From: George Murphy <GMURPHY10@neo.rr.com>

Date: Thu Sep 18 2008 - 08:47:41 EDT

Date: Thu Sep 18 2008 - 08:47:41 EDT

1) For the rocket problem I posed, it's a question of whether you're better off adding speeds or heights. The answer is speeds.

The problem is simplest if we assume that motion takes place near the earth so that the earth's gravitational field is approximately uniform. Note, however, that this doesn't change the qualitative answer but just makes the algebra a bit longer. I also assume that the burn of the 2d stage is short so that it doesn't move very far while the burn takes place.

Let the 1st stage burn out at height A when a speed V has been attained. If we let it coast till speed drops to zero it will rise a distance H above A given by energy conservation, MgH = (1/2)MV^2 , where g is the acceleration of gravity. Note that the mass m cancels out, as it does in all problems of motion under gravity, so that it doesn't matter whether the "dead weight" of the burned out 1st stage is carried along or not.

Now at altitude A + H add the kinetic energy of the 2d stage's exhaust, (1/2)mv^2, which will carry it to some final height L. If you work out the energy conservation equation you get L = A + (1/2)(V^2 + v^2) for the final height. This is just A plus the sum of the two heights to which the stages would coast.

The 2d way is to add the speed v as soon as the 1st stage has burned out & separated. This makes the energy at height A = mgA + (1/2)m(V + v)^2, and this will enable the 2d stage to rise to height L' = A + (1/2)(V + v)^2. This differs from the previous result by having the square of the sum of the speeds instead of the sum of the squared speeds, and will thus be greater because of the cross term 2vV.

2) On the balloon problem: Answers previously given are OK but there's a broader viewpoint - perhaps surprisingly, general relativity. According to Einstein's equivalence principle, an observer in an accelerated frame can always treat the effects of acceleration as an artificial gravitational field in his/her vicinity. In an accelerating car this gravity is toward the rear and we all know that balloons go up - opposite to the gravitational field - so in this case it will go forward.

Lest this seem too theoretical, I've done it. Einstein was right.

Shalom

George

http://home.neo.rr.com/scitheologyglm

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Received on Thu Sep 18 08:48:18 2008

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