grav potential energy

>: What is the equation relating potential energy to distance, once you allow
>: for the fact that gravitational attraction is itself dependent on distance?

The standard form is GPE = - G m1 m2 / r, where r is the separation of
the two bodies.

>: Specifically, does your potential energy increase or decrease as your distance
>: from the attracting body increases?  I know the simple equation for potential

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.

>I don't remember the formula either.  I think that there is a certain
>amount of energy which is enough to overcome ALL of the gravitational
>potential energy of the Earth.  If you assume that energy is kinetic

Yes, there is. If a ship's total energy ( KE + (-ve) PE ) is greater
than zero, the ship is not bound by Earth (assuming PE is measured with
respect to Earth).  The escape velocity is about 11 km/s--any ship
moving this fast with respect to Earth has enough kinetic energy to
escape to infinity--ie, it will never return.

>energy, then you can divide out mass from the equations and get an
>escape velocity.  I would say that for a jump drive to land you 1/2
>way between Sol and Alpha Centauri, it would have to expend more than
>enough energy to bring the ship to escape velocity.

Almost certainly. By that distance, the 1/r factor has made PE with
respect to Earth so small that almost any kinetic energy will tip the
balance.

There are two other very important factors, Sol and Alpha Centauri,
however.  At those distances, Earth is completely negligible. A fair
approximation is to assume 2 light-years is a sufficient approximation
to infinity, and call the potential energy zero. The energy required
to make the jump is then

	(final energy) - (initial energy) = 0 - (-GmM/r) = GmM/r,

where r is the initial separation. 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. 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.

	roald

- --
Scientific progress goes 'BOINK'?	-- Hobbes