Can Canceling Orbital Motion Improve Rocket Efficiency?

[Janus]

^re: C = A * ~2.99... = A * ~3 -- So where did I do my math wrong?

To do the problem correctly, you have to use conservation of energy. The equation for finding the gravitational potential per unit mass anywhere inside a sphere of uniform mass is
$$ E = \frac{2}{3}\pi G p (r^2-3R^2) $$
where p is the density, r is the distance from the center, R is the radius of the mass and G the universal gravitational constant.
Given that for a spherical body,

$$ M = \frac{4 \pi R^3}{3} $$

we can make this

$$ E = \frac{GM}{2} (r^2-3R^2)$$

Given that we are looking for the difference between the surface where r=R and the Center where r=0, we are looking for the difference between

$$E= -\frac{GM}{R}$$

and

$$E= \frac{3}{2}\frac{GM}{R}$$

For the Earth, GM = 2.987e14 m³/ s²

and if we calculate out the difference we get 31255879.6 joules/kg

Since this is the energy that is converted to kinetic energy for the falling body, and KE = mv^2/2,

Then

$$ v = \sqrt { 2 (31255879.6)} $$
( we can ignore the "m" as we are looking for "energy per unit mass".

which equals 7906.4 m/s.

orbital velocity for an object just at the surface of the Earth is

$$ v= \sqrt {\frac{GM}{r}} $$

which equals 7906.4 m/s

This is no accident as

$$ -\frac{GM}{R} - \left (- \frac{3}{2}\frac{GM}{R} \right ) = \frac{GM}{2R} $$

Thus

$$ v = \sqrt{ 2 \frac{GM}{2R}} = \sqrt {\frac{GM}{r}}$$

 

[metastable]

which equals 7906.4 m/s

Isn’t this the average speed from side to side through the earth? If it’s an average there should be a min velocity and a max velocity. The min is starting from 0m/s at the surface so if the avg speed from side to side is 7906.4m/s, then what is the maximum that gives that average?

 

[PeterDonis]

Isn’t this the average speed from side to side through the earth?

No, it's the maximum speed--the speed the object has at the instant it is passing the center of the Earth, when all of the potential energy it had at the surface (where it was at rest) has been converted to kinetic energy.

 

[metastable]

No, it's the maximum speed--the speed the object has at the instant it is passing the center of the Earth, when all of the potential energy it had at the surface (where it was at rest) has been converted to kinetic energy.

1/2 space station orbital period = 2700 seconds
1/2 circumference of Earth meters = 20037500 meters
A = ~7421.29 meters per second = 20037500 meters / 2700 seconds = rough circular orbit velocity
radius of Earth = 6371000 meters
duration of fall from across Earth diameter = 2291 seconds
duration of fall to center of Earth = 1145 seconds
avg velocity from non-rotating surface crossing center = 11127.56 meters/s = B
C = max velocity from non-rotating surface crossing center = assumption 2*B = 22255.12 meters/s

C = A * ~2.99... = A * ~3
A/C= 0.333... = ~1/3

which equals 7906.4 m/s.

space station orbital period: 92.68min
https://en.wikipedia.org/wiki/International_Space_Station
" The traveler would pop up on the opposite side of the Earth after a little more than 42 minutes."
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/earthole.html
42min = 2520 seconds
earth radius=3,963 miles
https://en.wikipedia.org/wiki/Earth_radius
earth diameter = 2 * 3,963 miles = 12755660.54 meters

I think I figured out where I went wrong, the roughly correct avg speed is:

12755660.54 meters Earth diameter / 2520 seconds across diameter = 5061m/s avg
^the right answer

The mistaken formula I had previously used was:

(2*radius of Earth = 6371000 meters) / 1145 seconds across radius = 11128.38m/s avg
^the wrong answer

 

[metastable]

I count at least 3 differences between the "galactic barycenter plunge" vs "fall-through-the-earth" scenarios:
-the galaxy is more of a disc than a sphere
-the matter within the solar systems radius with respect to the galactic barycenter is not a uniform density
-there is a disproportionately large over-density very close to the center of the galaxy in the form of a supermassive black hole

 

[Janus]

I count at least 3 differences between the "galactic barycenter plunge" vs "fall-through-the-earth" scenarios:
-the galaxy is more of a disc than a sphere
-the matter within the solar systems radius with respect to the galactic barycenter is not a uniform density
-there is a disproportionately large over-density very close to the center of the galaxy in the form of a supermassive black hole

While the visible matter of the galaxy is disk shaped, there is also the dark matter which is spherically distributed, and the amount of DM closer to the center of the Galaxy than the Earth is amounts to a significant fraction of the mass closer to the center.

The SMBH at the center is roughly 4 million solar masses, this is pretty small compared the mass of the galaxy as a whole. As others have already mentioned, its effects would not become prominent until you were almost all the way to the center. In which case, you could approach the speed of light if you were to swing in close enough to it, but that is true for any black hole and not just super-massive ones.

 

[PeterDonis]

you could approach the speed of light if you were to swing in close enough to it, but that is true for any black hole and not just super-massive ones

This is true, but the smaller the hole, the more precisely you have to aim to achieve a speed close to the speed of light without falling into the hole. A tangential flyby at less than $r = 3M$ will fall in (since at $r = 3M$ photons moving purely tangentially will remain in a circular orbit about the hole), and at a radius much larger than that your peak velocity will fall off fairly fast. So rough order of magnitude, for a 10 solar mass black hole you will have to aim with an accuracy of a few tens of kilometers. For the SMBH at the center of the Milky Way, the required accuracy is "only" a few tens of millions of kilometers.

 

[DaveC426913]

If you killed the Earth's galactic orbital velocity, how fast would it be moving when it reaches the center? . Oddly enough, the answer works out to be 215 km/sec, or its original orbital speed.

:slaps forehead:
Of course!
A satellite that takes 90 minutes to orbit the Earth will take 90 minutes - whether the orbit is circular or whether it is a degenerate ellipse (straight down, straight up, through a hole).

The velocity when it crosses the centre will be the same, no matter the shape.
So, in Earth's case, 7.7km/s (same as its orbital speed).
In the Milky Way's case, 215km/s (same as our orbital speed around the galaxy).

 

[metastable]

If there's an extra 215km/s worth of fuel on board the spacecraft , and I use it right after the initial burn to head straight down, I expect to have roughly 215km/s "exit" velocity (relative to the point in spacetime where I exit the solar system radius) after passing the barycenter & the closest approach to the black hole.

But suppose I take advantage of the Oberth effect in the vicinity of the super massive black hole, and use the 215km/s fuel on board as quickly as possible and as close as possible to the black hole (as close as possible without the spacecraft being destroyed by tidal forces, and as quickly as possible without being destroyed by acceleration forces), how much velocity will I have relative to the point in spacetime where I exit the solar system radius after passing the barycenter & the closest approach to the black hole?

 

[PeterDonis]

I expect to have roughly 215km/s "exit" velocity (relative to the point in spacetime where I exit the solar system radius)

You can't have a velocity relative to a point in spacetime. That doesn't make sense.

You also can't, in general, have a velocity relative to a distant object in a curved spacetime., as I explained before.

 

[metastable]

You can't have a velocity relative to a point in spacetime.

You also can't, in general, have a velocity relative to a distant object in a curved spacetime., as I explained before.

Let me rephrase: If we make the galaxy more like falling through the Earth with even spherical density, and I burn to cancel my orbital motion, and then burn 215kms straight down, I expect when passing near the Earth on the other side of the galaxy, a trapped electron in my craft will have roughly 215km/s * sqrt(2) relative velocity to a trapped electron on Earth's surface.

 

[metastable]

So I am asking if a different planet besides Earth (with trapped electron on its surface) also happened to be orbiting the galactic barycenter at 215km/s at the same circular distance from the barycenter as Earth, and my craft passed very near this planet on its outbound trajectory from sagittarius a* (assume for simplicity the milky way is spherically symmetrical and evenly dense except for the super massive black hole), and rather than using my 215km/s burn immediately after the orbital cancellation burn, I use the 215km/s burn as close as possible to sagittarius a* without destroying the craft from tidal forces or acceleration, what's the largest relative speed the electron on my craft can have with respect to the electron on the described planet at "outbound close approach?"

 

[DaveC426913]

I kind of wish you'd simply state what you're trying to accomplish in context, and let is figure out the best approach. Is this for a story plot? Are you planning an entangled electrons experiment at maximum separation?

Why do you keep referring to a trapped electron on the Earth's surface. That seems needlessly specific.
Surely just Earth is sufficient as your reference point.

 

[PeterDonis]

rather than using my 215km/s burn immediately after the orbital cancellation burn, I use the 215km/s burn as close as possible to sagittarius a* without destroying the craft from tidal forces or acceleration, what's the largest relative speed the electron on my craft can have with respect to the electron on the described planet at "outbound close approach?"

So, as you mentioned before, you want to use the Oberth effect. If the black hole weren't there, we could simply use the non-relativistic Oberth effect formula, which since the delta-v of the burn and the periapsis velocity are the same for this case, would (I think) simply multiply the delta-v of the burn by a factor of $\sqrt{2}$. So the "outbound close approach" velocity would be $215 \sqrt{2}$ km/s.

However, if we are using the black hole, we can achieve much larger periapsis velocity, and we also have to take into account relativistic effects. I don't have time to calculate that right now, and I can't find a quick reference online. But the basic idea would be to calculate the change in energy at infinity produced by the burn; because energy at infinity is conserved for geodesic motion, this change will show up as kinetic energy (per unit mass) of the object at "outbound closest approach" (because without the burn at periapsis kinetic energy would be zero at that point).

 

[metastable]

If you plan to coast to the galactic centre, you're still going to have a heckuva time dodging all the stellar gravity wells you pass through.

What if each star passing is thought of more as an "opportunity" for an oberth manuever?

I kind of wish you'd simply state what you're trying to accomplish in context, and let is figure out the best approach. Is this for a story plot? Are you planning an entangled electrons experiment at maximum separation?

Why do you keep referring to a trapped electron on the Earth's surface. That seems needlessly specific.
Surely just Earth is sufficient as your reference point.

Ethics aside, the spacecraft could be dropping off capsules of extremophile bacteria onto exoplanets along the way. The electrons are chosen as points of reference, since other particles on Earth could be moving quite fast.

 

[PeterDonis]

The electrons are chosen as points of reference, since other particles on Earth could be moving quite fast.

Huh? To a very good approximation for your purposes, every particle on Earth is moving at the same speed, including your electron sitting in a trap in someone's lab. Relative velocities between different parts of the Earth are no larger than a couple of kilometers per second; that's rounding error when you're talking about rocket burns with 215 km/s delta v.

 

[PeterDonis]

What if each star passing is thought of more as an "opportunity" for an oberth manuever?

Every such maneuver requires you to carry more fuel. How much total delta v budget are you planning for? You started with one 215 km/s burn (to negate the solar system's velocity relative to the galactic barycenter). Then you added a second 215 km/s burn. How many more burns are you planning?

 

[metastable]

How many more burns are you planning?

Sorry, I didn't mean to make the problem more complicated before we solved the simpler case.

 

[PeterDonis]

I didn't mean to make the problem more complicated before we solved the simpler case.

Well, your general question seems to be "what's the highest speed that can be achieved" (with some caveats required about what "speed" means). If you're going to allow yourself an unlimited number of rocket burns so you can do a gravity assist whenever you want, which means an unlimited amount of fuel, then the general answer is that you can get your speed as close to the speed of light as you want. There's no need for an elaborate discussion if that's the answer you're looking for.

 

[metastable]

Well, your general question seems to be "what's the highest speed that can be achieved" (with some caveats required about what "speed" means). If you're going to allow yourself an unlimited number of rocket burns so you can do a gravity assist whenever you want, which means an unlimited amount of fuel, then the general answer is that you can get your speed as close to the speed of light as you want. There's no need for an elaborate discussion if that's the answer you're looking for.

Sorry I didn’t want to diverge the topic from the 2 total burns, each 215km/s. I was suggesting a way to possibly overcome the other problem that was mentioned (somewhat unpredictable in advance gravity wells), by using additional fuel.

 

[PeterDonis]

I was suggesting a way to possibly overcome the other problem that was mentioned (somewhat unpredictable in advance gravity wells), by using additional fuel.

In a practical sense, the problem you have is not needing more fuel to deal with unpredictable gravity wells. It's that you need really, really, really, really precise aiming of your trajectory from the start (which is impossible without knowing in advance where all those unpredictable gravity wells are), if you want to have any appreciable chance of passing by the galactic center and coming out all the way to the "outbound close approach" point on the other side.

In any case, this is developing into an open-ended series of questions on your part, not a well-defined question with a well-defined answer. So this thread is closed.