grav potential energy

"C. Roald" [email removed] writes:
>
> The physics convention is that potential energy is NEGATIVE, so that it
> increases (gets less negative) as you increase the separation. At
> infinite separation, the PE rises to zero. (A common definition is that
> PE represents how much energy has to be added to bring two objects to
> infinite separation.

Ah, this makes things a little easier.  In effect, if you are a long way
from something, your potential energy due to its gravity is approximately
zero; if you're sitting on top of it, your potential energy is large and
negative - you have to expend a lot of energy to get away.  If you're at
distance 0, you're in big trouble, but that's really only possible if the
something that you're trying to get away from has radius 0, i.e. it is
probably a black hole.

>				     On the other hand, if you were to
> jump from a distance r from Sol to a distance r from AC, then the net
> cost of the jump is
>
> 	G m M(AC)/r - G m M(Sol)/r = G m [M(AC)-M(Sol)]/r.
>
> A first approximation calls M(Sol) about the same as M(AC), so the net
> cost of the jump is ~= 0, plus the drive energy to make a jump of twice
> the distance.

Why twice the distance?  (I assume that in this sentence, "distance" means
"distance from Sol to Alpha Centauri".)  If the energy require to make a
jump of this distance is E, then you're saying you need to expend 2*E.

>		On the other hand, GmM/r is the escape energy of Sol,
> which is the kinetic energy required to produce a velocity of ~30 km/s.
> For a ship weighing say, 1 T, that's 4.5e11 J, or 450 000 MJ. Not to
> be sneezed at.

A ship accelerating at 1-G will achieve that velocity in less than an hour,
using just its manoeuvre drive.  (To get to 30000 m/s using an acceleration
of 9.8 m/s^2 will take 3061 seconds, i.e. 51 minutes.)  Maybe not to be
sneezed at, but that's the sort of figure that all spacecraft play with
routinely - that's why you don't want to collide with anything!

So the ship expends its 4.5e11 J, and get into jump space.  One week later,
it emerges from jump space near Alpha Centauri, and gets 4.5e11J back -
what happens to this energy?  Presumably it gains this energy as it emerges
from jump space, and the energy ends up back in the ship's jump capacitors.

- --
 "Keyboard?  How quaint!" - M. Scott

 Adrian Hurt			     |	JANET:  [email removed]
 UUCP: ..!ukc!cs.hw.ac.uk!adrian     |  ARPA:   [email removed]

Adrian Hurt [email removed] writes:
>
> >		On the other hand, GmM/r is the escape energy of Sol,
> > which is the kinetic energy required to produce a velocity of ~30 km/s.
> > For a ship weighing say, 1 T, that's 4.5e11 J, or 450 000 MJ. Not to
> > be sneezed at.
>
> A ship accelerating at 1-G will achieve that velocity in less than an hour,
> using just its manoeuvre drive.  (To get to 30000 m/s using an acceleration
> of 9.8 m/s^2 will take 3061 seconds, i.e. 51 minutes.)  Maybe not to be
> sneezed at, but that's the sort of figure that all spacecraft play with
> routinely - that's why you don't want to collide with anything!

Perhaps that is one answer -- require all ships entering jumpspace to
have enough kinetic energy (speed) to escape from the local star's
gravity well.  Then, a ship jumping from Sol to empty space would end
up with (almost) no kinetic energy after the jump, then after jumping
to Alpha Centauri it would gain as much kinetic energy as potential
energy that it lost.

Of course, there will have to be extra energy expended for the jump
itself.  Probably that energy would be expended as lost heat energy
or something.

> So the ship expends its 4.5e11 J, and get into jump space.  One week later,
> it emerges from jump space near Alpha Centauri, and gets 4.5e11J back -
> what happens to this energy?  Presumably it gains this energy as it emerges
> from jump space, and the energy ends up back in the ship's jump capacitors.

As I mentioned, the energy could just as easily be transferred into
kinetic energy.  Or it could go into the jump capacitors.  As long
as you aren't getting free energy from the jump, I don't mind how
it is modelled.

Jerry Williams ([email removed])

Jerry Williams ([email removed]) writes:
> Adrian Hurt [email removed] writes:
> >
> > >		On the other hand, GmM/r is the escape energy of Sol,
> > > which is the kinetic energy required to produce a velocity of ~30 km/s.
> > > For a ship weighing say, 1 T, that's 4.5e11 J, or 450 000 MJ. Not to
> > > be sneezed at.
> >
> > A ship accelerating at 1-G will achieve that velocity in less than an hour,
> > using just its manoeuvre drive.
>
> Perhaps that is one answer -- require all ships entering jumpspace to
> have enough kinetic energy (speed) to escape from the local star's
> gravity well.  Then, a ship jumping from Sol to empty space would end
> up with (almost) no kinetic energy after the jump, then after jumping
> to Alpha Centauri it would gain as much kinetic energy as potential
> energy that it lost.

Of course, things don't balance out quite so neatly when the destination
star has more or less mass than the departure star.  Maybe part of the jump
calculation involves arranging that you exit jump space at a distance such
that the escape velocity from the destination star at that distance, is equal
to the escape velocity from the departure star at the distance at which you
entered jump space.  Using distance from the star to balance out the different
mass of the star ensures that the escape energy at each end is about the same,
and the jump drive doesn't get overloaded trying to compensate for the
different energies.

The jump drive would then convert the ship's kinetic energy into jump energy,
and back again when the ship exits jump space.  Kinetic energy would probably
be meaningless or totally different in jump space.

> Of course, there will have to be extra energy expended for the jump
> itself.  Probably that energy would be expended as lost heat energy
> or something.

That's accounted for by the huge amount of fuel which has to be burned by the
jump drive.

When you exit jump space ...

> As I mentioned, the energy could just as easily be transferred into
> kinetic energy.  Or it could go into the jump capacitors.  As long
> as you aren't getting free energy from the jump, I don't mind how
> it is modelled.

And now you know why you have to be at least 10 diameters away from the
planet before you jump, preferably 100 diameters.  (And probably some
significant distance away from the star, too.)  If you're too close to
a gravity source, the jump capacitors or jump drive get overloaded.  Boom.

And now for something completely different:

James T Perkins [email removed] writes:
> Subject: (3337) Comment on Classic Traveller
>
> A funny thought.  On my X-windows display, the configured icon image for
> my shell windows is an Imperial starburst, not a shattered sun ala
> MegaTraveller.

Use the shattered sun to indicate when something goes wrong.  :-)

- --
 "Keyboard?  How quaint!" - M. Scott

 Adrian Hurt			     |	JANET:  [email removed]
 UUCP: ..!ukc!cs.hw.ac.uk!adrian     |  ARPA:   [email removed]

Adrian Hurt [email removed] writes:

> Of course, things don't balance out quite so neatly when the destination
> star has more or less mass than the departure star.  Maybe part of the jump
> calculation involves arranging that you exit jump space at a distance such
> that the escape velocity from the destination star at that distance, is equal
> to the escape velocity from the departure star at the distance at which you
> entered jump space.  Using distance from the star to balance out the different
> mass of the star ensures that the escape energy at each end is about the same,
> and the jump drive doesn't get overloaded trying to compensate for the
> different energies.
>
> The jump drive would then convert the ship's kinetic energy into jump energy,
> and back again when the ship exits jump space.  Kinetic energy would probably
> be meaningless or totally different in jump space.

That was kind of what I was thinking.  If you think of stars and planets
as "gravity wells", then you would jump to a point with the same "depth"
as the other "well".  The 100-diameter rule would be to keep you clear of
planets and stars, no matter what their relative densities are.  I'm sure
you can still imagine jumping from an near a small star into the inside
of a red giant since the gravity is the same (has anyone read A Mote in
God's Eye?  Hope that black globe is working 8-).  The biggest worry I'd
have is being too close to the planet, and jumping out closer to the star
than I expected since the "gravity well" is the same depth there.

I like the idea of requiring that you be at escape velocity first, since
that means your ship is at "zero energy" overall.

Note, if you don't allow direct and immediate conversion of kinetic to
potential energy upon leaving jump space, then it would be impossible to
jump to a point of higher potential energy (further from the star), since
that would require you to add potential energy to the ship.  Also, if you
tried to jump to a point nearer to the star, there would be a great deal
of excess energy, which nature often expresses as heat.  Can you say BOOM?
And a vacuum is a great insulator!

(If you don't want Niven-style jump drives, then you still must account
for the extra energy.  In this case, simply start with enough kinetic
energy to reach escape velocity and end at near rest when well away from
any start, allowing kinetic energy to be converted to potential energy
by the jump.  Similarly, add kinetic energy when you jump towards a
star.  If you jump from deep space towards a large star, you'd end up
hurdling towards the star when you arrived -- don't arrive too close!
Of course if you use this, there is the problem of momentum -- are you
allowing ships to store up momentum in different directions?).

> > Of course, there will have to be extra energy expended for the jump
> > itself.  Probably that energy would be expended as lost heat energy
> > or something.
>
> That's accounted for by the huge amount of fuel which has to be burned by the
> jump drive.

Now, where does this energy go?  If it goes into the mysterious "jump
space", then jump drives are slowly leaking energy out of our universe.
I think more likely it would be radiated into space as heat.  Of course,
that heat had better be left behind, not brought with you (Have you ever
watched a ship enter jump space?  Well, the ship looks like it's
stretching for a split second as the jump drive activates, followed by
a great white flash when the ship disappears.  No doubt boldly going
"where no one has gone before". 8-)

Jerry Williams
[email removed]