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 mv^{2} = 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.
Answered by: Yasar Safkan, B.S. Phsyics Ph.D. Candidate, M.I.T.
'A theory with mathematical beauty is more likely to be correct than an ugly one that fits some experimental data. God is a mathematician of a very high order, and He used very advanced mathematics in constructing the universe.'