Gravitational Attraction
What would happen if two people out in space a few meters apart, abandoned by their spacecraft, decided to wait until gravity pulled them together? My initial thought was that …
A teachable moment...lost
So I just watched the Mythbusters episode where they recreate the bus jump from the movie Speed. They do two things: a miniature version and full-scale recreation. In their miniature version they scale down the bus by a factor of 12, very carefully building the model bus as closely as possible. Then they scale down the bridge by the same factor. They then point out that they can't scale down gravity without going to the moon. Technically, that would scale gravity by 1/6, not the required 1/12. You wouldn't even have to go nearly as far as the Moon to achieve this. Since force gravity decreases as the square of the distance away from the Earth (starting at 4000mi, the radius of the Earth), you would only have to go up this high:
compared to the 240,000 mi, that's a real bargain! But this is about 10,000 mi above the Earth, whereas the Hubble is less than 600 mi above the Earth, just to give some perspective.
So, without leaving the Earth, the NASA experts say that one can compensate by going faster. Mythbusters scrawls the analysis on the side of the buss and says "basically, what these hieroglyphics mean is to compensate for the physical impossibility of scaling gravity, the speed of our 1/12 scale bus has to be just over 20 miles per hour". What bothers me most about this is not that they don't really go through the analysis, but that they refer to basic math as hieroglyphics and they give no sense for why going a bit faster would compensate for gravity. I am going to include the full analysis here, but below I will also give a simpler explanation that they could have used, which only includes a small amount of math that would have easily fit on the side of the bus.
Analysis
Their analysis is equivalent to the following: the components of the speed off of a ramped angle are
and the x and y positions versus time are given by the standard motion equations
Here is the critical step. We solve for time, t, and get rid of it in the second equation. This way we have the shape of the entire trajectory in space, without any dependance on time.
or:
Now, what happens to this equation when we scale the distances down by a certain amount?
which is almost the same, except for one factor of gamma over the v^2^ term. Thus, if we replace the speed with
the trajectory of the new version is identical to the old version. Now, remember, that this doesn't include time: the scaled version, going a faster, will reach the destination sooner.
A Clearer Way
I certainly wouldn't expect the television audience to follow that analysis, although I wouldn't mind them showing it anyway (but more explicitly). It's the sort of thing where many would ignore it, but the ones who could understand it would get more out of the show. So let's see if we can put it a bit more clearly. I'd start, first, by scaling down the sizes by a factor of 16 not 12. That way I can take the square root more easily. Then there'd be two more facts about gravity that I would mention
- gravity doesn't affect motion horizontally
- vertically, if I throw something up at three times the speed, it will go up nine times the height (the square of the speed increase)
Scaling down just the size, but not the speed, by a factor of 16 would decrease the time by the same factor of 16. If we scale the speed down by a factor of 4, then three things happen: the height of the trajectory reduces by 16 (item 2 above), the time of flight reduces by a factor of 4, and thus the horizontal distance covered (speed times time) is reduced by a factor of 4x4=16. Notice that in doing so, the object trajectory is scaled in both the vertical and horizontal directions by 16, which is the goal of the scaling.
I think that this is clearer than the way presented by Mythbusters, and should have been covered in this way, or some similar way. It could have have been a good teaching moment!
