Re: a surprising escape speed problem

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

> I would also think that v is dependent on direction of launch. Let us
> imagine that the Earth is at rest and a rocket is launched at 11.2 km/s.
> When the rocket is at height of let's say 1000 km the Earth is suddenly
> pulled 500 km away so that the distance to the rocket is 1500 km. The
> gravitational attraction is suddenly less and the rocket arrives at
> infinity with a nonzero speed, i.e the escape velocity was less than
> 11.2 km/s.
>
> When the Earth is pulled in a circle, there will be a continuous
> 'yanking away' of the Earth from the rocket when it is launched
> vertically upwards from the 'midnight equator' so the escape velocity
> should be slightly smaller than 11.2 km/s. Launched from the noon side v
> should be slightly larger.

All of this is borne out by my simulation.  In my last message I said
that, to the extent that the radius of the Earth is NOT negligible,
the required launch speed depends on both the ratio R_earth/R_orbit
and the ratio v_orbit/v_esc.  I should have added that it also
depends on the direction of launch in exactly the way Tore has
indicated.

It should also be noted, however, that "escape velocity" in a case
like this does not produce "zero veocity at infinity" except,
perhaps, wrt the Earth at one point in its orbit.

--
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.