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Re: a surprising escape speed problem



"Carl E. Mungan" wrote:

METHOD #2. Formal derivation using the conservation laws.

System = earth + rocket. All velocities below are vector quantities
so that we can explicitly test the directional advantage (if any) of
launch.

energy conservation => (v + v_orbit)^2 - v_esc^2 = M/m*(v'_orbit^2 - v_orbit^2)

where the initial velocity of the earth is v_orbit and its final
velocity is v'_orbit (assuming the earth has not yet revolved out of
its initial straight line path by the time the rocket is far away).
Yes, I know it seems to strange to posit that earth's speed is
influenced by the gravitational pull of the rocket, but just bear
with me for now.

momentum conservation => m*(v + v_orbit) = M*(v'_orbit - v_orbit)

Take the dot product of both sides with A = (v'_orbit + v_orbit).
Substitute into the energy conservation equation. Now approximate A =
2*v_orbit. After all, it certainly is true that the influence on
earth's speed is small. Simplify the result to get my solution.


???? So if the earth were moving in a straight line through intergalactic space (no
sun) at 29.4 m/s, the escape speed would be (11.2^2 + 29.4^2)^0.5 ?????

Bob at PC

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.