Htrae is a planet with an abnormally strong gravity. In Htrae, the escape velocity is greater than the speed of light. Suppose a civilization like humans (call them snamuhs) evolves, with the unique ability to withstand Htrae's abnormal gravity. If the snamuhs have the ability to become as technologically advanced as possible without breaking the laws of physics, will they eventually be able to leave the planet Htrae and go into space?
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6$\begingroup$ I don't think such a planet can exist. Whatever it's gravitational attraction, you can create an engine that generates more acceleration in the opposite direction, guaranteeing escape. Remember, the formulas for escape velocity are approximations that don't take into account relativistic effects. $\endgroup$– user15334Mar 13, 2016 at 18:29
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214$\begingroup$ Such planet is more commonly referred to as a black hole. $\endgroup$– PlutoroMar 13, 2016 at 20:18
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21$\begingroup$ your question is unfortunately based on an incorrect understanding of escape velocity; escape velocity only matters for unpowered ("inertial") escape. If you can supply enough power to counteract local gravity even a tiny bit you will eventually escape -- assuming there is a way to counteract local gravity (e.g. you're not sitting at the event horizon of a black hole) $\endgroup$– KutuluMikeMar 14, 2016 at 0:42
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10$\begingroup$ @Mike The thing is, you are sitting at the event horizon of a black hole. The radius at which the escape velocity is equal to $c$ is commonly known as the Schwarzschild radius and is the definition of the event horizon. Powered escape using anything short of a perpetuum mobile is ouf of the question as well, since even with a photon rocket you can't have the requisite delta-v. $\endgroup$– Mike L.Mar 14, 2016 at 12:03
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9$\begingroup$ It really bothers me when someone accepts a incorrect answer just because it's what they wanted to hear, and not what's correct. $\endgroup$– MermakerMar 14, 2016 at 19:54
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10 Answers
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We'll be assuming your planet somehow does exist as it is.
Short answer: technically, yes, but that depends on what you call "space" and what you want to do with it.
On Earth, space starts (conventionally at least) at the Kármán Line, or 100 km altitude. Simply going past that line does not require escape velocity. Escape velocity (11.2 km/s for Earth) is what you need to reach to break free of your planet's gravitational field.
If you want to put things in orbit, you'd need to reach orbital velocity (7.9 km/s for Earth). Reaching escape velocity would actually be the opposite of helpful, because then your satellite would just break free and go away.
You could potentially send things into orbit (if your orbital velocity isn't also faster-than-light). You wouldn't be able to send them to other planets or what have you, unless you can justify going at FTL speed.
answered Mar 13, 2016 at 19:29
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24$\begingroup$ Orbital velocities would also be above the speed of light, until you are 1.5 times further than the point at which the escape velocity drops below light speed. This planet is surrounded by an event horizon, so it has already been crushed to a singularity. $\endgroup$– James KMar 13, 2016 at 21:46
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8$\begingroup$ @MichaelKjörling: James is speaking the truth. You know that equation for calculating orbital velocity? You have to throw it away when you're near a black hole, and use a relativistic version instead. To give you an idea of how wrong Newtonian physics is near a black hole, imagine standing on the surface of our hypothetical planet. Which direction is up? Towards space? No such direction even exists. $\endgroup$ Mar 13, 2016 at 23:16
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9$\begingroup$ @QPaysTaxes: It's not that "space isn't there", but more that all directions lead towards the singularity. $\endgroup$ Mar 14, 2016 at 7:13
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7$\begingroup$ @QPaysTaxes It's not easy to understand if you don't have a good understanding of general relativity. Basically, with a non-rotating, non-charged black hole, from behind the event horizon, there is no path back outside ("up" on our planet), other than back in time. All due to the spacetime distortion. Newton mechanics only work in flat space - the surface of the hypothetical planet is anything but. Your intuition is mostly based on Newton mechanics, because that's what you see working in your daily life (to a useful approximation). $\endgroup$– LuaanMar 14, 2016 at 11:45
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5$\begingroup$ @CaptainCranium Yes, if relativity is the model for gravity. Outside a black hole it is impossible to stop moving forward in time. Inside a blackhole it is equally impossible to stop moving towards the singuarity. In fact the singularity is more like a point in time than a point in space. There can be no platform, no stable surface inside a black hole. It is possible that something happens on a very small scale that stops an actual singularity forming, but this is beyond relativity, and there is no settled science on this. $\endgroup$– James KMar 14, 2016 at 17:28
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No, they will not be able to reach space. At least if we assume that our understanding of physics is correct. Since you do not state anything to the contrary, that's an assumption I am willing to make.
Basically, what you have is a (very small) black hole. A black hole is a mass that is so dense that the escape velocity becomes greater than the speed of light. In order to get that, you need absurd densities; for comparison, if our moon were to somehow magically collapse into a black hole of identical mass, it would have an event horizon the size of a grain of sand. At these scales, many of the equations we can use to describe our everyday world (including Newtonian mechanics and some of the simpler solutions to special and general relativity) are no longer valid. See also Are black holes very dense matter or empty? over on the Physics SE.
Because nothing can go faster than light, and the escape velocity of their world is greater than the speed of light, your species cannot accelerate beyond the escape velocity of their world, meaning they cannot leave it. Apparently (see the discussion in the comments to AmiralPatate's answer to this question) they won't even be able to establish a stable orbit around their planet, because the orbital velocity only drops below $c$ well beyond the event horizon, and being able to establish a stable orbit seems the lowest usable definition of "reach space", let alone leave the planet (suborbital spaceflight has very few applications that atmospheric flight cannot cover at a significantly lower cost).
Actually, though, it's worse than that. When dealing with the absurd gravities of black holes and similar objects, gravitational spaghettification becomes one of the things that you need to worry about. Basically, the gravitational pull is so intense that the difference in gravity is noticeable along macroscopic lengths, which destroys the matter that makes up objects of interest to us. Hence, even if we ignore the issue of real estate prices on such a tiny world, such beings could not possibly evolve, because there is no matter (as commonly thought of) that could come together to form these beings and remain in a coherent shape under the gravitational stress!
And in a way, it's even worse than that. If the planet is dense enough to have an escape velocity greater than the speed of light, I would love to learn more about the star it is in orbit around, because stars tend to be vastly more massive than their planets. For comparison, in our solar system, the Sun is approximately 1,047.8 times more massive than Jupiter, or 333,000 times more massive than Earth. If the planet has an escape velocity greater than the speed of light, that makes me wonder what its insolation from its sun is like...
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4$\begingroup$ Just because the star is more massive does not mean it is a black hole. An object collapses into a black hole when it is dense enough, not when it is massive enough. $\endgroup$ Mar 13, 2016 at 23:22
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45$\begingroup$ It's not reminiscent of a black hole. It is a black hole. $\endgroup$ Mar 14, 2016 at 1:58
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2$\begingroup$ @DietrichEpp: that's a play on words. The only known way to achieve "enough density" is by being massive enough. $\endgroup$ Mar 14, 2016 at 6:29
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7$\begingroup$ @MartinArgerami: No play on words. I am not sure what definition of "density" you are going by, but you can also increase the density by decreasing the volume. According to theory, any object compressed within its Schwarzschild radius collapses into a black hole. There is no minimum mass for a black hole. Example: A0620-00 is a black hole candidate estimated at 11 solar masses. There's a long list of more massive objects out there which are not black holes, like η Carinae A (which is about 10x as massive as A0620-00). $\endgroup$ Mar 14, 2016 at 7:20
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11$\begingroup$ @MichaelKjörling: You are correct that you usually need large mass in order to create a black hole, but the question assumes that the black hole already exists. There is no question about whether a planet with superluminal escape velocity would be a black hole since "superluminal escape velocity" is basically a three word definition of "black hole". $\endgroup$ Mar 14, 2016 at 10:13
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Your 'planet' is what's known as a black hole, so not only can nothing escape from the planet, such a planet can't exist. At least according to current physics, since everything that falls into a black hole contracts to a point. (Of course I'm simplifying a bit: see e.g. Misner, Wheeler, and Thorne's book "Gravitation".)
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3$\begingroup$ I have to say, MTW isn't the greatest book for easy casual reading on gravity. $\endgroup$– HDE 226868 ♦Mar 13, 2016 at 21:45
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$\begingroup$ @HDE226868 Hey, if you want to get it right, you need to make sacrifices. Better start early... $\endgroup$– userMar 13, 2016 at 21:58
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$\begingroup$ @HDE 226868: Yeah, that was supposed to be a joke. Like calling it a little light reading. For those who've never seen the book, it's 1279 pages, and weighs in at (per Amazon) 5.6 lbs - in paperback :-) $\endgroup$– jamesqfMar 13, 2016 at 22:55
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2$\begingroup$ Cons: You'll be dead. Pros: Since time is broken inside the black hole, you may outlive everyone outside the black hole, in a certain sense. $\endgroup$ Mar 14, 2016 at 2:28
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$\begingroup$ The usual joke about MTW is that it's a very useful book for those studying gravity, because as well as containing useful information it also generates a strong gravitational field. $\endgroup$ Mar 15, 2016 at 16:36
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The fact that -
(1) the planet exists without collapsing
(2) the inhabitants exist at all
already violate so many laws of physics that we might as well have violated a few more. But keeping in spirit with your question, how about we reformulate and ask: "Can this civilization reach space without breaking additional laws of physics?"
If the planet itself has a non-trivial, non-approaching-zero radius (as it would if it were a black hole), it means that matter does not undergo gravitational collapse. That might also mean that, say, even though escape velocity cannot be reached by means of aeronautics, why not build themselves out?
Construct a skyskraper or a pyramid of some sort until they reach the event horizon. Build a platform on top of that - boom, conventional space travel.
Alternatively, if we assume infinite technological advancement, you can allow FTL travel by way of the assumed space distortion following massive amounts of energy or exotic energy, and in this manner be able to reach the event horizon with a spacecraft.
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1$\begingroup$ Also, wormholes. +1, especially for the "...without breaking additional laws..." part, though we've already thrown all four fundamental forces out the window so I'm not sure what's left. $\endgroup$– thanbyMar 15, 2016 at 19:09
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$\begingroup$ If there is actually an event horizon, then it can’t be reached with skyscrapers either. You possibly assume that General Relativity is broken and the “planet” hasn’t a horizon around it. $\endgroup$ May 17, 2016 at 12:05
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1$\begingroup$ Why couldn't it be reached? Spacetime is not infinitely curved outside of the singularity, and since the singularity cannot have volume, a planet existing inside the black hole must per definition be larger/outside of the singularity, and thus, not be subject to the infinite curve problem. With finite curve, there's nothing (bar the gravitational pull itself, which will also not be infinite and which we from the question must assume is not so strong as to be impeding) us from reaching the event horizon. $\endgroup$– VegardMay 18, 2016 at 6:56
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$\begingroup$ You can't build your way out because no material, even in principle, can withstand such a force. General relativity places fundamental limitations on material strength. If you move one end of an infinitely strong rod then the other end moves instantly, violating the speed of light information transfer limit. Any material must compress sightly as pushed, transmitting the push slower than light. $\endgroup$ Jun 17, 2016 at 20:03
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1$\begingroup$ Well, the question supposes that such materials can exist, given how these non-flat organisms are alive on a non-crushed planet. $\endgroup$– VegardJun 18, 2016 at 23:18
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Let's go back to the definition. The escape velocity is the speed you need to escape the gravity field of an object, under a free fall trajectory.
i.e. a object that does not have any means of propulsion needs to be pushed to that speed to escape. But a spaceship can continue to accelerate.
Escape velocity depends on how far you are from the center of mass of the body you are escaping. 11.2km/s for Earth is valid at ground level. If you are already as far as the Moon is, it is only 1.4km/s.
Basically, you can escape the gravity field of earth at any speed, provided you have the means of propulsion to go against gravity for a long time.
To give a simplistic example, it is as if one were to say that the velocity needed to send a pinball from New York to L.A. is 3000km/h: a car can get there going no faster than 100km/h
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$\begingroup$ See the pilot episode of Salvage 1 - Here is a link to the relevant hilarity on youtube. $\endgroup$– user15970Mar 14, 2016 at 4:05
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$\begingroup$ The problem is that you can't have a means of propulsion to resist gravity. At the event horizon, you need infinite force, and therefore infinite energy, just to stay stationary. So you can't have "a little more than infinite" propulsion to eventually escape the black hole, even in a very large black hole with negligible tidal effects. $\endgroup$– MichaelSMar 14, 2016 at 14:00
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1$\begingroup$ @MichaelS that's true, but not really relevant to the SciFi world of the orignal question. Since the snamuhs already have evolved to withstand the local gravity, clearly they can "stand up" and move around. Thus, they have some way of harnessing enough energy to move upwards. And that's all it takes: moving upwards slowly for a long time, to get to space. They don't need to hit escape velocity. $\endgroup$ Mar 14, 2016 at 19:13
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Not only is it impossible to reach space, it is impossible to move any distance up from the surface.
The escape velocity is actually a statement about energy. It is, and is computed as the velocity at which an object's kinetic energy equals its gravitational potential energy. What is an object at the speed of light's kinetic energy? Infinity. You see the problem?
Moreover to move any distance up, the amount of work you need to do is related to the escape energy by the proportion DISTANCE/RADIUS OF PLANET. So that is infinite also.
If this planet exists at all, then they are literal flatlanders, stuck to the surface of a smooth sphere. There is no up. There is certainly no space.
Some life might be possible if your gravity doesn't work in the conventional inverse square fashion, but the fact that escape velocity is c implies there is an impenetrable gravitational barrier at some distance from the planet that simply cannot be crossed, no matter what sort of drive system you use.
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Moreover to move any distance up, the amount of work you need to do is related to the escape energy by the proportion DISTANCE/RADIUS OF PLANET.Can you explain this in more detail? I can't see it intuitively. $\endgroup$– DevsmanMar 14, 2016 at 15:35 -
1$\begingroup$ Look at en.wikipedia.org/wiki/… For moving up by a distance you'd derive work done as GMm(r1-r0)/r1r0, compared to escape energy's GMm/r0, which is a factor approximately equal to Distance/Radius. Of course this isn't considering General Relativity which makes things... complicated, but it does demonstrate that escaping is not unrelated to 'merely' going up. $\endgroup$– FhnuzoagMar 14, 2016 at 16:13
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$\begingroup$ MichaelS's comment to January First of May's answer is probably a bit more rigorous. $\endgroup$– FhnuzoagMar 14, 2016 at 16:25
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Such a hypothetical planet would be roughly 200 million times the mass of Earth and only just over a quarter of the radius. In order for a ship to leave the gravitational pull of such a body it would have to reach a velocity of over 669,600,000 miles per hour. Assuming Einstein was correct and at the speed of light: a: Time stands still and b: Energy is converted to mass
Then although you may be able to lift off from the surface given enough thrust, you will always fall back to the surface of this planet, becasue as you apply more energy to break free of the gravitational pull, and you reach the velocity at which you can, your energy would be converted to mass, increasing your gravitational pull and consequently making the escape velocity higher. Net effect, you are going nowhere.
Now if Einstein was wrong......
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$\begingroup$ Wrong.You can't lift off from the surface or even stay on it. If you can lift off any distance at all then you can release a smaller rocket when you get there. As you managed to go up and the smaller rocket is further from the source of gravity, it should be able to go up. Their could either be some magic ceiling, the possibility of total escape, or accept relativity. Staying still at the surface of a black hole is equivalent to going at the speed of light. $\endgroup$ Jun 17, 2016 at 21:10
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I just did this calculation quickly so I won't be humiliated if someone points out an error.
Escape velocity can be calculated by:
$v=\sqrt{\frac{2GM}{r}}$
where $G$ = gravitational constant, $6.67 * 10^{-11} m^3/(kg*s^2)$
$M$ = mass of planet
$r$ = radius of planet (or wherever we're starting our trip, but lets assume the surface)
So for $v=c$, where $c ~=3 * 10^8 m/s$, and assuming we somehow have a planet approximately the same size as the Earth, let's say $r= 6 * 10^6 m$
$3*10^8m/s = \sqrt{2 * 6.67* 10^{-11} * M/ 6 * 10^6}m/s$
$M = 4.0 * 10^{33} kg$
Wow! The gravity on this planet would then be:
$g = \frac{GM}{r^2}$
= $\frac{6.67 * 10^{-11}m^3/(kg*s^2) * 4.0*10^{33}kg}{(6*10^6m)^2}$
= $7.4 * 10^9 m/s^2$
Acceleration of gravity on the surface is 7.4 BILLION meters per second squared. Compare that to Earth's 9.8 meters per second squared. This is 700 million g's!
It's hard to imagine how anything remotely like human life or human civilization could exist in such conditions.
Of course it's hard to see how anything remotely resembling an Earth-like planet with a radius of 6000 km could exist under those conditions, and at these numbers I think there would relativistic effects so my Newtonian equations are probably inaccurate.
I'm not prepared to say that no technological civilization would be possible. Who knows? This is so far outside our experience that it's hard to know where to begin the discussion.
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$\begingroup$ I was calculating the same, but from the density point of view. Will post as an answer anyway. $\endgroup$ Mar 14, 2016 at 13:13
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2$\begingroup$ The error is that you're using Newtonian math in a relativistic setting. "Surface gravity" at the black hole is infinite in terms of force. Because time passage goes to zero as acceleration goes to infinity near the event horizon, there's a sort of $0\cdot\infty$ thing where it doesn't look like infinite surface gravity to an observer at the horizon (other objects would appear to accelerate past them at finite speeds), but it would certainly feel like infinite gravity if an unobtanium floor stopped them at the horizon. $\endgroup$– MichaelSMar 14, 2016 at 14:13
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$\begingroup$ @MichaelS Sure. I said something like that in my second-to-last paragraph. It's an awkward scenario, where the question essentially posits Newtonian physics in a situation where they don't apply. If dogs could talk, would German shepherds speak German? How do you answer such a question? I was trying to point out that the premise is ... if not impossible, at least requires some discussion of how to interpret it. $\endgroup$– JayMar 14, 2016 at 18:36
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$\begingroup$ Of course, even if Newtonian physics did hold, 7.4 billion ${m/s^2}$ would accelerate objects to the speed of light rather quickly, which obviously won't work. $\endgroup$ Mar 14, 2016 at 18:45
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1$\begingroup$ Definitely. And those atoms would get crushed down as well. Point was: you'd never be able to lift anything off the surface as you'd have to overcome a non-trivial amount of that acceleration. $\endgroup$ Mar 14, 2016 at 19:06
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Short answer would be Yes. It is possible to reach to space.
If said simply escape velocity the speed with which you need to throw a stone so that it can reach space.
If need little technical explanation it can be said like- The kinetic energy at the surface should be equal to potential energy at the edge of planet after which you can go to space. i.e.
$ 1/2 mv^2 = mgh $ or $ 1/2v^2 = gh $
Assuming g is same at all heights(for simple calculations).
Now for v(velocity) to be very large(like greater than light), either g(gravity) or h(radius) of planet should be very large.
Since you said that snamuhs are capable of surviving in such high gravity
So for escaping to space his ship/rocket need to generate force enough to counter gravity. Escape velocity has no role to play here.
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5$\begingroup$ This formula is incorrect, the correct formula is $E^2 = m_0^2 c^4 + p^2 c^2$, and this will diverge when you set v=c. The formula in this answer only works for low velocity (relative to the speed of light). $\endgroup$ Mar 14, 2016 at 10:00
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I'll start out: there are, unfortunately, some relativity-based considerations that make it impossible according to actual laws of physics, even ignoring the whole "how the heck Htera isn't collapsing" part.
(TL/DR: the so-called space curvature will mean that objects sufficiently far away are literally unreachable in any possible way. If light can't reach it, nothing else can reach it either. This is called the event horizon.)
However, this does not necessarily mean that the surface gravity is going to be extremely large. Or that space exploration inside the event horizon is impossible (but we'll get to that).
You know what else has a reasonable surface gravity but a huge escape velocity? A Dyson sphere. (I mean the outer-facing kind, not the silly inner-facing version that has everything falling on the central star.)
Specifically, the law of gravity being what it is, for the surface gravity to be the same, the mass of the object must be proportional to the square of its radius.
Plugging this proportionality (specifically, $M={gr^2\over G}$, which follows from $g={GM\over r^2}$ for surface gravity) into the equation in Jay's answer, we get $v=\sqrt{2gr}$, which means $r={v^2\over 2g}$, or $g={v^2\over 2r}$.
This is actually fairly easy to navigate - if $g$ is Earth gravity, $c^2\over 2g$ is approximately ${(3\times10^8)^2\over 20}$, or $4.5\times 10^{15}$ meters - or 4.5 trillion km, or 4500 billion km, or 30000 AU (a bit over a tenth of a parsec, a bit under half of a light year). This gets proportionally smaller if surface gravity can be higher.
So yes, if Htera is sufficiently large (about a light-year across for Earth-equivalent gravity), it can have reasonable surface gravity while having an escape velocity higher than the speed of light.
From the locals' perspective, it will essentially be a giant flat plane with constant gravity (this might not actually be true, due to space-time warping, but I'm not enough of a physicist to be sure); it will certainly be possible to travel up by way of balloons, airplanes, or even the occasional rocket (which will, of course, fall right down the way a cannonball does, because the orbital velocity is also huge, but with sufficient starting speed can stay up for minutes or longer).
With occasional - probably laser-based - acceleration along the way, it should probably even be possible for a single ship to go all the way around the planet without going below a certain height (say 1000 km). It would take an awfully long time though (about 500 years for a ship moving at 2000 km/s).
answered Mar 14, 2016 at 11:46
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3$\begingroup$ The relativistic escape velocity is $v=\sqrt{2\Phi-({\Phi\over c})^2}=\sqrt{{2GM\over r}-({GM\over rc})^2}$. The surface gravity formula gets messy; after a bunch of math, you get $a=\sqrt{r\over r-r_s}{GM\over r^2}$. As $r\rightarrow r_s$, the root blows up to infinity. So you can't actually make it work. $\endgroup$– MichaelSMar 14, 2016 at 13:47
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$\begingroup$ @MichaelS "Escape velocity greater than the speed of light" means that Htera's surface is inside the black hole event horizon. I thought that, as far as they are concerned, this means that the weird effects from the whole "being a black hole" thing would not happen, or at least not locally (relative to a light year); otherwise the "are we inside a black hole" question often asked regarding our own universe would not make any sense. It could be that, in fact, these effects become weirder (and result in gravity being a complex number or something like that). Really not a physicist. Sorry. $\endgroup$ Mar 14, 2016 at 19:28
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$\begingroup$ The universe isn't a black hole because the mass outside our part of the universe counteracts the mass inside it. Even though all the local mass is trying to turn us into a black hole, the non-local mass keeps space relatively flat. In this case, there's nothing to counter-balance the black hole. That said, I did come up with a thought experiment involving an unobtanium spherical shell with $r<r_s$. The gravity inside the shell would be zero, gravitational stresses provide plenty of energy to jump-start life, and you can't escape. But such an infinitely-strong shell isn't really possible. $\endgroup$– MichaelSMar 15, 2016 at 0:49