Re: Goddamn fscking ZFFDY ZFQR
- From: Steve VanDevender <stevev@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: 16 Sep 2005 19:49:06 -0700
Kenneth Brody <kenbrody@xxxxxxxxxxx> writes:
> Anthony de Boer - USEnet wrote:
> >
> > Roger Burton West posted thus:
> > >What I want is to have the hardware to let me send Certain Products into
> > >the local fusion lamp in a personal way.
> >
> > That's a tricky one; Earth-escape velocity just gets you off this rock
> > for starters, and then you get to discover you're still in approximately
> > the same 1-year orbit around the Sun and you need one hefty deorbit burn.
>
> Wouldn't that depend on the direction applied to the velocity
> vector? You'll get totally different orbits depending on whether
> you aim it parallel to Earth's orbit, "anti-parallel", directly
> towards the Sun, directly away, or any one of the other infinite
> directions for the vector.
If the object achieves exactly escape velocity from Earth in any
direction, then it will be in (very roughly) a circular orbit around the
Sun similar to the Earth's, since as it gets farther away from the Earth
its relative velocity to the Earth approaches zero, and so it then
carries just the Earth's orbital velocity relative to the Sun.
> Now, whether any of these intersect the Sun's surface, or at least
> get close enough to vaporize your payload, I don't know.
>
> And, who said it wasn't $ESCAPE_VELOCITY + $POSITIVE_DELTA ?
It will have to be a substantial positive delta, quite a bit above
Earth's escape velocity of about 8 km/s from the surface. The Earth's
orbital velocity around the Sun is about 30 km/s, so the most efficient
way of dropping something into the Sun directly from Earth would involve
launching it in a direction directly opposite the Earth's orbital motion
and achieving a velocity close to 30 km/s relative to the Earth, hence
~0 km/s relative to the Sun, after escape. You actually only need to
achieve an elliptical orbit whose perihelion is at the Sun's surface,
but the Sun's radius is quite small compared to the Earth's orbital
radius so the delta-v for that orbit is not much smaller.
A gravity-assist trajectory might be cheaper but take quite a bit
longer. That's how they're getting the MESSENGER probe to Mercury. But
it doesn't arrive until 2010.
--
Steve VanDevender "I ride the big iron" http://hexadecimal.uoregon.edu/
stevev@xxxxxxxxxxxxxxxxxxxxxxx PGP keyprint 4AD7AF61F0B9DE87 522902969C0A7EE8
Little things break, circuitry burns / Time flies while my little world turns
Every day comes, every day goes / 100 years and nobody shows -- Happy Rhodes
.
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