B Oberth Effect Near Sag A*

[Devin-M]

TL;DR Summary

How efficient is the oberth effect near Sag A*?

It's possible.

^Referencing this quote from a different thread…

Given a Falcon 9 sized craft with a total delta-v of 0.33c, a total thrust time of 7 minutes, delta-v of 0.1c already consumed heading towards Sag A* (0.23c delta v remaining), how close to the event horizon would you need to get such that the oberth effect boosts the efficiency of the 0.23c delta v burn such that the outbound craft, as it reaches intergalactic space, has 0.95c? Or is it impossible?

 

[Devin-M]

If it makes the problem more tractable, the thrust could be treated as a semi-instantaneous impulse, such as an explosion, rather than 7 minute sustained burn. I’m mainly concerned with yes / no is 0.95c possible with 0.33c delta v near sag a* by taking advantage of the oberth effect.

 

[pervect]

I can quote some of the equations you need for a Schwarzschild black hole, I don't know how good an approximation that is for Sag A*.

From "Oribits in Strongly Curved Space-time", https://www.fourmilab.ch/gravitation/orbits/, which is the same eqautions as MTW's textbook "Gravitation" we have
$$\left( \frac{dr}{d\tau} \right) ^2 + \left( 1 - \frac{2M}{R} \right) \left( 1 + \frac{L^2}{r^2} \right) = E^2 \quad L = r^2 \frac{d\phi}{d\tau} $$

The geometric units , where c, the speed of light, and G, the gravitational constant, are both 1 may be confusing.

It's easiest if we use the fact that the event horizon r_s is at R=2M in these units, so we can replace $\frac{2M}{r}$ with $\frac{r_s}{r}$.

I was thinking we'd have dr/dtau = 0 at the burn point, but I realized this is wrong :(. More thought is needed here. The goal is to get E up to the desired value. One problem is we'll need to make sure we can escape - while having E > 0 gives you enough energy to escape to infinity, orbital mechanics for a black hole are different and you won't escape inside to infinity the photon sphere via angular velocity alone. I'm not sure how to compensate for this - if you don't do something involving thrust away from the black hole, the closest you can escape with a burn would be at r>3M.

As far as velocity at infinity goes, that would be given by the relationship $\gamma = E = 1/\sqrt{1-\beta^2}$, where $\beta = v/c$. In geometric units $\beta$ would be v, but this is an attempt to give the highlights of geoemtric units.

Sorry I don't have the time to make these musings more intelligible to someone not already familiar with the equations and unit system, but those are my first thoughts. And I might have made a few more mistakes besides thinking (for some reason) that dr/dtau = 0 at the burn point.

 

[PeterDonis]

I was thinking we'd have dr/dtau = 0 at the burn point, but I realized this is wrong

Not if you approximate the burn as an instantaneous impulse, as the OP suggested in post #2, which should be a reasonable approximation for this discussion, and makes the calculation much simpler. Basically you're switching at periapsis from one hyperbolic orbit to another.

 

[PeterDonis]

From "Oribits in Strongly Curved Space-time", https://www.fourmilab.ch/gravitation/orbits/, which is the same eqautions as MTW's textbook "Gravitation" we have

I don't think this potential equation is necessary. I think all we actually need is equation 25.18 from MTW, which tells us that, at any point in the orbit,

$$
\frac{dt}{d\tau} = \gamma = \frac{E}{1 - 2M / r}
$$

At $r \to \infty$, this gives us $\gamma = E$, as we already know; but we can also get it to tell us $\gamma$ at periapsis. If periapsis is at $r = 3M$, then we have $\gamma = 3 E$ there.

We can then calculate as follows:

Coming in, at infinity, we have $v = 0.1$ (I'm using units where $c = 1$ throughout this post). That gives $E = 1.005$.

So at periapsis, we have $\gamma = 3.015$, which equates to $v = 0.9434$.

We then use the relativistic velocity addition formula, with $\Delta v = 0.23$, to find that after the instantaneous impulse burn at periapsis, we have $v = 0.9642$. That translates to a $\gamma = 3.771$ after the burn.

Dividing that new $\gamma$ by 3 gives us the new energy at infinity $E = 1.257$, which equates to a new $v$ at infinity, on the way out, of $0.4033$. [Edit: should be 0.6059, see post #9.]

So I think the answer to the OP question is no, that amount of delta-v is not enough. The Oberth effect did give something of a boost--the $v$ at infinity went up by a little more than $0.3$ with only $0.23$ delta-v applied at the burn--but not enough to get to the OP's desired value.

 

[Devin-M]

I’m a bit confused.

In this section of the wikipedia article on the Oberth effect, it says:

“So if a spacecraft is on a parabolic flyby of Jupiter with a periapsis velocity of 50 km/s and performs a 5 km/s burn, it turns out that the final velocity change at great distance is 22.9 km/s, giving a multiplication of the burn by 4.58 times.”

Why is the boost so small with the black hole considering a flyby of Jupiter can give a multiplication factor of 4.58?

Was this assuming the inbound velocity towards the BH was 0.1c before burn, then 0.23c burn at periapsis? How did we calculate the optimal altitude of the periapsis?

The Oberth effect did give something of a boost--the v at infinity went up by a little more than 0.3 with only 0.23 delta-v applied at the burn--but not enough to get to the OP's desired value.

 

[PeterDonis]

Why is the boost so small with the black hole considering a flyby of Jupiter can give a multiplication factor of 4.58?

Because, first, you're now dealing with relativistic velocities, which don't add linearly, and second, you're diving much deeper into the gravity well, so you have to climb much farther back out after the burn, which decreases the gain you get. [Edit: the second part here is not correct--see post #13.]

Was this assuming the inbound velocity towards the BH was 0.1c before burn, then 0.23c burn at periapsis?

Yes.

How did we calculate the optimal altitude of the periapsis?

We didn't.

I calculated what the gain would be for a periapsis of $r = 3M$, the limiting case. There might be some larger $r$ where the gain would be larger. I haven't calculated that. But you should be able to adapt the calculation I did to any $r$, to see how the gain behaves as a function of $r$.

However, I think it's extremely unlikely that you'll find any periapsis altitude that will give you enough of a gain to go from 0.1c at infinity all the way to 0.95 at infinity with only a 0.23c delta v burn at periapsis.

 

[Devin-M]

For a destination such as Andromeda (2.5 Mly) would it be quicker to take this detour?

It's possible.

Seems 0.4c final velocity craft with oberth effect would indeed arrive out to the distance of Andromeda Galaxy before the 0.33c direct route craft, despite the extra distance and time spent getting to Sag A* first at 0.1c before the final burn 0.23c burn.

 

[PeterDonis]

a new $v$ at infinity, on the way out, of $0.4033$.

I just rechecked this in the course of running a Python script I wrote to do the computation I did in post #5 for various values of $r$, and the value I posted is wrong. It should be $0.6059$. I've edited post #5 to note this.

 

[PeterDonis]

How did we calculate the optimal altitude of the periapsis?

We didn't.

But now I've checked numerically using the Python script I mentioned in post #9 just now, and $r = 3M$ does indeed give the largest final $v$ for the given parameters. As $r$ goes to very large values, the final $v$ asymptotically approaches the relativistic velocity sum of $0.1$ and $0.23$, i.e., no gain over the provided delta-v.

For more on how all this ties in with the Wikipedia page, see my next post.

 

[PeterDonis]

Why is the boost so small

It isn't really. It seems small because the scenario the Wikipedia page is talking about is not quite the same as the one we've been talking about in this thread.

On looking at what you quoted from the Wikipedia page, I noticed something that I'd missed before:

if a spacecraft is on a parabolic flyby of Jupiter

A parabolic flyby means that the starting velocity relative to Jupiter is zero. For that case, all of the final speed at infinity is a gain in speed over the condition coming in. The ratio of 4.58 that the Wikipedia page computes is the ratio of the final speed at infinity (i.e., after climbing out of the gravity well) to the delta-v applied at periapsis.

But here we have a hyperbolic flyby, i.e., the initial speed at infinity, coming in, is not zero, it's 0.1c. The gain in final speed is 0.5059c (with the corrected final speed I gave in post #9), which is a little more than twice the delta-v applied at periapsis. Still not a gain ratio of 4.58, but that boost ratio assumes a parabolic flyby, which we aren't computing.

If I compute a parabolic flyby with 0.23c delta-v at a periapsis of $r = 3M$ using the same Python script I referred to before (and you can check this easily enough by using the formulas I gave in post #5 with an initial speed of $0$ instead of $0.1$), I get a final speed of $0.6003$---in other words, almost as fast as the final speed we computed before! In other words, you're better off spending your rocket power on the periapsis burn than on trying to increase your initial speed going in. In fact, if we do that for your scenario, taking our entire delta-v budget of 0.33c (the 0.23c delta-v you used, plus the 0.1c delta-v needed to get to the initial speed you use), and use that at a periapsis of $r = 3M$, we get a final speed of $0.6940$.

 

[PeterDonis]

you're now dealing with relativistic velocities

To illustrate this effect, when I use my Python script to compute a parabolic Jupiter flyby, the boost ratio decreases as the delta-v goes up. At a delta-v of 10 km/s, it's 4.2; at 20 km/s, it's 3.1; at 50 km/s, it's 2.1; at 100 km/s, it's 1.6; at 200 km/s, it's 1.4.

 

[PeterDonis]

Regarding my statement in post #7 about going deeper into the gravity well, I was wrong about that (and I've edited the post to note that). Going deeper into the gravity well helps, as what I posted in post #10 should make clear.

 

[Devin-M]

It should be 0.6059.

I assume this is treating the burn as an instantaneous impulse. Would extending the burn over time result in higher speed? In this case, during the burn, a portion of the fuel would acquire more kinetic energy before being utilized.

 

[PeterDonis]

I assume this is treating the burn as an instantaneous impulse.

Yes.

Would extending the burn over time result in higher speed?

I don't think so. Ultimately you're converting fuel energy to energy at infinity of the spacecraft, and the fuel energy is the same. In practice, of course, you can't do the burn instantaneously, and there are plenty of details of how the burn is done that can make a difference to how efficiently you convert fuel energy to energy at infinity of the spacecraft--but in this discussion we're ignoring all that and assuming that there are no losses anywhere in the process and that the burn is done with perfect efficiency, i.e., the only limitation is the laws of physics themselves.

In this case, during the burn, a portion of the fuel would acquire more kinetic energy before being utilized.

But it would also have momentum--in the wrong direction. You want the rocket exhaust to have momentum in the opposite direction from the rocket. Starting from fuel that has momentum in the same direction makes that harder.

 

[Devin-M]

Hmm, I thought some of the fuel getting more kinetic energy before being used might be helpful, since the wikipedia article for oberth effect says this:

“ In terms of the energies involved, the Oberth effect is more effective at higher speeds because at high speed the propellant has significant kinetic energy in addition to its chemical potential energy.[2]: 204  At higher speed the vehicle is able to employ the greater change (reduction) in kinetic energy of the propellant (as it is exhausted backward and hence at reduced speed and hence reduced kinetic energy) to generate a greater increase in kinetic energy of the vehicle.[2]: 204 ”

 

[Devin-M]

And this:

“ Oberth effect also can be used to understand the behavior of multi-stage rockets: the upper stage can generate much more usable kinetic energy than the total chemical energy of the propellants it carries.[2]”

 

[PeterDonis]

I thought some of the fuel getting more kinetic energy before being used might be helpful

Those statements are very heuristic. They're talking about why the Oberth effect works at all, i.e., why it's better to use your rocket power at periapsis than at infinity. They're not talking about the difference between an impulsive burn and a burn spread over time given that both burns are done at the same point in the orbit.

Oberth effect also can be used to understand the behavior of multi-stage rockets: the upper stage can generate much more usable kinetic energy than the total chemical energy of the propellants it carries

I think this is an example of why Wikipedia is not always a reliable source. The advantage of multi-stage rockets is that you can discard useless dry mass in the spent lower stages so that the rocket in the upper stage is pushing a smaller payload, hence gives you more acceleration for the same thrust.

 

[Filip Larsen]

Would extending the burn over time result in higher speed?

You guys have already discussed this, so I will just add quick comment (if I can figure out to do so without LaTeX rendering for me in preview).

For classical (non-relativistic two-body) orbits its easy to see that the hyperbolic excess speed after making an impulsive speed increase at periapsis in a parabolic orbit is $v_\infty = \sqrt{w^2+2 w v_p}$, where $w$ is the burn delta-V and $v_p$ is the periapsis orbital speed. This means the first order sensitivity on excess speed vs orbital speed for impulsive burns near periapsis can be found as $\frac{d v_\infty}{d v_p} \approx \frac{w}{v_\infty}$ which is positive indicating that if we, say, make two impulsive half burns at a slightly lower orbital speed on each side of periapsis (keeping the total speed change the same), then the hyperbolic excess speed must be less.

 

[Devin-M]

Would you get the same boost at the edge of the photon sphere with any black hole or is this specific to Sag A*?

 

[PeterDonis]

Would you get the same boost at the edge of the photon sphere with any black hole or is this specific to Sag A*?

Look at the formulas in post #5. The only parameter that's specific to the black hole is $M$, and it only appears in one formula--and not by itself, only in the ratio $M / r$. What does that tell you?

 

[Devin-M]

I will try to work this out later when I have a chance, but on the basis of the photon sphere being smaller on small black hole, and therefore light has to orbit in a tighter radius, I would assume space is more curved at the photon sphere of a smaller black hole. It might take me a day or 2 to run the numbers.

 

[Ibix]

I will try to work this out later when I have a chance, but on the basis of the photon sphere being smaller on small black hole, and therefore light has to orbit in a tighter radius, I would assume space is more curved at the photon sphere of a smaller black hole. It might take me a day or 2 to run the numbers.

The photon sphere lies at $r=3M$. That should shorten the calculation considerably!

Of note, though, is the size of your ship. A solar mass black hole has a Schwarzschild radius of 3km, so a large ship would experience noticeable tidal gravity. An SMBH with a Schwarzschild radius in the millions of kilometres would reduce that issue significantly.

 

[PeterDonis]

on the basis of the photon sphere being smaller on small black hole, and therefore light has to orbit in a tighter radius, I would assume space is more curved at the photon sphere of a smaller black hole.

None of what we're discussing in this thread has anything to do with space curvature. You'll notice that that did not appear anywhere in my calculations in post #5.

 

[PeterDonis]

Of note, though, is the size of your ship. A solar mass black hole has a Schwarzschild radius of 3km, so a large ship would experience noticeable tidal gravity.

For this discussion I would recommend not even trying to introduce such complications. I would assume a small ship, something like the Apollo CSM, which was about 11 meters long, small enough for tidal gravity to be negligible at $r = 3M$ even for a solar mass black hole. Of course a ship of that size capable of a delta v of 0.33c will have a propulsion system much more sophisticated than the actual Apollo CSM. But again, for this discussion I would recommend idealizing away such things.

 

[Ibix]

For this discussion I would recommend not even trying to introduce such complications. I would assume a small ship, something like the Apollo CSM, which was about 11 meters long, small enough for tidal gravity to be negligible at $r = 3M$ even for a solar mass black hole. Of course a ship of that size capable of a delta v of 0.33c will have a propulsion system much more sophisticated than the actual Apollo CSM. But again, for this discussion I would recommend idealizing away such things.

The maths is much more complex, yes - you'd need to look into geodesic deviation and that's probably A level. Qualitatively, however, it's a reason to prefer a larger black hole.

 

[PeterDonis]

Qualitatively, however, it's a reason to prefer a larger black hole.

Sag A certainly qualifies as "larger", so for this discussion I think it's definitely not necessary to go into such complications.

That said, I'm not sure what effect tidal gravity would actually have. Heuristically, at periapsis it's going to slightly squeeze the ship, since the ship will be oriented tangentially. But I'm not sure what, if any, effect that would have on the delta-v burn.

 

[Devin-M]

So in the formula in post 5, is y the lorentz factor, r the distance in meters from the center? Is M the mass of the black hole in kg? What is E? Energy in Joules?

 

[Nugatory]

So in the formula in post 5, is y the lorentz factor, r the distance in meters from the center? Is M the mass of the black hole in kg? What is E? Energy in Joules?

The definition of $\gamma$ is right there in the post: $\frac{dt}{d\tau} = \gamma$. The Lorentz factor from special relativity is a special case of this definition - but because this is a special case from SR the SR formulas can't be assumed to apply elsewhere and should not be used unless you know what you doing.

The units can be whatever you want as long as you are consistent. Choose convenient units for length, time, and mass and use them everywhere. Once you've made that choice the other units will fall into place: for example energy has dimensions $ML^2/T^2$.

 

[Ibix]

So in the formula in post 5, is y the lorentz factor, r the distance in meters from the center? Is M the mass of the black hole in kg? What is E? Energy in Joules?

$E$ is the kinetic energy total energy per kilogram at infinity - i.e. the kinetic energy you have out in deep space plus your rest mass energy, measured relative to a frame where the black hole is at rest. $M$ is the mass of the black hole. $r$ is the Schwarzschild radial coordinate. It turns out not to be meaningful to refer to "distance to the center" of the hole, so $r$ is related to the area, $A$, of a spherical shell "centered" on the black hole by $r=\sqrt{A/4\pi}$. In Euclidean space this would just be the familiar radius, but here it's not. To get the measured radial distance between two radii $r_1$ and $r_2$ you need to evaluate $\int_{r_1}^{r_2}\dfrac{dr}{\sqrt{1-r_s/r}}$ where $r_s$ is the Schwarzschild radius.

Units don't matter as long as you use a consistent system. Note that Peter used a system where $c=G=1$. The easiest way to deal with this if you aren't familiar is to replace his $M$ with $r_s/2$ and go from there.

 

[Ibix]

Heuristically, at periapsis it's going to slightly squeeze the ship, since the ship will be oriented tangentially.

"Slightly" depends on ship size and black hole size, so if @Devin-M is considering a flight path via other BHs then that might affect his thinking. That's the only point I was making.

 

[Devin-M]

E is the kinetic energy at infinity

kinetic energy per Kg?

 

[Ibix]

kinetic energy per Kg?

My apologies - it's the total energy per kilogram.

I'll edit my post above.

 

[Devin-M]

Sorry it’s taken me so long. I’m planning to run the numbers with a 5 solar mass black hole and a 1 solar mass black hole - in a conversation I had with Grok, it told me that the 5 solar mass BH could give 0.95c and the solar mass black hole could give 0.99c at a large distance in this scenario (with a .1c approach speed and 0.23c burn). Does that seem reasonable? I’m very curious if when I run the numbers that will indeed be the case.

 

[PeterDonis]

in a conversation I had with Grok, it told me that the 5 solar mass BH could give 0.95c and the solar mass black hole could give 0.99c at a large distance in this scenario (with a .1c approach speed and 0.23c burn). Does that seem reasonable?

Not even close, based on calculations posted earlier in this thread.

 

[PeterDonis]

I’m planning to run the numbers with a 5 solar mass black hole and a 1 solar mass black hole

Note that in the calculations posted earlier in the thread, the mass of the central object (whether it's a black hole or the planet Jupiter or anything in between) doesn't matter in itself. All that matters is the ratio $M / r$ at periapsis.