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To: dr_lew
I think your example confuses weight and mass. The initial speed imparted to a craft by the suggested mousetrap mechanisms depends on the mass of the craft, in this case ~100kg. The height attained due to this speed is then independent of the mass. As Galileo noted, any given mass may be considered as two smaller masses conjoined, and hence the motion of objects in a gravitational field must be independent of mass ( in so many words ! )

It still would not take much impetus to move it off the surface. Yes, it has mass, but not too much force is needed to move it. Question then is how much force is stored in the mousetrap spring and how fast can it be applied? It may not need to be very much. The acceleration of gravity on that comet is merely 1.5mm/sec^2. They need a small bounce of a few meters to get the lander out into the sunlight for a longer exposure for each "day." How far? I certainly don't know. Maybe they need a "better mouse trap." By-the-way, those were rat traps. . . a bit larger than meese traps. heheheheh

i just looked it up and you can get a little over 3 Joules of energy out of a Rat Trap Spring at the lever end. . . hmmmm that's almost 10 Joules from three of them. Although I was posting facetiously there just might be enough there to get something accomplished. . . Perhaps bigger rat traps would have been enough. LOL!

44 posted on 11/13/2014 8:26:26 PM PST by Swordmaker (This tag line is a Microsoft insult free zone... but if the insults to Mac users continue...)
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To: Swordmaker
i just looked it up and you can get a little over 3 Joules of energy out of a Rat Trap Spring at the lever end. . . hmmmm that's almost 10 Joules from three of them. Although I was posting facetiously there just might be enough there to get something accomplished. . . Perhaps bigger rat traps would have been enough. LOL!

It is a very low-g environment, and your rat traps are definitely not out of consideration.

If we take 10 joules at face value, this translates to v = sqrt( 2E/m ) = sqrt( 20/100 ) m/sec = .44 m/sec, and the height attained would be 1/2 X .442/10-3 m = 96.8 m

Correct me if I'm wrong! I mean that seriously.

47 posted on 11/13/2014 8:59:46 PM PST by dr_lew
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