Date: 6/13/96 at 9:50:25 From: dhautree Subject: Escape velocity How did Einstein work out how fast you need to go to get off the earth?
Date: 6/13/96 at 21:1:12 From: Doctor Luis Subject: Re: Escape velocity This is not really a problem Einstein worked, rather it is a well known result in Physics. It can be derived as follows: The "escape velocity" of an object can be found by finding the work done by a gravitational field on a particle: As you may know, the gravitational force field can be alternatively represented by the negative gradient of a special function, called the potential function F = GMm/r^2 (m:particle's mass; M: planet's mass; r:dist from ~ center) = - grad p (p:potential function) Now, the work it takes for a particle with mass m to go from a pt A to a pt B is the line integral of the dot product of the force with the differential displacement around the path AB b W = S F * dr ( the "*" means dot product) a ~ ~ Since only the radial component of the displacement contributes to the work, b W = S Fr * dr , where Fr is the radial component of the force a ~ ~ Now, Fr = - mg (R/r)^2 ~ where r is the distance of the particle from the planet, R is the planet's radius, and mg is the weigth of the particle at the surface of the planet (g is the acceleration of gravity at the planet's surface) So, b W = -mgR^2 S (1/r^2) dr a = -mgR^2(-1/b + 1/a) Now, setting b at infinity (this can be interpreted as: what is the energy required to move a particle from pt A out of the planet, i.e., to a point where the gravitational attraction is negligible) and setting a (the initial position) as R ( the radius of the planet): W = -mgR^2( 0 + 1/R) = -mgR Now, the work done is equal to the change in kinetic energy (mv^2)/2 So that (m(vf)^2)/2 - (m(vi)^2)/2 = -mgR now, vf = 0 (initial vel.) so that, - m(vi)^2 = -2mgR (vi)^2 = 2gR or vi = sqrt(2gR) And this is the (treshold, should I say?) velocity required for a particle to overcome the planet's gravitational field. -Doctor Luis, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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