny Small]

And I expect the answer would back up your answer and not my earlier answer as it will be quite a shallow angle.

[robertinventor]

The ISS is in free fall around the Earth. So you can't "just drop it". It is already falling freely under gravity. Does that answer you?

[Tiny Small]

I know. When I said drop it, I meant accelerate it towards earth using only the force of gravity. And with regards to the angle, it would be around 9.95 degrees measured from the plane of lateral travel... which is very shallow so yes, I think it would enter an orbit rather than falling towards earth in the 19.4 degrees of orbit travel.

[robertinventor]

Okay gotcha, from the point of view of the ISS, the Earth's surface, and relative to a straight line path tangential to the orbit, should seem to fall away at exactly the same rate that the ball (or the ISS) accelerates towards it.

[Tiki]

Anyway my intuition would be that if you threw it in the right direction you could surely make its orbital period the same as the ISS. But intuition is easily lead astray as we see.

There are two important parameters here: the angular momentum (L) of the ball relative to the earth's centre and the balls total energy, E  (ie. KE + PE). No matter how you throw the ball at least one of the parameters (E or L) will change and a different orbit will result. E and L completely define the orbit.

If you really had the throwing business down to a fine art you could ensure that the ISS and the ball would be able to meet up periodically. A small enough throw and it might appear to return but it would drift away in the long term. P^2=k*a^3 (Kepler III) 'P' is the period and 'a' is the semimajor axis with 'k' a constant that depends upon the units.

[Chris Rowland]

If the ball were thrown sideways at exactly the correct direction the energy and so the orbit period would not change. There would be an orbit plane change and it would pass the ISS twice each orbit. Getting this exactly right would be difficult. Any error would mean that there would be an energy change and so an orbit period change. It might still pass the orbit but the ISS would not be there.

I tried the original experiment in Orbiter using the Delta Glider.  I gave it about 1 m/sec movement towards the earth.  It moved away from the ISS, got about 5km away, and then returned. But the period was slightly different so the closest it got was about 700 metres.

This is not at all intuitive. as has been demonstrated here.

Chris

[robertinventor]

Chris, thanks for trying the experiment with Orbiter and Delta Glider. So, you would need to throw it really really slowly to hit the ISS on its way back. I suppose that also explains how the lost tool bag missed the ISS next time around.

So, for an elliptical orbit with the same period as the ISS, you want one which has the same semimajor axis. If the ISS is in a circular orbit, then you need to send your ball into an orbit that never travels further from the Earth than the ISS (because semimajor axis just means it's longest radius, so if it travels further away than the ISS it will have a larger semimajor axis).

So, throwing a ball towards the Earth will never succeed in getting an orbit with the same period as the ISS.. The only way to do it is to throw the ball gently backwards along the direction of motion of the ISS. It would, I think, seem to arc away from you towards the Earth, and eventually half an orbit later, it would arc back up again from the direction of the Earth and you could catch it again.

Throwing the ball forwards along the direction of the ISS orbit would not work because that would send it into a higher orbit.

I think that's right now. Agreed?

[robertinventor]

There I meant of course, to throw it backwards in the diametric opposite direction to the motion of the ISS relative to the Earth. You might like to try that with Orbiter, and see if it works . Shouldn't matter how fast you throw it, would always end up with same period as the ISS if I understood it right, unless of course you throw it so fast that it falls towards the Earth far enough so that it hits the Earth or gets into significantly thick areas of its atmosphere. If it goes just a short way towards Earth perhaps it won't matter as the atmosphere is of similar thickness, and the ISS is also spiraling towards the Earth slowly (until its next orbit lifting maneuvre)

[robertinventor]

Oh just realised that's not quite right either. Because it's an ellipse with the Earth's centre at one of its foci, so it won't come back to the ISS exactly until a full orbit later. And it's semi-major axis would be less than for the ISS so it would go into a shorter period orbit.

To deal with that you would need to throw it back along the orbit of the ISS but also towards the Earth as well, enough to increase its semi-major axis - and then it will cross the ISS orbit  twice in quick succession so first time comes back towards you so you can catch it,, let it pass by and it will come back to you again from behind you (as facing in the direction of the throw from the ISS).

As for exact angle to throw it depending on the velocity, I don't know you'd have to work it out in detail. But - I think that is about right now. Anyone else here a whiz kid at celestial mechanics able to just see the answer or work it out quickly?

[Chris Rowland]

This is not the sort of problem that can be solved by rhetoric.

There might be velocities with which a ball could be thrown that would cause it to return but it is basically unstable. Tiny differences in speed, direction or position will accumulate.  After all a one second difference in orbit period means that one orbit later the objects will be about 8 km apart.

Chris

[robertinventor]

Okay, good point, but your one second difference in orbit means a difference in orbital velocity of a bit over 5 km/ hour. So for a gentle lob, you might not need to be that accurate for the ball to come back again to the ISS, if you know the right rough direction to throw it.