Thanks warsprite. I didn't know about that little gem. And the fact that there is no wind or other means of eroding them (tides on a beach) means they stay around until some meteorite strike on or near them and obliterates them.

A bloke in a bloody heavy spacesuit on the moon will exert a similar ish force on the surface as a bloke in the nude on dear old Earth, yes. My argument was not with what you said about footprints, but your insistance on using e=mc^2 in your justification. It is a pet hate of mine to have that equation chucked about in places where it doesn't belong and is completely meaningless :p

I have no doubt that the astronaughts gave it some welly a couple of times for the sake of the cameras and the sheer unadulterated fun of it, but for the most part they walked about in a rather boring and unimpressive fashion, which is what most of the footprints will have been made by.

I'm sure they mostly took photos of the more impressive impressions (including the first one).

In many cases they walked in quite a shuffling way, because the suits were quite cumbersome, and which you can see in some of the footage. This is one of the reason that the trails of footprints show up so well in the recent photos.

But it does apply. Drop an object of mass X in 1G at a height of Y inches. Now drop an object of mass 3X in .16G at a height of Y inches. Because the second object's hang time is longer, it actually moves faster than the expected 1/6 of the first object, increasing the inertia it has on impact to be roughly equal. It's just that the factors involved are different factors to get the same rough effect. And in walking (not jumping) the foot is pressed down, providing more energy into the equation. The equation always applies, I just used it loosely to explain. But it wasn't outright wrong because it does still apply to the situation. E=MV^2 applies to all motion and energy computation in some way (C is a constant meaning E=MC^2 doesn't apply but it's logically derived function does).

How does E=Mc^2 have anything to do with it at all??? The energy described is not transfered or lost or anything at any time during walking or jumping or sleeping or dancing the can-can.

@ Octavious I understand what your saying. But after accumative dose of a week plus, 24hrs a day of standing in that machine, you would not be feeling to good. Than how would you fake the broadcast using the first hand held TV camera. Anybody could intercept it who had the equipment and pointed it at the Moon. Also the equipment was hand made from military grade parts, not your typical consumer grade, and the older eguipment because of it's nature was more resistant to radiation damage.

@Octavius - Is there movement? Is there mass? Then there is energy. Is there an impact between a moving object and a (realively speaking) non-moving object? Then there is elastic energy transferrence and displacement of mass. And please quit saying E=MC^2. It's E=MV^2 for our needs. The speed of light (C) doesn't have squat to do with this. The Velocity and the Mass are the factors. And the contribute to the displacement of the mass (leaving a foot print).

Kinetic energy is not equal to MV^2. It is confusion over this, I think, which has led to this argument.

I keep forgetting to divide by 2. Damn! E= (M/2)V^2 Better?

Delighted :p.