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The USS Enterprise encounters an anomaly 100,000 km off its bow. It makes one complete circular orbit around the anomaly and finds it has traveled about 630,000 km in the orbit. The captain gets wary and moves to 250,000 km away from the anomaly, and makes another complete circular orbit around the same anomaly. Oddly the length of the orbit is still about 630,000 km! What is going on?

Hint/Clarifications (from comments) :

- The orbit and the anomaly are in the same 2D plane
- The anomaly did not move. Only the Enterprise moved
- The second orbit is around the same anomaly

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closed as too broad by manshu, question_asker, LeppyR64, Deusovi, Engineer Toast May 24 '16 at 19:26

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ There is definitely a distortion in space-time making the travel distance not depend on radius. $\endgroup$ – fffred May 23 '16 at 13:18
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    $\begingroup$ @Marius - yes, it is relevant. $\endgroup$ – Jiminion May 23 '16 at 14:15
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    $\begingroup$ This isn't really a puzzle, just "what physics concept would explain this". $\endgroup$ – question_asker May 23 '16 at 17:04
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    $\begingroup$ So... turns out this is not only not lateral thinking, but I was exactly right in it just being a physics problem. $\endgroup$ – question_asker May 24 '16 at 15:37
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    $\begingroup$ Yeah, definitely not lateral thinking in the way I understand that phrase (or the way the tag is defined here). Also I would argue that general relativity is quite a lot of notches past "What colour is the bear". I studied maths at university including a module on relativity and I am still not convinced that the question/answer is correct. $\endgroup$ – Chris May 24 '16 at 16:10

21 Answers 21

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I think the easiest and simplest answer is:

a neutron star or a black hole. Image courtesy by http://i.stack.imgur.com/Ste0l.gif Original from: http://imagine.gsfc.nasa.gov/docs/teachers/blackholes/bhm/images/st_diagram.gif

Imagine both orbits inside the funnel, the first slightly "above" the neutron star and the second 150.000km further up the funnel, but still with almost the same orbit length.

Explanation:

A strong point of gravitation will curve space-time.
Imagine a very heavy weight on an elastic plane, it will create a deep funnel.
If the enterprise is close to the object, it will orbit around it inside of the funnel. If it moves away from the object, the enterprise will "climb up" the inside of the funnel for 250.000 km, but will still be inside of the funnel.

From a top down view the enterprise will have almost the same distance to the black hole in both cases, since it moved along the curved surface "upwards" away from the anomaly, but the funnel will have almost the same radius, so the orbit has almost the same size (it will be a bit bigger, but the numbers are ~about the same)

From their personal viewpoint they will measure their distance along the wall of the funnel "down" to the black hole and by that measurement will have a distance of 250.000 km in the second case and keep this distance while moving in orbit. - Only once they are out of the extremely curved space-time around the black hole the relation of radius and orbit length will get back to expected values.

So, is this physically possible? It seems that it is.

Suppose our anomaly has a mass of about 30,000 solar masses. Its Schwarzschild radius will then be about 100,000km. Let the object be just barely too big to collapse into a black hole; then for a large range of radial distances from its surface, the circumference of a circle concentric with the object hardly changes.
(In the limit where the size of the object equals the Schwarzschild radius, it becomes a black hole; its surface is the event horizon; and the event horizon is infinitely distant from any point outside, as viewed by a stationary external observer.)
So now we begin 100,000km away (radially) from the surface of the anomaly; the fact that our orbit's circumference is about $2\pi$ times this distance is a mere coincidence; what's physically significant is that it's about $2\pi$ times the Schwarzschild radius of the object. We move to 250,000km away, but provided we put the anomaly surface close enough to the Schwarzschild radius this is still close enough that the orbit circumference is approximately $2\pi r_s$.

The term "orbit" is a bit misleading here, because

there are no actual stable orbits this close to a very massive object. (There are none closer than $3r_s$.) So the Enterprise must actually be taking active measures to avoid being pulled in by the anomaly.

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    $\begingroup$ If you start to distort space-time that much, it does not make much sense to speak about distances of two points in "space", because there is no global time. Only distances between space-time events are well defined as the maximum proper time of a connecting geodesic. $\endgroup$ – M.Herzkamp May 24 '16 at 12:17
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    $\begingroup$ Is this solution actually compatible with the circumference being approximately 2 pi times the first distance to the anomaly, as stated in the question? $\endgroup$ – Gareth McCaughan May 24 '16 at 16:28
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    $\begingroup$ Does this make the orbit still in the same plane? What does "plane" even mean in this situation? $\endgroup$ – paste May 25 '16 at 1:31
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    $\begingroup$ Meh, actually I guess it kinda-obviously does work. We take $r_s$ approx 100,000km, we make the anomaly a teeny-tiny bit bigger than the Schwarzschild limit, and now by making that teeny-tiny bit very small we can make the distance you can go from the surface while still keeping $r$ very close to 100,000km as large as we like. So make the teeny-tiny bit small enough that you can go out 250,000km with negligible change in radius, and we're done. $\endgroup$ – Gareth McCaughan May 25 '16 at 10:46
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    $\begingroup$ @Jiminion: The problem with this answer is when you get close enough for the "approximately" to have the effect you want you're far too close to orbit at all. $\endgroup$ – Joshua May 27 '16 at 20:06
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Is the anomaly

An Ellipse

Illustrated:

Orbits

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  • $\begingroup$ No. The anomaly can be considered spheroid. $\endgroup$ – Jiminion May 23 '16 at 14:48
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Q is hanging around and in a mood. Between the completion of the first orbit and the beginning of the second, he snapped his fingers and changed the value of pi.

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    $\begingroup$ "Q! Where are my pants?! Bring back my pants!!" $\endgroup$ – Jiminion May 23 '16 at 15:54
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    $\begingroup$ "Really, you humans and your pants. How childish." $\endgroup$ – Irishpanda May 23 '16 at 16:02
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    $\begingroup$ and just like that, he disappeared, called away to a cartoon pony convention $\endgroup$ – question_asker May 23 '16 at 16:17
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    $\begingroup$ Sans pants, of course. $\endgroup$ – Irishpanda May 23 '16 at 16:18
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This is a great example of classic "lateral thinking" puzzle misdirection, sending everyone into more and more convoluted spacetime-based answers, when the true answer is actually much simpler, if you just stop and read the puzzle carefully.

Highlighting the important words from the puzzle:

The USS Enterprise encounters an anomaly 100,000 km off its bow. It makes one complete circular orbit around the anomaly and finds it has traveled about 630,000 km in the orbit. The captain gets wary and moves to 250,000 km away (...) Oddly the length of the orbit is still about 630,000 km! What is going on?

Therefore:

The Enterprise's distance from the anomaly hasn't actually changed between the two orbits.

The puzzle only tells us that the Captain moved further away from the anomaly after the first orbit (in a shuttlecraft of some sort, one assumes?). The Enterprise is therefore presumably still at a 100,000 km distance from the anomaly, and repeats the same circular orbit that it did before.

It is therefore unsurprising that the distance traveled during that second orbit remains roughly 2 * pi * radius, approximately the same 630,000 km value given in the original question; no complicated theoretical spacetime geometries required.

QED.

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I notice that the word “about” is used a couple of times in the question, so I assume that we’re allowed to use some approximations in our answers.  And, of course, the conventional answer is that the circumference (the length of the orbit) is $2 \pi r$.  For $r=250,000$, this formula gives us a circumference of $2 \times \pi \times 250000 = 1570796 \text{ km}$.  Now suppose that

the helmsman was in a funny mood (perhaps inebriated?) and decided to report the length of the orbit in base 12: \begin{align}1570796_{(\text{base 10})}=639038_{(\text{base 12})}\end{align}

so, in that sense, the length of the orbit was “about” 630,000 km.

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The object was

A quickly rotating planet - the orbits were measured in terms of the planet's coordinate space rather than that of the Universe.

This works because the second orbit:

started at one of the poles, let's say the North pole, and went around the planet to the South pole and back up again to the North - a full circle relative to the North pole.

But the first orbit:

started at some point on the equator, and went around the equator in the same direction as the planet's rotation. In order to get to the same point above the planet's surface, they had to double up the path to keep up with the rotation, thus taking a longer path in the planet's coordinate-space, offset by the fact that they were closer, so they just happened to be the same distance traveled in Universe-space.

Illustration:

enter image description here

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Is the anomaly a

Wormhole?

Consider the following

(The above image may have been copied from one of the following sources: 1, 2, 3, or 4.)

Then

The Enterprise starts off adjacent to the blue side, orbits the hole there then moves along the illuminated path in the picture, then orbits the hole on the yellow side.

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  • $\begingroup$ Your answer cannot be correct, since in your answer, the USS Enterprise is not '250,000 km away from the anomaly' when starting the second orbit. $\endgroup$ – jarnbjo May 23 '16 at 14:29
  • $\begingroup$ I was thinking about that, but a wormhole wouldn't be recognised by an observer as an anomaly. It would be just normal continuation of space-time, although curved. $\endgroup$ – fffred May 23 '16 at 14:29
  • $\begingroup$ No, but very creative answer! The (only) problem with this answer is that the Enterprise is travelling from one anomaly to another (the other side of the wormhole.) In the puzzle, they moving away from a single anomaly. $\endgroup$ – Jiminion May 23 '16 at 14:31
  • $\begingroup$ This is on the right track, in a general sense. $\endgroup$ – Jiminion May 23 '16 at 14:32
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    $\begingroup$ +1 for the picture. Where I it from? $\endgroup$ – BmyGuest May 23 '16 at 21:38
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Here's a somewhat silly answer to this question:

The USS Enterprise is 150,000km long (bow to stern), and they simply turned the ship around so the bow is 250,000km away. The orbit will still be the same length of travel

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  • $\begingroup$ I think the USS Enterprise is smaller than that: gizmodo.com/… $\endgroup$ – Marius May 23 '16 at 14:13
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    $\begingroup$ The Constitution-class Enterprise (TOS) is about 288 meters long. $\endgroup$ – Jiminion May 23 '16 at 14:13
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    $\begingroup$ He never said it was the USS Enterprise from Star Trek, that was just assumed (though yes it is silly for a ship to be this large) $\endgroup$ – Gordon Allocman May 23 '16 at 14:14
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This can certainly happen if the Enterprise and the anomaly are in a

very highly curved region of spacetime.

Here's how it works "one dimension down", on a 2-dimensional surface.

Imagine a planet that, for some reason, is far from spherical: it's "squashed" so that its pole-to-pole distance is much shorter than the distance across its equator. Specifically, its pole-to-pole distance is 350,000km but its equatorial radius is quite a lot bigger.

The anomaly is at the south pole. The Enterprise is near the south pole, 100,000km away from it. Because of the odd shape of the planet, a circular orbit at that distance from the south pole (on the surface of the planet) isn't much shorter than the $2\pi$ times 100,000km you'd get in a perfectly flat space. That's the initial "about 630,000km".

Now the Enterprise moves north until it's 100,000km from the north pole of the planet. It's therefore 250,000km away from the south pole. But its new orbit is the exact same length as before.

To make this work in space, we need to arrange that a 2d "slice" of space containing the anomaly and the Enterprise's two orbits has the same sort of geometry as the surface of our planet. That would require some pretty severe gravitational fields but hey, it's an anomaly.

PS.

The bear was white.

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    $\begingroup$ This is basically correct. (The first tag, anyway.) The second tag is a bit more off the mark -- the overall situation is much much simpler than this. $\endgroup$ – Jiminion May 23 '16 at 15:19
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    $\begingroup$ Well, it is assuming a planet with a radius much larger than Jupiter's and it is assuming it is highly elliptical in shape, which is also without precedent. $\endgroup$ – Jiminion May 23 '16 at 15:53
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    $\begingroup$ The planet is only for illustrative purposes: it's intended to describe the geometry of a slice of spacetime near the anomaly. I'm not proposing that there's an actual enormous planet anywhere around; just that whatever gravitational weirdness may be going on has the effect of curving spacetime in a manner similar to the surface of the planet. $\endgroup$ – Gareth McCaughan May 23 '16 at 15:55
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    $\begingroup$ No one said the anomaly was the sole cause of the gravitational weirdness. Perhaps the thing at the south pole is just very sparkly or something. $\endgroup$ – Gareth McCaughan May 23 '16 at 16:38
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    $\begingroup$ @GarethMcCaughan SPOCK: "Captain, sensors are picking up something very sparkly, dead ahead." KIRK: "Would you say it's sparkly, or shiny?" SPOCK: "Wrong franchise, Captain". $\endgroup$ – Todd Wilcox May 23 '16 at 17:12
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The Enterprise and the anomaly are

on the surface of a planet (an oblate spheroid) with a circumference of 700,000 km.

When the Enterprise is 100,000 km from the anomaly

it does a circuit of the planet, centered on the anomaly. At this point, the radius of the circle is 630,000km.

When the Enterprise is 250,000 km from the anomaly

it has crossed over the equator (relative to the anomaly) and is on the other side of the planet, 100,000 km from the place on the planet opposite the anomlay. Here, the circumference of a circle around the planet is 630,000km, same as it was before.

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Was the object . . .

A ring?

So if it was, then . . .

They started on the inside of the ring that is 450,000 km in diameter, and they were 100,000 km straight into the ring. They orbited and moved 250,000 km to the other edge of the ring and orbited around the opposite edge. The ring would be of constant thickness so the orbits were equal.

Picture

orbits

Picture Explained

They would start at A (100,000 km from edge), orbit around the nearest point on the ring, move to B (100,000 km from opposite edge), and orbit around the nearest point on the ring

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  • $\begingroup$ OP already mentioned in another comment that the anomaly can be considered spheroid. $\endgroup$ – paste May 24 '16 at 4:18
  • $\begingroup$ ... and has reiterated "The enterprise moved 150,000 km and determined they were 250,000 km away from the anomaly." I don't see how your answer is consistent with that. $\endgroup$ – Peregrine Rook May 24 '16 at 6:35
  • $\begingroup$ Also this doesn't account for the total distance traveled remaining constant. $\endgroup$ – Daniel Arnett May 24 '16 at 11:41
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The Enterprise is in

A game of 'Asteroids'. It completes an orbit 100km around an asteroid that isn't moving. Then it travels 150,000km to the next 'screen'. Now the Enterprise is simultaneously 250,000km away from the asteroid in the previous screen and 100,000km from the very same asteroid in the current screen. They orbit the asteroid and travel the same distance as the previous orbit.

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Are they

Just inside the event horizon of a supermassive black hole (but somehow not destroyed by tidal forces)?

-

The Enterprise orbits once at the first perceived distance, then attempts to move away but only actually moves infinitesimally closer to the event horizon. Then it orbits at the second perceived distance. Gravity must be affecting their sensors.

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  • $\begingroup$ This answers the question by saying "suppose the problem statement is actually wrong". In the absence of any indication in the puzzle that that might be so, I don't think it really counts as an answer. $\endgroup$ – Gareth McCaughan May 23 '16 at 15:43
  • $\begingroup$ The enterprise is not inside an event horizon. $\endgroup$ – Jiminion May 23 '16 at 15:50
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    $\begingroup$ The problem statement must be wrong on its face. The measurements can't be both correct and incorrect at the same time, and there's no way to specifically determine what's going on from the information given. You could say that the Enterprise is hitting micro-warp-bubble-o-matic-trons around the anomaly and popping in and out of existence so that it only records 630,000km of movement without noticing the gaps it jumped. Given no further restrictions, I'd have to say this is too broad. $\endgroup$ – Epicedion May 23 '16 at 15:59
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    $\begingroup$ Then we're looking at a region of space where the relative spatial dimensions are curved so sharply that all directions away from the anomaly are along the interior surface of a sphere with its center at the anomaly; i.e., the interior surface of the event horizon of a black hole, or something very much like it. $\endgroup$ – Epicedion May 23 '16 at 16:22
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    $\begingroup$ No need for an event horizon. (My answer does describe a spacetime where you can't escape to infinity away from the anomaly, but it's easy enough to change that: imagine a surface whose cross-section is something shaped a bit like a capital-Omega instead of an O, where the "feet" of the Omega go off to infinity, and now arrange for slices of spacetime through the anomaly to look like that instead.) $\endgroup$ – Gareth McCaughan May 23 '16 at 16:41
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The Enterprise moved

In parallel to the orbit's axis

So they moved further away and they could perform

A halo orbit around the axis and "above" the anomaly

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  • $\begingroup$ The enterprise moved 150,000 km and determined they were 250,000 km away from the anomaly. $\endgroup$ – Jiminion May 23 '16 at 16:00
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Is it

a star shrinking? 530,000km is 76% the radius of the Sun. If it's a star that rapidly collapses from a radius of 530,000km to 380,000 km then the Enterprise would not need to move.

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    $\begingroup$ Interesting answer. $\endgroup$ – Jiminion May 23 '16 at 22:52
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Space in this area has some pretty absurd curvature not reducing to flat as we get farther away within the domain here. I can't find what's going on but I can prove all geometries radially symmetric around the "anomaly" and reduce asymptotically to flat in all directions are disallowed. Therefore, the real anomaly is not the stated one or the wormhole answer is correct. Since the wormhole answer wasn't accepted, we are dealing with spacetime curved back on itself.

The simplest solution involved is a hypersphere with circumference of ~ 4 * (100,000km + 250,000km) / 2 ~= 700,000km, with the anomaly at one pole. But this leaves no path home.

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  • $\begingroup$ That simplest solution is basically the same one I (and later Joe M) proposed, and neither has been accepted so presumably Jiminion has something else in mind. Can you sketch the proof that one can't have a radially symmetric spacetime like this that is asymptotically flat? (I handwavily claimed in comments that it ought to be possible, but without doing any calculations; if you've done the calculations, I bet you're right.) $\endgroup$ – Gareth McCaughan May 24 '16 at 10:40
  • $\begingroup$ It's a really simple proof based on the inverse square law combined with knowing that cylindrical spacetime closed at one end is not an allowed spacetime. $\endgroup$ – Joshua May 24 '16 at 15:17
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New answer, best/worst I could come up with:

The anomaly is causing the Enterprise to go much faster in its orbit the further away it is. If it's in non-relativistic-land at 100,000km then at 250,000km the Enterprise is traveling about 0.92c relative to the anomaly, give or take.

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This is perfectly possible:

The first orbit is calculated from the edge of the anomaly. The second orbit is calculate from the center of the anomaly. The anomaly itself is 150,000 KM in radius. So the Enterprise location hasn't actually changed, just the way the orbit is calculated.

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  • $\begingroup$ the Enterprise has to be at 100 000 km of the center of the orbit/anomaly in order to have an orbit length of 630 000 km. So the anomaly can't have a 150 000 km radius $\endgroup$ – Fabich May 24 '16 at 12:37
  • $\begingroup$ @Lordofdark You're right. My mistake. $\endgroup$ – Nzall May 24 '16 at 12:44
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Perhaps "bow" is important here:

If the Enterprise is on the inside of a 350,000 km hollow sphere.

The Enterprise starts with the bow pointed toward the anomaly 100,000 km away and while keeping the bow pointed at the anomaly it does one orbit...

on the inside of the sphere.

Then, backing up 150,000 km, it does another orbit with the bow still pointed at the anomaly, turning at the same rate...

causing it to orbit on the same orbit but facing in the opposite direction. The stern is now 100,000 km away from the anomaly but the distance away is measured from the bow, and thus the reason why it was important to add this distinction in the puzzle description.

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The Enterprise and the anomaly are

on the surface of a curved fourth dimension with a circumference of 350,000km, assuming the anomaly is basically a point in space. The Enterprise began 100,000km away from the anomaly in one direction, and 250,000km away from it in the exact opposite direction.

So when the Enterprise moved 150,000km, and was ostensibly 250,000km away

it actually achieved a maximum distance of 175,000km before it started getting closer again from the other direction. Thus, when it was time to make another circular path, they took the shorter route.

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    $\begingroup$ I wouldn't quite put it that way. $\endgroup$ – Jiminion May 23 '16 at 20:21
  • $\begingroup$ @Jiminion Was it my numbers that were the issue, or something else? $\endgroup$ – Joe M May 23 '16 at 21:07
  • $\begingroup$ Before the move, it was 100,000 km away from the anomaly; after the move, it is 250,000 km away. $\endgroup$ – Jiminion May 23 '16 at 22:39
  • $\begingroup$ P.S. This answer is just a rewording (arguably, a more understandable one) of Gareth McCaughan's answer. $\endgroup$ – Peregrine Rook May 24 '16 at 6:24
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As the enterprise moves away,

the anomaly gets smaller and the enterprise doesn't move; therefore the radius to the orbit's center hasn't changed but the distance to the anomaly's edge has.

In this case it would seem

the only way to escape the anomaly's effects would be to move far enough away so the anomaly vanishes into nothingness... Looking out the window should demonstrate that positions of the stars relative to the enterprise's position don't move as the enterprise "moves away" from the anomaly.

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  • $\begingroup$ The tardis answer. $\endgroup$ – Jiminion May 23 '16 at 22:42

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