Re: Goddamn fscking ZFFDY ZFQR

Kenneth Brody <kenbrody@xxxxxxxxxxx> writes:

> Steve VanDevender wrote:
> > 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.
>
> But, reaching 8 km/s from Earth's leading edge results in a 38 km/s
> Sun-relative velocity, and from Earth's trailing edge results in a
> 22 km/s Sun-relative velocity. Certainly, these two trajectories
> will result in substantially different trajectories.

Err, maybe I wasn't clear.

About 8 km/s is escape velocity from Earth's surface to infinity. So if
you start out with the ability to achieve a delta-v of 8 km/s relative
to the Earth near its surface (being in low Earth orbit isn't much
different from being on the surface for this purpose) then Earth's
gravity reduces you to to zero relative velocity with respect to the
Earth as you move away. And you lose most of that relative velocity
before getting all that far away from the Earth, so no matter which
direction you go, you're left carrying the 30 km/s relative to the Sun
that you started with, and your orbit around the Sun is some ellipse
crossing (or at least touching on) Earth's orbit.

So to drop something into the Sun, you need the capability to give it 38
km/s delta-v relative to the Earth, the first 8 km/s to escape the
Earth, the other 30 km/s to cancel Earth's orbital velocity with respect
to the Sun once you're away from the Earth. I'm pretty sure that the
minimum-energy trajectory is to achieve 38 km/s relative to the Earth in
the direction opposite Earth's orbit around the Sun. I don't think you
can achieve an orbit that intersects the Sun's surface in any direction
making an angle of less than 90 degrees with the Earth's instantaneous
orbital motion vector, and other trajectories between 90 and 180 degrees
will take more delta-v to counteract Earth's orbital motion.

--
Steve VanDevender "I ride the big iron" http://hexadecimal.uoregon.edu/
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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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