Does Homer Simpson have to come and remind you all about the Air In Space Museum?

"Well, if you jumped off anything in space, wouldn't you eventually land back on it again, theoretically?"

Not if you jumped off at escape velocity or higher. Not gonna happen off the earth or moon (you'd need to jump at thousands of miles per hour), but you could totally escape-jump off of smaller things like an asteroid or space station, which are not massive enough to have escape velocities higher than jumping speed.

Stefan Jones is right, too: If you're in orbit around a massive celestial body and jump off something else that's also in orbit, you and the other thing would keep going around in orbit, but in slightly different orbits. Depending on the relative masses, and in the direction you jumped off (along the direction of the orbit? Perpendicular to it, to the side? Or in the up/down direction?), you could even end up hitting your platform after as little as half an orbit, if the resulting orbits cross each other and have similar inclinations and apogees/perigees.

"...given that your weight on the moon is roughly one sixth that of your weight on the earth, does that necessarily mean you'd be able to jump six times higher on the moon than you can on earth? I have a hunch that's not necessarily so"

The acceleration due to gravity is one sixth what it is on earth. So if you jump up at the same speed (ignoring cumbersome spacesuits and whatnot), then you spend six times longer in the air (or in the vacuum... six times longer off the ground). In that time, your velocity profile over time is the same as it would be on earth (ignoring air resistance), just stretched out in time 6 times. If you're going the same speed (or the same speed profile) but for six times as long, then you're off the ground six times higher.

The height reached by a projectile is

1/2 of V2 over A

or one half of the speed squared divided by g. ("A" is the acceleration which is G). So if g is divided by six, then the height gets multiplied by 6.

Proof of the above formula:

Time to maximum height = time until speed goes to zero = speed divided by acceleration = V/A (e.g. if you jump off at 32 feet per second, it will take one second at Earth gravity before you've lost your speed and are at the top of the jump)

Max height Y = 1/2 of A times T squared (e.g. if you jump at 32 feet per second and smoothly decelerate to 0 feet per second, then on average over that time you were going up at 16 feet per second. Acceleration times time gives total change in speed. Divide that by two and you get the average speed, assuming the final speed (when you reach the top of the jump) is zero. Multiply that by time and you get the distance).

So if T=V/A and Y=1/2 AT2, then A=V/T and y = 1/2 VT = 1/2 VV/A

which is half of the square of your jumping speed, divided by the acceleration of gravity.

So with one sixth the gravity you jump six times as high. But if you could just jump six times as fast (even here on Earth) you'd go up 36 times higher...

Edited by author 01-29-2007 02:56 AM
I did some calculations and got I could jump 7.5 times as high on the moon.
Assumed:
a)leg recoil before jump = 0.2 m
b)I can jump = 0.4 m
c)legs apply a constant force during the jump

d) F(legs) + F(gravity) = ma
e) get lift off velocity by integrating over my leg recoil distance
f) Use energy conservation, knowing everything but F(legs)
g) On earth this gives my leg force to be about 3 g's worth
e) assume same leg force on moon
f) assume same prejump recoil distance
g) get new lift off velocity on moon
h) Use conservation of energy to get height = ~ 3m = 7.5 earth distance

And yes, I realize that in my example, if you jumped at 32 feet per second, that would mean your feet are 16 feet in the air, which not even Michael Jordan could do. But it was just an example to illustrate how the speed of the jump has to be divided by g to get the time in the air (since acceleration is change in speed per unit time).

Did no one else here take high-school physics? ;]

This was too interesting to pass up.

Given v[0] (initial jump velocity) and v[1] (apex, v=0), and uniform acceleration (a), we can use the equation

t = (v[1] - v[0]) / a

to get our time. So a jump on the moon will take six times longer than on earth. For distance we can use:

d = v[avg] * t
v[avg] = (v[1] - v[0]) / 2

Therefore our jump will distance (height) will increase six times on the moon given an equal initial velocity.

But will our initial velocities be equal? Solving for v[0] is simply v[0] = a * t, but solving for those two variables proves impossible. The pre-liftoff time varies dramatically because of the decreased load (weight), and calculating muscle speed at varying loads is really a matter of anatomy and physiology. The pre-liftoff acceleration also varies largly, especially because the "escape acceleration" (acceleration > g in order to lift off) does not scale linearly as mass and strength vary.

So to answer the problem, if you jump the same speed as on earth, you will jump six times as high. But in reality it matters how much you weigh, how strong you are, how quickly your muscles move, etc...

I saw one that said goldfish only have a 3 sec memory but thats wrong its about 18 months

If you jump in orbit, you'll just end up in a more elliptical orbit, which you'd experience as a rise followed by a fall back towards the item you jumped from.

Is this before or after eating a happy meal?

The claim on the McDonald's bag isn't right even on the moon. You can't actually jump six times as high as on earth, you can only raise your center of gravity by six times as much as on earth, which makes for a much smaller jump, as the center of gravity is somewhere in your chest or abdomen when standing.

Arthur C. Clarke did a interesting lecture, or possibly article, in which he debunked this, and calculated approximately how high the current world record holder for the high jump could jump on the moon. Can anyone find a link to the transcript on the web somewhere?

And we all know that no man ever set foot on the moon.


THE DELUSIONS END NOW!

Good catch. My husband and I noticed something similiar on the children's channel, Noggin. They suggested that bison and squirrels hibernated during one of their interval songs. What kind of information is that to be giving to a 2 year old?

Edited by author 01-29-2007 06:53 AM
Bisons and squirrels can sing before hibernating? Quality!

My son's book, "Machines As Big As Monsters," claims that work on the Space Station is taking place "millions of miles" out in space.

Err, shes clearly jumping off the moon in the illustration. I think that the connotation is obvious. When did you become the web's Andy Rooney?

You *can* jump six times higher in space.

Consider a person who can jump 1 foot high on earth. If that person were to jump from a space platform, assuming the platform itself had no gravity, then at some time t0 they would be at 0f, at t1 they would be at 1 foot, just like on earth, on up to t6, when they would be at 6ft, or 6 times higher than their earth jump.

The bag doesn't say anything about what happens after time t6.

Connotations are bad science, no?

If the bag said "on the Moon" instead of "in space", the complaint would be a lot more pedantic.

"On the Moon" is a subset location of "in space". Everything is technically "in space". Obviously the writer meant "on the Moon" and "on Earth", but didn't say that, hence the bad science.