Ask a Question
or
Create a Poll
  1. Home /
  2. Questions /
  3. Science
  4. / Physical Sciences
  5. / Physics
  6. / Gravity

by JP1967 on November 20th, 2003

JP1967

Question

Help answer this question below.

What is "escape velocity"?

Share:
Other

Answers. 5 helpful answers below.

  • by Dan Whitacre on December 28th, 2003

    Dan Whitacre

    Escape velocity is the velocity needed by a projectile on launch to exit the gravitational potential well of the object it is leaving.

    It is a mistake to think that escape velocity is needed to get to space. Actually much less is needed, since most of our space-destined objects are meant to stay in the gravity well of the Earth, in orbit, usually.

    Technically, since gravity is infinite in its reach, it takes an infinite time and energy to escape an objects pull. Practically, what needs to be done is to get an object away from the launching body enough that the launching body's gravity becomes negligable.

    Even if you get off the Earth, you still have the sun's gravity to overcome, so really we are in a hole in a bigger hole. To get away from the entire solar system has been done only twice, by the Voyager I and II spacecrafts.

    Incidentally, when a body is so dense it's escape velocity (outside it's surface) is the speed of light in a vacuum (3*10^8 m/s) it is, by definition, a "black hole".

    • Like
    • Report

    3 comments | Post one | Permalink

  • by James Beatty on April 21st, 2004

    James Beatty

    "Escape velocity" is a misnomer, because it doesn't refer to a velocity, but rather a speed. It is properly "escape speed".

    The escape speed for a mass is the speed required to travel an infinite distance from said object. In other words (and taking Re to be the radius of the Earth), escape speed for Earth is the speed at r = Re such that speed at infinite r is 0. Obviously, that would take an infinite amount of time, but it is a useful theoretical exercise.

    As speed at infinite r is 0, kinetic energy is also 0. Gravitational potential energy is defined to be 0 at infinite distance, so at infinite distance, the total energy of the object is 0. Due to conservation of energy, that must be the energy of the object on the surface of the Earth.

    Therefore: E = (1/2)*m*v ^ 2 - (G*M*m)/Re = 0,
    when E is total energy, m is the mass of the object, v is the escape speed, G is the gravitational constant, M is the mass of the Earth and Re is as above.

    Solving for v, we find:
    v = sqrt( (2*G*M) / Re )

    With the known values of G, M, and Re, v is found to be 11.2 km/s. It is worth noting that this speed will work no matter what direction the object moves (obviously, it can't go through the Earth).

    Of course, this doesn't account for the effects of gravity from other bodies, such as the Moon, the Sun, or any of the other planets, but in most cases such an effect would be more or less negligible.

    • Like
    • Report

    2 comments | Post one | Permalink

  • by Stephen Z on March 2nd, 2004

    Stephen Z

    Escape Velocity is quite a misnomer in that it has nothing to do with escaping a planetary body but has rather to do with entering orbit around it. The escape velocity is actually the speed required for any object to maintain orbit from a given altitude around a given mass.

    This escape (or orbit) velocity decreases with distance from the object and increases with the mass of the object or body
    .
    This term must have caused untold confusion over the years and should perhaps itself be projected away from the earth at whatever velocity it requires to escape it.

    As far as the velocity required to escape the earths atmosphere... well I hate to de-sensationalise this but walking speed would do just fine if maintained for an appropriate duration.

    If we explore this example, the misuse of the word 'escape' can perhaps be more easily excused: Maintaining a constant walking speed in the upwards direction would require constant power to be applied to the object. In reality, it iwould be more practical to reach a given speed and then terminate power. Escape Velocity in this sense tells us the minimum speed at which the object will no longer fall directly back to earth when power is cut at a given altitude.

    N.B. - This does NOT mean the object will have escaped from the earths gravitational pull. Gravity decreases but never ceases...

    • Like
    • Report

    4 comments | Post one | Permalink

  • by JP1967 on November 20th, 2003

    JP1967

    Escape velocity is the launch speed that is necessary for an object launched from the surface of an object (usually a planet) to escape the gravitational pull of that planet.

    For the earth, this speed is approximately 11 kilometers per second. That is to say, if you could launch something (a couch for instance) from the surface of the Earth at 11 km/s, it would escape the Earth's gravity and continue on into space. Of course, this is ignoring wind resistance... so the real speed would have to be even faster.

    • Like
    • Report

    1 comment | Post one | Permalink

  • by Aminor on June 16th, 2005

    Aminor

    Nothing ever escapes from gravity, any gravity. It's the literal truth that you're feeling the gravitational influence of Pluto right now, and it's also sensing you -- though not very much. So what really happens when something gets into orbit?

    You probably know that a bullet fired parallel to the earth starts falling as soon as it's out of the muzzle, and falls to earth in a long curve. Suppose you were on a long, sloping hillside whose curve exactly matched the bullet's descending curve. Now the bullet stays the same height above ground for as long as the slope continues (which can really happen). And for the time that it's in flight, a bug traveling on it (ignore air resistance and any spin -- maybe make it a cannon ball) would, after the initial bang, feel weightless. Since the cannon ball is falling freely (for as long as the ground falls away beneath it), the bug would have no sensation of weight, but would be like a passenger on one of those rides that just takes you up a tower and drops you.

    Now remove the earth's atmosphere and make the descending curve not just a local hill, but the curve of the earth itself. If the cannon ball is moving fast enough, could its descending curve exactly match the earth's, and thus just go on and on? Absolutely -- it would be in orbit (of course, it also has to have a clear path -- difficult for an orbital altitude of maybe four feet). But we can't remove the earth's atmosphere, so instead we go high enough that its last traces can be neglected -- it's air resistance we're escaping, not gravity. We then arrange to travel "forward" -- parallel to the ground -- at a speed such that our falling curve exactly matches the earth's -- and now we are in free fall, which all the early sci fi writers knew was the accurate description of an object in orbit.

    And how fast is that? For low earth orbit, just far enough out of the atmosphere, about 17,000 miles an hour -- the approximate equivalent of Mach 25 if Mach numbers meant anything in space (but those numbers are from memory -- I haven't checked them against the other answers here). That's what's usually meant by "escape velocity," which has proved to be kind of a confusing term. For one thing, it seems to reinforce the idea that we escape from gravity when we get into orbit. The fact is that we give in to it completely and just fall -- but keep moving forward fast enough that the earth falls away beneath us at the same rate. For another, it apparently suggests to some that that's the speed required just to get outside the atmosphere, whereas any speed will do if you can just continue it long enough -- that's an engineering issue.

    So why this somewhat non-standard description, instead of the usual ball on a string? Certainly the facts are the same, and I wouldn't quarrel with any of the answers that talk, one way or another, about the cetrifugal force of inertia balanced by the centripetal force of gravity. But if I were a passenger in a ball being whirled on a string, I'd expect to be thrown against the outside wall, to experience an artificial gravity. I have to massage the image somewhat to remember why that doesn't happen, and whatever explanation I devise runs the risk of sneaking in the notion of "escaping" from gravity.

    And who first predicted south Florida as a logical site from which to launch things into orbit? Jules Verne.

    No comments. Post one | Permalink

Want to attach an image to your answer? Click here.

Did this answer your question? If not, then ask a new question or create a poll.

More Questions. Additional questions in this category.

You're reading What is "escape velocity"?

Follow us on Facebook!

Related Ads

Related Questions

Professionally Researched

ANSWERBAG BUZZ

What is escape velocity
What is meant by escape velocity
Escape velocity air resistance
Physical science escape velocity earth atmosphere
How to measure escape velocity

Answerbag

a Demand Media company