Escape Velocity Question
Escape Velocity Question |
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Message boards : SETI@home Science : Escape Velocity Question
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Escape velocity has always puzzled me and when I've asked the following question to people I've never gotten a satisfactory answer. | |
ID: 215071 |  | |
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Escape velocity is different from Orbital velocity. | |
| ID: 215174 | | |
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In that case I am seeking an answer to escape velocity. Using my limitless supply of fuel why is it not possible to slowly thrust away at an incredibly slow speed? | |
| ID: 215187 | | |
In that case I am seeking an answer to escape velocity. Using my limitless supply of fuel why is it not possible to slowly thrust away at an incredibly slow speed? It is perfectly possible, and is exactly what would happen. I think the confusion here is over the definition of escape velocity. Escape velocity is the speed an object must be at so that it will escape an objects gravity without the need for thrust of any sort. What this means is once you reach excape velocity, you can turn off your engines and you will never fall back to the ground, nor will you end up in an orbit. You'll just keep getting farther away. As for your question, the reason it works is that escape velocity depends on your distance from the object you're trying to escape. The farther you are away from it, the slower you need to be moving. So, if you stay at 1000 miles per constantly, you'll eventually be far enough away that if you turned off your engine, you would never fall back down. In the case of the Earth, this would require you to be about 2.5 million miles away. ____________ | |
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Hi Robert, | |
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The escape velocity is related to the kinetic energy to obtain a stable orbit. To answer your question, yes your solution is fine. Someone correct any mistakes I make. | |
| ID: 215232 | | |
The escape velocity is related to the kinetic energy to obtain a stable orbit. To answer your question, yes your solution is fine. Someone correct any mistakes I make. Stable orbits can be achieved over a rather wide range of energies. Escape velocity is, in fact, related to the minimum energy at which a stable orbit cannot be reached. That is to say, escape velocity is the speed you need to move at so that you have to much energy to stay in a stable orbit. Let's start with some definitions. Just noting that this should say that it is in meters v = velocity of object relative to earth In the literature, this is usually "g", and its units are m/s^2 You can use force or energy: Again, escape velocity is the speed needed such that you won't fall all they way around, but will keep moving away. The downward potential energy is for a fall around the earth is: Energy of any sort has no direction. What you're talking about is simply the potential energy of an object. Also, your formula for it is not correct. If dealing with a situation where everything stays near Earth's surface, we generally write this as: V = m(object)*g*h, where h is the height aboves Earth's surface. However, this form is really not useful for the escape velocity problem. So, we turn to the more general formula: V = -G*m(object)*m(earth)/r, G is Newton's constant (6.67*10^-11 m^3/(kg*s^20)) Don't let the fact that this is everywhere negative trouble you too much, as the only thing we can every actually measure is a change in potential energy; so, we could, if we wanted, add a constant to that equation (which would set potential energy to 0 somewhere that isn't an infinite distance away) without changing any physics. The momentum energy for a stable orbit is when the velocity perpendicular to the downward force produces the same force to counter act it. The way I like to think of it is "how much energy is it to fall around the earth 1 time and return to your starting point?" Momentum and energy are two very different ideas. I think what you are trying to talk about here is kinetic energy (energy associated with motion). There are quite a few problems with this part of your discussion. First, velocity does not produce a force. Forces are interactions between two (or more) objects. Furthermore, if something were causing a force opposing gravity, then the object the force was exerted on would not stay in orbit. An orbit is a curved path, but objects will only travel in straight-line paths if there are no net forces on them (i.e. all the forces present are cancelled by other forces). The other large problem here is that in a stable orbit you neither gain nor lose energy. The problem here comes with the way you define work. Work is correctly defined: W = Force*distance travelled in the direction the force acts In the case of an orbit, you end up back where you started. So, since the gravitational force points downward, but you end up at the same altitude you started at, there's no work done on you by gravity. My last comment here is that the kinetic energy an object has is given by: T = 1/2*m(object)*v^2 so when you put The manipulations you've done do lead to the correct answer, but that seems to be a convenient accident. Here's how to use energy to get escape velocity. And object's total mechanical energy, U, is the sum of its kinetic and potential energies: U = T + V = 1/2*m(object)*v^2 - G*m(object)*m(earth)/r Since potential energy is negative, it is clear that the object could have a negative total energy. Imagine that we took such an object and let it move around without changing its total energy. Since the potential energy increases with increasing distance from the earth (eventually reaching 0 an infinite distance away), if we let our negative energy object move farther away, we would find that its kinetic energy (and, hence, its velocity) would decrease to compensate for its increasing potential energy. Eventually it would reach a point where its potential energy was equal to its total energy; so, it not be moving, since it couldn't have any kinetic energy. If to get to this particular distance, the object must stop moving, it is clear that, without adding energy, the object could never move farther away from the earth that this particular distance. The situation for escape velocity, then, is the case where the distance that the object can't pass is infinitely far away (in other words, it can go anywhere). Infinitely far away the potential energy is 0. So, the object's total energy there (and, thus everywhere else) must also be 0. We can, then, plug in 0 for U in the total energy equation and solve for v. This gives: v_escape = SQRT(2*G*m(earth)/r) So, if you are a distance r away from the center of the earth, you must be travelling no slower than v_escape to be able to get any arbitrary distance away from the earth without the aid of some sort of engine. ____________ | |
| ID: 215271 | | |
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1. kinetic energy does have direction | |
| ID: 215318 | | |
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Hey solomon, you seem to know a few things as I read your previous posts. i'm posting a thread in a moment on entanglement and teleportation. A friend has been asking me about it but I really have no opinion on it and thus don't know much about it, I'd like your input if you know what it is and how it's been treated in popular literature. | |
| ID: 215332 | | |
1. kinetic energy does have direction No. Velocity has direction. Energy does not. If you were travelling north as 50 mph, you would have the same energy as if you were travelling south at 50 mph, east at 50 mph, or even straight up at 50 mph. All it depends on is how fast you're going, not which direction. 2. negative values literally depend on how you define your axis. I'm not really sure what you mean by axis here. Assuming you mean how I define my coordinates, this relates back to my first comment. Energy does not have a direction associated with it. It is a scalar quantity like temperature or mass. In the potential energy, the negativity is expressing the fact that the total mechanical energy of an object in a gravitating system is smaller than its kinetic energy. The conclusions derived from this simply don't work if we interchange positive and negative without redefining the quantities we're talking about. 3. convenient accident. lol. I just approach things differently. A stable orbit is a significatly different situation from escape. The easiest way to treat the orbit problem (at least if we restrict ourselves to the simple case of circular orbits, rather than the more general problem of elliptical ones) is to look at Newton's second law: F = m*a - the total force felt by an object is equal to its mass times its acceleration. The gravitational force on an object, orbiting or not, is: F = -G*m(object)*m(earth)/r^2 When an object moves in a circle, its direction is constantly changing; so, even though it keeps moving at the same speed its velocity changes, because velocity cares about direction. In this case, the acceleration is always centripetal, meaning perpendicular to the direction of motion. The general form of a centripetal acceleration is: a_c = v^2/r So, adding a minus sign to account for the centripetal acceleration being downwards (the same direction as the gravitational force), Newton's second law gives: -G*m(object)*m(earth)/r^2 = -m(object)*v^2/r We can solve this for the orbital velocity (that is, the velocity such that this relation will hold at a given radius r): v_orbital = SQRT(G*m(earth)/r) This means that the escape velocity at a given value of r is a factor of SQRT(2) greater than the orbital velocity at that same position (v_escape = SQRT(2)*v_orbital). 6. if you are a teacher or professor and a student has a positive equation instead of negative. Instead of it being wrong for being positive, it should be wrong for not defining the axis. But the other way to deal with positve and negative values is to understand that the left side of the equal sign is negative to the right side and visa versa. All valid mathematics. Convention and preconceptions trap the mind. I am a teacher (well, a TA). And, valid mathematics is not necessarily valid physics. ____________ | |
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Thanks...I knew it was something silly I was overlooking. | |
| ID: 215377 | | |
1. kinetic energy does have direction Forgive this, i'm about to try to rationalize some ideas here. I'm confounded. I understand what you are saying and perhaps I did say it as it is vector force and and momentum that is of concernt but then how can you add or subtract it from other energy? Generally when I think of energy, it results from a sum of area or volume and thus has no direction but how about the gradient. Direction is only introduced when there is a change. Thus if your r vector is in the opposite direction that you difined your positve direction to be. If E(kinetic) = (1/2)<v>^2 where <v> is a vector, then E(kinetic) is a vector itself and has direction beacause vectors contain directional information. In particular I think of an object gaining or losing energy. One will be positive and one will be negative because there is direction to the energy flux. It would be equivelent to lifting an object against gravity compared t allowing it to lose some of it's potential by lowering with gravity. Or another example is the Poynting vector. You can use it to calculate the Energy per Unit time (Watt) passing through the surface of a wire. This is not a direction in space but a direction in time. Of which, time is often omited, not to my dismay, from the vectors and kept inside the equation. Looking for the question that I need to answer: If then it's defined as a quality of a field, then it has no direction. Fields are fields because they influence pieces of energy and cause action which obtains direction. The direction has to come from a gradient of energy which defines the field. Flux defines the changes in the qualities of the fields or energy densities. Flux is typically over time. Since Energy experience a flux, it flows and thus must change over time and consequently must have direction either in space or time or both. Have I missed something? ____________ TEAM LL | |
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Thanks...I knew it was something silly I was overlooking. Sleestak and Solomon's debate notwithstanding, popular literature (from science fiction to sloppy textbooks) make it sound like escape velocity is needed to leave a gravity well. Escape velocity is what's needed at a given starting point so that you have enough kinetic energy to never fall back to your starting level. If one were to fire a bullet at the sky, it would need to exceed escape velocity or else it would eventually fall back to Earth. If the projectile has its own source of acceleration, such as the rocket in your example, then it merely needs to exert an acceleration greater than the gravitational acceleration trying to pull it down. Earth pulls down (at the surface) at about 9.8m/(s^2). If your rocket engines push with a force even slightly greater than what's needed to hover then you would slowly leave the Earth's gravitational pull. Since the escape velocity decreases with distance, eventually you'd be able to turn off the rocket engine and drift outward. Whether orbit is an "escape" or not is a definitional issue. You've managed to not fall back to Earth, but you haven't "escaped" the gravity well either. Orbit is the boundary between bodies that fall back to Earth and those that leave and never return. Objects placed in orbit are not shot from a cannon, but if they were they'd be shot at a precise[\\i] velocity (and inclination) very close to the escape velocity. It is much easier to build a vehicle that could escape, but then fires maneuvering thrusters to nudge itself into an orbital path. ____________ [i]No animals were harmed in the making of the above post... much.  | |
| ID: 215722 | | |
1. kinetic energy does have direction This is pretty astute. The gradient of a scalar field is a vector field, always pointing in the direction of the greatest change in the scalar field. The gradient of potential energy, for example, is simply minus the force associated with it. Thus if your r vector is in the opposite direction that you difined your positve direction to be. It seems to be worth stopping and talking about vector multiplication. There are three ways to multiply vectors: 1) Scalar product (or dot product, or inner product) The scalar product between two vectors is a scalar which depends only on the magnitudes of the two vectors and their relative directions: A.B = A*B*Cos(/theta), where \\theta is the angle between the two vectors. This is like taking the product of the projections of the two vectors into a common directions. In particular, the scalar product of a vector with itself is simply its magnitude squared: A.A = A*A = A^2 2) Vector product (or cross product) This gives a vector perpedicular to both vectors multiplied, and proportional to the parts of the two initial vectors perpendicular to each other. |AxB| = A*B*Sin(/theta) The direction of the resulting vector is perpendicular to both initial vectors. The cross product of a vector with itself is 0, as it has no part perpendicular to itself. 3) Tensor product The tensor product is what you get by multiplying each component of one vector with each component of another vector. This gives an object with 9 components, which is clearly not what we're looking for with energy; so, I'll leave this be. Only the scalar and tensor products of a vector with itself are non-zero; and energy is clearly not a nine component object, so the only vector multiplication that makes any sense for the v^2 in kinetic energy is the scalar product, which just gives the squared magnitude of velocity. In particular I think of an object gaining or losing energy. One will be positive and one will be negative because there is direction to the energy flux. It would be equivelent to lifting an object against gravity compared t allowing it to lose some of it's potential by lowering with gravity. When talking about flux (Poynting flux or otherwise), you are implicitly specifying a surface through which the flow passes. Surfaces are orientable, that is, they have direction, so in talking about flux there is a direction involved. A related idea is talking about a current. Currents (whether electrical, or fluid, or otherwise) have a direction associated with the flow; however, what is flowing need not have any direction inherently associated with it. Water has no direction, but it can flow in a directions. I hope this clears things up a little. ____________ | |
| ID: 215768 | | |
Message boards : SETI@home Science : Escape Velocity Question
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