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Re: a surprising escape speed problem



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

The flaw here is in assuming that v_infinity = 0.

THANKS! John M gets a gold star. If I add in v_orbit^2 to the
right-hand side and continue the derivation as before, I now indeed
get v = v_esc. So I suppose that what I should do in general is add
in some v_f^2 and then choose v_f so as to minimize v.

However, I don't think this invalidates the general procedure I used.
So I would like to go back over the dropped ball problem again:

Suppose I drop a ball from rest from a height of h = 5 m. Take g = 10
m/s^2. Then the rock hits the ground with a speed of v = 10 m/s,
right?

Now view the situation from the perspective of a man in an elevator
rising upward at a constant speed of say u = 5 m/s. He says the
initial speed of the ball is 5 m/s and its final speed is 15 m/s.
Also, he agrees the ball got 5 m closer to the earth.

He is chagrined to discover that 2*g*(5 m) is not equal to (15 m/s)^2
- (5 m/s)^2 and announces that he has discovered a violation of
energy conservation. Why should the Nobel prize committee hesitate
before making him an award?

Here are two ways of answering this question.

METHOD #1. Apply the work-KE theorem to the ball alone.

The man says the ball falls a distance of not h but h+ut where t=v/g
is its flight time (which everyone agrees on). Hence the work done by
gravity on the ball is mg(h+ut). Set this equal to 0.5m[(v+u)^2 -
u^2] to correctly find v = sqrt(2gh).

METHOD #2. Apply the conservation laws to the system of ball plus
earth. Earth has mass M.

conservation of energy => 0.5mu^2 + 0.5Mu^2 + mgh = 0.5m(v+u)^2 + 0.5Mu'^2

conservation of momentum => mu + Mu = m(v+u) + Mu'

Solve the second for u' (final speed of earth) and substitute into
the first. Drop the smallest term (mv^2/M) to again find v =
sqrt(2gh).

The first method emphasizes that the work done by the earth on the
ball is frame dependent. Note that in the elevator frame, the ball
also does work on the earth (namely -mgut, which "cancels" the extra
work the earth does on the ball).

The second method emphasizes that we cannot neglect earth's recoil.
True, earth doesn't move much, but it doesn't need to in order to
make a contribution.
--
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.