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a surprising escape speed problem



Gene Mosca is responsible for causing me to lose a lot of sleep last
night mulling over an interesting problem about escape speed.

I present a simplified version of the problem I have been toying
with. Suppose we get rid of the sun and instead put the earth at the
end of a string (whose length equals the usual earth-sun distance)
and whirl it around so that it matches earth's usual orbital speed of
v_orbit = 29.8 km/s. Also suppose that the earth does not spin on its
axis.

Got it? Okay here's the question: With what speed (relative to the
earth) must you launch a stone so that it just escapes to infinity?

Think you know the answer? Well, let me give you some WRONG answers first:

WRONG ANSWER #1.

The escape speed from an isolated stationary earth is the usual
textbook answer of v_esc = 11.2 km/s. But objects on the earth are
already going 29.8 km/s. Consequently anything not bolted down on the
earth immediately flies off to infinity.

Maybe because the sun is not around to "hold them down" you say?

Sorry, wrong. In the worst case, suppose everything not bolted down
gets "centrifuged" around to the far side of the earth. These things
will not fly off if the gravitational force of attraction on them due
to the earth exceeds the centripetal force on them due to their
orbiting. Go ahead and work it out. Nothing flies off spontaneously.

WRONG ANSWER #2.

Let v be the required launch speed relative to the earth. Look at
things from earth's reference frame. Initially the stone has speed v.
At infinity it has zero speed. Now jump back to a solar reference
system. We have to add v_orbit to the initial and final speeds of the
stone.

(I assumed we launch the stone in the direction of the earth's
motion, ie. tangentially forward. You can try other directions if you
like. It'll still be wrong. Also, the fact that earth orbits around
so that by the time the stone gets to infinity it's not necessarily
going zero speed relative to earth's new direction of travel is
another red herring.)

Energy conservation now implies (v_orbit + v)^2 - v_esc^2 = v_orbit^2
where the middle term represents the initial gravitational potential
energy and half the stone's mass has been divided out of every term.

Solve to find v = 2.0 km/s. Well, at least we got a positive answer.
I defer an explanation of what's wrong with this to the appendix.

RIGHT ANSWER.

I find v = sqrt(v_orbit^2 + v_esc^2) = 31.8 km/s no matter what
direction you launch it! And yes, this is the speed relative to the
earth not relative to the sun. Can you confirm this answer? Gene and
I would really like to know if this is right.

APPENDIX.

The following might give you some idea about what's wrong with answer
#2. It won't help you discover the key insight needed to get the
right answer, but anyways. Suppose I drop a rock from rest from a
height of 5 m. Take g = 10 m/s^2. Then the rock hits the ground with
a speed of 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 5 m/s. He says the initial
speed of the rock is 5 m/s and its final speed is 15 m/s. Also, he
agrees the rock 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?
--
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.