Okay, I see there have been some interesting messages so far. Let me
indicate two ways of getting the answer I got. Then you can see if
there's a flaw in it.
METHOD #1. Use the result for escape from the solar system and simply
"turn off" the sun's contribution.
For escape from the solar system, the escape speed is given by
Problem 5.17 in Barger and Olsson "Classical Mechanics" (2nd ed.) to
be:
v = (v_earth,esc)^2 + (v_sun,esc - v_earth,orbit)^2 = 16.7 km/s
relative to earth. Here v_sun,esc = 42.1 km/s. Now set v_sun,esc = 0
to get my previously quoted answer.
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.
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.
FOOTNOTE. Use the above method on another problem as a check.
Note that if you apply energy and momentum conservation in this same
way, it clears up the apparent violation of energy conservation in
the Appendix of my previous message about a rock dropped 5 m on earth
as viewed by someone traveling upward at 5 m/s:
cons. of E => v^2 + 2vu = 2gh + M/m*(u^2 - u'^2)
where v = 10 m/s of rock relative to earth after dropping 5 m, u = 5
m/s of earth initially relative to the observer, and u' = speed of
earth finally relative to the observer.
cons. of p => v = M/m*(u - u')
Substitute the second into the first to find v = sqrt(2*g*h). Sorry,
no Nobel prize.
--
Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu http://physics.usna.edu/physics/faculty/mungan/
This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.