Re: a surprising escape speed problem

[Tore Ottinsen <tottinse@CHELLO.NO>]

> 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.

Looks wrong to me. If the Earth is isolated in the universe and moves in
a straight line then of course v=11.2 km/s. In your problem it moves
almost in a straight line because the centripetal acceleration is on the
order of 0.01 m/s^2, very much smaller than g. So my intuition says that
v is still approximately 11.2 km/s.

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.


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.