There’s a campaign ad out right now where Chuck Norris endorses Mike Huckabee. In it, Huck claims that “when Chuck Norris does a push-up, he isn’t lifting himself up: he’s pushing the Earth down.”
Technically, both are true, right? Conservation of momentum means that as Chuck Norris pushes himself up the one foot or whatever, the Earth is actually moving away some infinitesimal amount. What actually is that amount? Obviously, the practical answer is “zero.” But I’m more interested in a theoretical answer.
Please show your work.
I’ll help you get started with some assumptions. You’ll have to find any others you need: CN height = 5’10. (Remember, his feet will remain on the ground during this pushup.) CN weight = around 90 kg (Extra points for being able find a more accurate weight.) CN arm length = 2’ (Total guess. Extra points for finding an average push up height, or an accurate height-to-arm length ratio.) CN pushup time = 1 second Earth mass = 5.9742x10[sup]24[/sup] kg (Yes - I know there are significant figures problems. Don’t worry about it.)
In round figures, Chuck weighs about 1.5 x 10[sup]-23[/sup] of the Earth’s mass. So if Chuck moves 1 metre. the earth moves about 1.5 x 10[sup]-23[/sup] metres the other way (on an average: in reality, since the earth is not perfectly rigid, a sound wave is sent through the earth’s crust, which soon disappears in all the other noise thravelling through the crust).
The point is that the center of mass of the Earth-Norris system doesn’t move. So you first need to figure out how far apart these two bodies move during a pushup - it will depend on where Chuck’s CofM is located relative to his arms. This isn’t specified in the problem, but call that TotalDist, and assume its about 60% of the length of his arms.
These equations give the essential relationships:
TotalDist = NorrisDist + EarthDist
EarthDist * EarthMass = NorrisDist * NorrisMass
Solving these for EarthDist yields
EarthDist = TotalDist / (1 + EarthMass/NorrisMass)
The result of this is that the earth moves about 5.5*10[sup]-24[/sup] meters.
Of course, in practice it moves in a complex spiral. The Earth orbits the Sun, with the orbit perturbed by the gravity of the Moon, and in turn the Sun orbits the centre of the Milky Way Galaxy.
I don’t get it. How can I push something that I am in already? To be able to push an object my feet have to be on a surface other than that of the object, against which I can apply the force that will then get transmitted to the object that I trying to push.
Very easily, Take a ball in your hand, jump up, and while in the air throw the ball. The ball moves away from you, and to a lesser extent you move away from the ball. The same thing happens when Chuck Norris does push-ups, except that the ball that he is pushing against is the Earth, reducing the imparted velocity somewhat, and increasing the gravitational attraction as well.
Can anyone address the fact that Chuck’s feet remain on the ground, so he’s actually making the Earth pivot around his feet? Shouldn’t the answer be in degrees or radians?
Assume his arms make a right angle with his body. The two legs of the triangle are the length from his shoulders to his feet and the length calculated earlier in the thread. By the small angle approximation the angle in degrees is the same as the distance found earlier.
Careful, this was why Copernicus and Galileo got into so much trouble with the Vatican.
I think that you’re taking the right tack though; I’d rather have a buch of scientists and old guys in Italy on my bad side than get into an argument with Chuck.
It would seem, then, that the strength of the pusher-upper is irrelevant. Given that my mass and arm length are about the same as Chuck’s, my flabby Freddy the Pig push-up moves the Earth exactly as far as his manly-man Chuck Norris push-up. Is this correct?
For example, we buy soda pop by the 2-liter–except when we’re buying it by the 16 or 20 ounce.