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Escape Velocity


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/   
    
Associated Topics:
High School Physics/Chemistry

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