Fuel conservation near a Schwarzschild black hole

1. Feb 20, 2014

yuiop

In another thread we determined that the proper total force acting on an orbiting object in the Schwarzschild metric, is given by:

$f_{total} = \frac{Mm}{r^2\sqrt{1-2M/r}} - \frac{mv^2}{r}\left(\frac{1-3M/r}{(1 - v^2)\sqrt{1-M/r}}\right)$

One interesting aspect of the equation is that when $r=3M$ the force becomes independent of the instantaneous local orbital tangential velocity measured by a stationary observer at r. This means the thrust required by a rocket to maintain an orbital radius of $r=3M$ is also independent of the tangential velocity. It also means the consumption of fuel, in litres per hour for example, is independent of the tangential velocity. However, this is the fuel consumption as measured by an observer on the rocket.

Now consider the following scenario. We have two identical rockets and one is hovering at $r=3M$ and the other is orbiting with a tangential velocity of v. If when they initially pass each other they have equal fuel loads, then on the next pass, the orbiting rocket will have more fuel left in its tank than the hovering rocket. This is basically due to time dilation. In this sense the hovering rocket is using more fuel than the orbiting rocket. If the number of litres per hour of fuel per unit thrust is denoted by K, then the fuel consumption (in Schwarzschild coordinate time) can be quantified by multiplying the thrust by the time dilation factor $\sqrt{(1-v^2)(1-2M/r)}$ and K to obtain:

$\frac{mK(M-v^2(r-2M))}{r^2\sqrt{1-v^2}}$

For a hovering observer, we can asymptotically approach the event horizon as close as we like and the fuel consumption, in coordinate time, will asymptotically approach $K\frac{Mm}{r^2}$, in line with the Newtonian expectation and does not increase without bound as you might expect it to. Interestingly the quantity $\frac{Mm}{r^2}$ is the surface gravity of the black hole that is often referred to in connection with Hawking radiation.

The last term of the equation at the top of this post is the centrifugal acceleration and reverses direction at $r=3M$ and starts acting inwards. However, there is still some benefit in having orbital velocity, as far as conserving fuel is concerned, if the orbit is larger than 2.5M. This critical radius can be calculated by taking the derivative wrt v of the equation above to obtain the gradient and then solve for v when the derivative is zero to find the turning points. There are 3 solutions:

$v=0$ and

$v = \pm \sqrt{\frac{2r - 5M}{r-2M}}$

This is the optimal tangential velocity to maximise coordinate time at a given radius with a given fuel load. For $r<2.5M$ the only real result is $v=0$. For $v>=2.5M$ the square root solutions are real and local minimums and the $v=0$ solution is now a local maximum.

At radii greater than 3M the optimal tangential velocity is greater than c according to the above equation, so the equation is not valid in that region. The optimal in that region is of course the geodesic orbital velocity (where no fuel is required) which is given by:

$v = \sqrt{\frac{M}{r - 2M} }$

It is easy to see that the above equation goes to $v=1$ when $r=3M$ and below that radius the result is imaginary and invalid, so we revert to the equation detailed above.

All the above assumes we are trying to maximise coordinate time, but if the objective is to maximise proper time at a given radius for a given fuel load, then for any radius $r<=3M$ the optimal fuel efficiency is to hover and not orbit. To use a black hole as time machine (maximise time dilation) the best path would seem to be just outside the photon orbit radius at the maximum achievable orbital velocity in a geodesic orbit.

Last edited: Feb 20, 2014
2. Feb 20, 2014

tiny-tim

hi yuiop!
r = 3M is the photon sphere, at which photons (light) can orbit forever (at the speed of light, of course)

i think anything with mass would also have to orbit at the speed of light … which it can't

3. Feb 20, 2014

PAllen

I assumed OP meant circular, powered, trajectory rather than orbit (else the whole post was not sensible, as noted). In that case, there is nothing preventing circular, powered trajectory at 3M at any speed between (0,c).

4. Feb 20, 2014

yuiop

Yes, that is what I meant. The required proper thrust is quantified by the first equation and is the same for any orbital velocity (0,c) at r=3M. The fact that any mass cannot orbit at c is the reason fuel is required to keep any massive object in a circular orbit at r=3m.

5. Feb 20, 2014

PAllen

Then please stop calling it orbit. It confuses people. Orbit means a free fall trajectory. I guessed your meaning, but all confusion would be avoided by saying powered circular trajectory.

6. Feb 21, 2014

Mentz114

I agree with this. If a friendly passing ship gives your rocket a tangential bump and also a torque so that it rotates with angular velocity ω=1/T, (T being the time taken for a complete circle) then the ship will always be pointing its thrust away from the centre ( ie radially) and should continue in a circular path until the fuel runs out.

Calculating which rocket runs out of fuel first looks non-trivial so I'll pass on that for now.