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What is escape velocity?
Asked by: Michael Metzger
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.
Answered by: Paul Walorski, B.A. Physics, Part-time Physics Instructor
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)
v = sqrt(2gR)
Where g is acceleration of gravity on the earth's surface.
The value evaluates to be approximately:
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.
Answered by: Yasar Safkan, B.S. Phsyics Ph.D. Candidate, M.I.T.
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