

Escape Velocity
Name: Proctor G.
Status: other
Age: 30s
Location: N/A
Country: N/A
Date: 20002001
Question:
The following question and answer were found by a web
search for escape velocity. Steve L. has asked the same question which I
am interested in, and have found various ambiguous answers to.
Is escape velocity the velocity necessary at any given altitude (distance
from the earth's (or other body's) centre) to propel it out of the earth's
gravitational field WITHOUT the use of rockets or other machinery? In
other words, if there was an explosion on the earth large enough to propel
an object at a velocity of 25 000 mph straight outward from the earth,
that object would escape the earth's gravitational field. I am assuming
that this is true, in which case we come back to Steve L.'s question. IF
that object has instead some form of propulsion other than the large
explosion, why must it reach 25 000 mph to escape the earth's
gravitational field. If it had enough fuel why could it not propel itself
at a constant 5 mph for instance, until it was far enough from the earth
so as not to be affected by its gravity? And if this is true, then why
does a rocket have to reach an "escape velocity" of 25 000 mph when by the
time it has attained that speed it would be considerably further from the
earth's centre than where it started (at the surface)? Isn't the escape
velocity less the further you are from the centre of mass? Therefore
isn't it possible that by the point at which the rocket reached a velocity
of 15 000 mph it would be distant enough from the earth to have attained
escape velocity FROM THAT POINT?
Replies:
You are absolutely correct in your analysis. If you are interested in
the mathematics, the equation is:
E = GMm/r + mvv/2
where E is the total energy, G is Newton's gravitational constant
(6.67 x 10E11 Nmm/(kg kg), M is the mass of the earth (6 x 10E+24 kg),
m is the mass of the satellite, r is the distance of the satellite
from the center of the earth, and v is the speed of the satellite. The
first term is the gravitational potential energy and the second is the
kinetic energy of the satellite.
When E=0, the satellite has reached escape velocity. Then vv = 2GM/r.
Notice that when r is infinite, v=0 is the escape velocity. This
makes sense, since the satellite has escaped at that point.
On the other hand when r is the radius of the earth (6.5x10E+6 m), the
escape velocity is 11,000 m/s or about 25,000 mi/hr. Once the rocket
achieved that velocity (and was outside the atmosphere so there was no
drag), it could shut down its engine and it would continue to infinity
(and arrive there at zero velocity).
You could, however, keep the rocket engine running so the rocket speed
was always 5 mi/hr and still get to infinity. It would, however, take
a long time and an enormous amount of fuel. If you slowed down just 5
mi/hr so the rocket was at rest, you would burn ALL your fuel without
moving a bit. Going 5 mi/hr is a little better, but not much!
Best, Dick Plano
Note: 10E+24 is 1 followed by 24 zeros and vv is v squared (v times v).
What you are suggesting is true. The further from the earth's center the
lower
the escape velocity. Thus, if you moved upward at a constant velocity of
5 mph
you would eventually escape the earth's gravity. Initially it would require a
large rate of fuel consumption but by the time you escaped earth's gravity you
would not be using any fuel at all.
The physics and economics of the situation are such that it is easier and
cheaper to build a rocket that accelerates rapidly to the escape velocity
early
in the flight than it would be to build a rocket that accelerates to a lower
velocity and holds that speed for a LONG time to escape earth orbit. Control
issues are more critical for low velocities as well.
Greg Bradburn
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Update: June 2012

