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Re: a surprising escape speed problem



Tore Ottinsen wrote:

I still think that the launch velocity should be dependent on the
launching point and direction, and it should be slightly different from
11.2 km/s.

You're right; it does. And the difference is far from "slight" when
the ratio R_earth/R_orbit is large. But the difference gets smaller
and smaller with the ratio R_earth/R_orbit for any *given* orbital
velocity.

To see this clearly, let us whirl earth-on-a-string so fast
that its centripetal acceleration is equal to 9.8 m/s^2. Then the force
from the ground on a rocket standing on the far side of the earth is
zero, and the rocket is on the verge of being thrown off. Its escape
velocity is zero. At the point where the string is attached, the escape
velocity will be much larger then 11.2 km/s.

Right. For any *given* ratio R_Earth/R_orbit, faster "whirling" will
indeed lead to larger deviations from a uniform, position-independent
escape speed. This is because larger "whirling" speeds will
eventually cause any specific Earth radius to become nonnegligible.
But pick any whirling speed you want and I can then pick an R_Earth
small enough to render position-dependent variations in the escape
velocity arbitrarily small. (I'm assuming here that we have fixed
the Earth's mass rather than its density.)

One way of getting a handle on the effect is as follows: In the case
of the actual Earth-Sun system the "acceleration due to the Sun's
gravity" at the position of the Earth is about 6 mm/s^2. So in our
Sunless version, the surface gravity is effectively enhanced by 0.06%
on the Sunward side of the Earth and diminished by the same amount on
the antiSunward side. An effect, yes, but not a very big one.

--
John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm

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