For a rocket launch from Earth , the velocity is 11.2 km/sec…around 25,000 mph.
Suppose, however, earth has constructed a space elevator from the surface to geostationary orbit, and the car can ascend to the heavens at a lesiurly 10-15 mph.
Could, theoretically, a similar mechanical method be used to escape a black hole?
What material do you build your tether out of which is strong enough to resist a pull that prevents light from escaping? And what kind of engine do you use to drive the car up the tether given the same force that is pulling it back?
Valgard is right, the main issue is how are you going to move the object up the space tether.
Rather than thinking of escape velocity it is useful to think of escape energy. You can achieve escape by either having enough velocity at your surface so that your kenetic energy is equal to your escape energy, or you can slowly add potential energy to your object by pushing it up to the top of the tower so that once your at the top, you don’t require much additional energy to escape.
The problem is that in the black hole, your escape energy is infinite so can’t push hard enough to get out.
I don’t think this would make escape from beyond the event horizon possible, even if we posit magically-strong materials. Once you’re close enough to the black hole, space is warped in such a way that the direction ‘away’ doesn’t really exist any more.
Exactly. If I remember my GR right, once you get sufficiently close to a black hole, the radial direction becomes time-like, meaning you can only travel along it in one direction with no going back on yourself.
Does the temporal dimension become spacelike? (Is that question well formed?)
If you’re using the right metric (mathematical description of space-time), then yes. For instance, if you assume the black hole is rotating (which it probably is), and use a Kerr metric, then yes, there is a surface, which defines what is known as the ergosphere, where the time-like co-ordinate becomes negative and behaves, mathematically at least, as a space-like co-ordinate.
No. Escape speed is a purely ballistic quantity; that is, the amount of speed you would need (neglecting air resistance) to have at ground level (or whatever your reference value is) to achieve a trajectory that does not remain in orbit of the planet. At a higher altitude (or in orbit) the required escape velocity is less. If you have powered flight, you can go at any speed at a constant vector away from the planet and eventually you’ll get to a point where your speed exceeds the requirement for escape velocity. You can even generalize this further (and the reason for this will become clear in a moment) to a kinetic energy requirement that is independent of speed or momentum vector, provided your path doesn’t actually intersect the surface of the planet.
A black hole is a special case however; while all masses distort the fabric of spacetime a little bit, a black hole stretches it enough that at some point it actually twists back upon itself such that the future of all possible paths either returns to the starting point or falls further toward the singularity. The most simple way to look at this is by energy, as above; the amount of kinetic energy required to escape this is greater than the sum of all energy in the system, and expending more energy (say, in terms of impulse from rocket thrust) actually makes the system become more massive, paradoxically pulling you inward. Another way to look at it is in terms of the Lorentz contraction (or in this case, expansion); the further down in the gravity well you are, the longer the radial geodesic curve (the “straight line” that leads directly away from the singularity) becomes, and at the event horizon, it becomes infinity (and after that will actually wrap back upon itself).
Note that as a passenger on the craft you won’t notice any of this; as long as the gravitational gradient, and in the case of a rotating black hole, the tidal forces, don’t create significant shear stresses, you’re just going to see light progressively become a little brighter and higher frequency as you pass through the event horizon (assuming that you aren’t moving fast enough yourself to introduce additional relativistic effects). The universe doesn’t wink out, or explode, or whatever, and nobody stops you to collect a toll. You just keep on going, oblivious to the fact that you’ve entered a small pocket of the universe from which there is no exit. Attempting to escape, however, will just cause you to fall inward faster. Even if you have a straight and perfectly flagpole to climb up, the underlying fabric of space is going to be distorted such that you’ll never be able to climb “up” it; it will either become (from your point of view) infinitely long, or bend back around the event horizon such that you can’t ever climb above it.
Stranger
No matter what coordinate system you use, you’ll have one timelike and three spacelike coordinates at every point in spacetime where those coordinates are defined (both inside the horizon and out). It isn’t necessarily always the same coordinate that’s the timelike one, though. In the usual coordinate system, r is a space coordinate outside and a time coordinate inside, and t is a time coordinate outside and a space coordinate inside, but there are also coordinate systems you can set up which are perfectly well-behaved at the horizon, and for which the time coordinate always carries the same label.
Thinking of the escape velocity aspect of the OP rather than the black hole aspect, perhaps escape velocity gets too much attention. I grew up watching the space age unfold, and learned like everybody else that it was impossible to leave a planet at less than the escape velocity, but that’s not right at all. The escape velocity is the minimum velocity for coasting away from a planet (with no friction losses), not the minimum for climbing away.
Though I never heard one way or the other, I figured that rockets require much more fuel to climb out more slowly than they do. Obviously, if you slow the rocket to a standstill, it still uses fuel pretty rapidly, so the efficiency goes all the way to zero; starting out at some altitude and working at 0.99 g would have the rocket drifting back to earth, so the efficiency could even be negative.
It seems like the optimum rocket speed will depend on several subtle things, such as the relative importance of aerodynamic drag versus gravity. So, I think it’s implausible that the escape velocity is the optimum velocity, or has a very simple relation to it. Anybody know what the importance of escape velocity is to the business of getting off our planet?
Can I attempt to answer Napier’s question?
(I’m learning to be a Physics teacher, so any criticism gratefully received)
Escape Velocity is the initial velocity you would have to give an object if you wanted it to exit the influence of the Earth’s gravitational field (ie to get to an infinite point away). Imagine it as firing the object out of a cannon at this speed. Already we are talking about a fairly notional figure.
Add to this that cannon firing is not our preferred method of getting into orbit or beyond. Unlike a cannon (where an object immediatly decelerates once it is shot out) a rocket provides constant acceleration (as long as it is firing). As long as your acceleration is a tiny bit above 9.81m/s/s (the acceleration due to gravity*) you’ll keep going up (the mass of the object is irrelevant because, very simply, it is being accelerated both up and down**). Note also that as you rise the notional escape velocity (the speed you’d need to be going if you turned off the rocket and wanted to keep on going) keeps going down.
Thus fuel does become the limiting factor. Keeping the rocket going costs fuel. More fuel means more mass, and this means the rocket is harder to accelerate (so therefore meaning more fuel needed - a feedback loop). The best way to work out the balance off fuel, mass and amount of acceleration is to ask a rocket scientist!
How’d I do?
*the value of g will also be faling as you rise of course
**for the upward force to be positive ma>mg, or just a>g. Similarly the escape velocity can be worked out by transferring all the gravitational potential energy of an object to kinetic energy. As both these depend on an object’s mass, it cancels out. Escape velocity only depends on the mass of the planet.
Napier is closer and your explanation misses the crucial point.
Escape velocity is the ballistic velocity, i.e. the velocity achieved when no more acceleration is added to the system. If you had an infinite amount of fuel you could proceed infinitely far even at less than escape velocity. (Which is the principle behind the space elevator.)
Napier is right with the word coasting. Rockets are different from cannons in that the fuel is expended over a longer period, but otherwise the system is the same in principle. Fuel is expended for a finite period and the velocity at the end of that period determines whether it has achieved escape velocity.
You can get a rocket off the planet without achieving escape velocity. Fine for orbital or moon missions. But so far all our rocket designs to reach escape velocity just use more fuel for longer to give it that extra oomph. We have no way of adding more fuel once the rocket has taken off, so it’s imperative to burn away all that dead weight and turn it into velocity. The amount of fuel depends on the speed you want and whether you’re leaving the earth’s gravity well (the voyage and destination). Escape velocity is just one of the factors you need to take into account.
Not too bad; the fuel mass (and the dead weight of tankage, controls, actuators, nozzle, and in the case of liquid propellants, plumbing) limits the amount of extra impulse that can be applied to accelerating the payload. One way of comparing the efficacy of different types of launch systems is to look at a ratio of overall energy expended versus the energy requirement to get a common payload to a specified orbit or pierce point, which is more useful than comparing specific or total impulse between otherwise unlike systems. This has different nomenclature and slightly different formulation for different types of propulsion systems, but is generally known as something like propulsive efficiency or net (payload) energy efficacy. This is the reason that all current to-orbit launch system use multiple stages/boosters or (in the case of the original SM-65 Atlas and derived vehicles) drop engines; the propulsive efficiency of conventional chemical rockets just isn’t high enough to take the entire vehicle to orbit, and adding more propellants (and more dead weight tankage to carry the extra propellant, and more engines or motors to increase the total thrust level) is a negative sum game. For any chemical rocket system configuration, there is a limit to which it can be scaled to carry a higher payload, and you have to add additional stages in parallel to achieve more payload capability; hence why the Titan IV, Atlas V, Delta IV, Ariane 5, and other conventional heavy boost systems use large parallel stage boosters or multiple common core rockets rather than making the vehicle larger and longer along the Saturn family concept. There are other reasons for parallel staging as well–it keeps the L/D ratio manageable, makes it cheaper to build processing facilities, easier to transport stages, et cetera–but propulsive efficiency is a large driver in why this has become the predominant configuration in heavy lift boosters.
But interplanetary space launch vehicles don’t generally reach the escape speed at the Earth’s surface (~11.2 km/s). Instead, they pop up into an elliptical orbit, wait until they’re at the near the perigee, and then fire an payload assist/escape stage like the Centaur or the Inertial Upper Stage, which is usually distinct from the ascent stages to used to achieve orbit, although the Saturn V S-IVB third stage was used for both orbital injection and trans-lunar injection. In any case, in order to achieve escape from the sphere of influence (SOI) of a celestial body, you have to achieve the escape speed at some point, but at high orbit it will be significantly less than at the surface of that body; indeed, at the boundary of the SOI it will be almost zero by definition, or at least smaller than the other gravitational influences in the area. You can crawl up to that point; it’s just really inefficient to do so with a chemical rocket.
Stranger
One other clarification: An object shot out of a cannon does not immediately decelerate, it immediately stops accelerating. Deceleration is a decrease in speed, not a decrease in acceleration. (Yes, external factors may cause deceleration, like air resistance, but this is the same for anything attempting escape.)
Sure. A rocket can carry a fixed amount of fuel and that fuel will burn for a specific amount of time. The more fuel, the longer it will burn, but the heavier the rocket will be.
A modern rocket cannot carry enough fuel to acheive completely powered flight like an airplane traveling across country. It would be too massive to be practical. So in order to achieve orbit or to escape Earth’s gravity, a rocket must be traveling fast enough when it runs out of juice that it’s momentum will continue to carry it where it needs to go before gravity pulls it back to Earth.
Technically, it experiences “negative acceleration” if I remember physics class correctly.
msmith: Yes, mostly. Negative acceleration is simply acceleration in the negative direction, for however you choose to define positive and negative. A car can accelerate to 30 kph, upon hitting a brick wall it will decelerate to 0 kph in a very short amount of time. A cannonball will be rapidly accelerated as it is launched, and … ah. I see now what Petrobey Mavromihalis meant. If I am firing my cannon on Earth (or inside any gravity well), then the acceleration of gravity will immediately act on the projectile (this is the definition of ballistic motion). If, as people often do, we define down as the negative direction, then this would be a negative acceleration or a deceleration.
I would like to rephrase my initial objection: the only error Petrobey made was not being clear about how “negative” was defined.
From a physics class calculation standpoint, it makes no difference if it is “positive” in a vector 180 degrees to the direction of travel or “negative” in the direction of travel. But basically what you said is correct. Once the projectile leaves the barrel it is traveling with a velocity v[sub]0 [/sub] in whatever direction the barrel is pointed. Ignoring air resistance for simplicity, v[sub]t[/sub] will be the cumulative effects of v[sub]0 [/sub] and an acceleration vector g pointing from the projectile towards the center of the Earth at any point in time t.
To achieve orbit, v[sub]0 [/sub] has to be such that the projectile never falls back to Earth. By the time it has fallen back along the x axis it has already moved along the y axis such that it misses hitting the ground. That’s why an object in orbit is said to be in “freefall”. If that makes sense the way I explained it.
In other words, one of the many things that can go wrong is running out of fuel for the last burn at less than the local escape velocity? They sure made it seem more central to the plan than that. For example, I can’t remember how many times I saw a figure for what the escape velocity would be at the Earth’s surface, which wouldn’t be relevant for rockets at all, right?
Some rocket launch systems will have a coast phase, in which there is a delay between end of burn (EOB) of one stage and ignition of the second, allowing the vehicle to use its acquired momentum to “coast” up to a higher altitude or flatter trajectory where the upper stage will have more efficiency (due to less drag or better trajectory shaping) even though it loses some momentum in the process. This gives an overall improvement of net energy efficacy or payload to orbit or whatever metric you want to use. Running out of fuel before achieving an orbit his highly unlikely short of critical engine failure; extensive studies are run that statistically predict motor performance by varying a wide array of parameters like burn rate variability, nozzle throat erosion, propellant combustion performance, atmospheric drag, et cetera, that assure successful performance out to 99.9% (three standard deviations, in statistical terms) or better. There have been cases where a lack of propellant performance characterization (specifically, with solid propellants) has led to such gross underperformance, or some kind of unintended energy-wasting vehicle dynamics have resulted in less useful impulse than necessary has occurred, but that is very, very rare and usually the result of unexpected phenomenology with a nascent launch system.
Stranger
Rockets aren’t bound by escape speed, but it’s certainly relevant for them. You get more efficiency by burning your fuel as quickly as possible, so if you’re doing it right, you’re still pretty close to the surface when you finish off your fuel.
Of course, most rockets don’t need to reach escape speed at all, since they’re just going into orbit around the Earth. But orbital speed is proportional to escape speed, so it’s still relevant.