anyone can explain about escape velocity ? - expecting from experts in aeronautical?

Answer 1

If you throw an object straight up, it will rise until the the negative acceleration of gravity stops it, then returns it to Earth. Gravity's force diminishes as distance from the center of the Earth increases, however. So if you can throw the object with enough initial upward velocity so that gravity's decreasing force can never quite slow it to a complete stop, its decreasing velocity can always be just high enough to overcome gravity's pull. The initial velocity needed to achieve that condition is called escape velocity.

From the surface of the Earth, escape velocity (ignoring air friction) is about 7 miles per second, or 25,000 miles per hour. Given that initial speed, an object needs no additional force applied to completely escape Earth's gravity.

Escape velocity is defined to be the minimum velocity an object must have in order to escape the gravitational field of the earth, that is, escape the earth without ever falling back.

The object must have greater energy than its gravitational binding energy to escape the earth's gravitational field. So:

1/2 mv2 = GMm/R

Where m is the mass of the object, M mass of the earth, G is the gravitational constant, R is the radius of the earth, and v is the escape velocity. It simplifies to:

v = sqrt(2GM/R)

or

v = sqrt(2gR)

Where g is acceleration of gravity on the earth's surface.

The value evaluates to be approximately:

11100 m/s
40200 km/h
25000 mi/h


So, an object which has this velocity at the surface of the earth, will totally escape the earth's gravitational field (ignoring the losses due to the atmosphere.) It is all there is to it.

Answer 2

In any trajectory affected only by gravity, the total energy of the object stays constant. The only thing that changes is the balance between kinetic energy and potential energy.

Kinetic energy = 1/2 mv^2 - where m is mass and v is velocity

Potential energy = GMm/r - where G is universal gravitational constant, M is big object, m is mass of moving object, and r is distance from the big object's center of mass.

Total energy:

E = 1/2 mv^2 - GMm/r

If you change the velocity, it causes a change in radius to keep E constant.

If E is negative, the object's trajectory is a closed orbit around the big object.

If E is 0 or positive, the object's trajectory is affected by the large object, but it will escape. An energy level of 0 is the minimum energy level required to escape from the big object's gravity.

To find the minimum escape velocity, rearrange the energy equation. Normally, the mass of the moving object is omitted and specific energy per unit of mass is used.

E = 1/2 v^2 - GM/r
let E = 0

1/2 v^2 = GM/r
v^2 = 2GM/r
v = sqrt (2GM/r)

The escape velocity depends on the distance from the center of the Earth. The typical problem is to calculate escape velocity from the surface of the Earth, but the downside is students tend to get the impression there's a single escape velocity for the planet Earth. Escape velocity for a spacecraft in orbit would be less than for a spacecraft on the surface of the Earth.

Answer 3

Escape speed is the speed at which something has a kinetic energy equal to its gravitational potential energy with respect to the object that it is trying to escape from. As long as no matter gets in the way (like the planet's atmosphere, or the planet itself), making the getaway does not depend on which way the velocity is directed, so it's better to say "escape speed" than "escape velocity."

Of course, if after leaving the planet you have some specific destination in mind, it does matter what the velocity is after you've fallen into an elliptical orbit with respect to the sun. The trick to getting the velocity right is part of the subject of transfer orbits.

Answer 4

It's actually pretty simple.

Every heavenly body has gravitational pull. In order to completely leave the orbit of a planet, star, etc. you must reach "escape velocity". The equation for this is actually very easy.

V = sqrt(2*mu/R)

where,

mu = specific gravitational constant = G*M (where G is Newton's gravitational constant, and M is the mass of the body whose orbit you want to leave.)
R = radial distance from the center of the body (or radius of the body + altitude)

Essentially, once escape velocity is reached you will leave the orbital influence of the body, and continue out in a straight line from where escape velocity is reached until you come under the gravity influence of another body.

If this wasn't exactly the answer you wanted, just shoot me an email and I'll happily clear any questions you have up.

Answer 5

Hi Senthil,
As far as I know, escape velocity is the speed required by a vehicle to escape the gravitational force of the earth. You can find the escape velocity in more detail, that is about the scientific detail, visit the wiki encyclopaedia.
And also, one more thing, right now I am trying to research or at least a literature review about anti gravity. I am going to research on this in the future.. please write to me about any ideas that you have... like anti gravity materials or opposing and using gravity to our advantage...
Waiting for your reply...
I basically gave this answer as you I thought you might be interested in research of changing the escape velocity using gravity and its theory....
Ok bye for now...

Answer 6

It is the velocity needed to overcome the gravitational pull and friction due to air and make an object just move out into space is escape velocity.

Answer 7

Is the minimum velocity that must be acheived by any object to escape the gravitational pull of the earth.

Answer 8

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Answer 9

What specifically would you like to know?
http://en.wikipedia.org/wiki/Escape_velocity