Re: a surprising escape speed problem

[Tore Ottinsen <tottinse@CHELLO.NO>]

 John Mallinckrodt wrote
> The flaw here is in assuming that v_infinity = 0.  In fact,
> you'd have to put in some *extra* effort to make the final
> velocity vanish because you'd need to cancel out all that
> unnecessary orbital velocity.  Orbital velocity, in this
> Sunless system, is an
> *arbitrary* parameter and it can trick you into analyzing the
> problem from the wrong reference frame.  The way to obtain
> minimal launch speed is to let the final velocity of the
> probe equal the (arbitrary) orbital velocity of the Earth.
> And that, in turn, means that you should simply make the
> launch velocity equal to the Earth escape velocity.
>
> Again, all of this only applies rigorously in the limit that
> the radius of the Earth is negligible wrt the distance to the
> Sun.


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. 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.

It is probably difficult, if not impossible, to obtain an analytic
expression for the escape velocity.


Tore Ottinsen
Hellerud high school
Oslo, Norway

Orbit Xplorer - The educational orbit simulator.
http://www.ottisoft.com

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