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Law of Gravitation- Find an expression for the magnitude of the initial velocity that a projectile must possess in order to leave the earth when air friction is neglected.
The gravitational constant for g is, G = 6.67428*10^-11 m^3 Kg^-1 s^-2

Please answer the question fully and please give any helpful websites, thankyou.

Also I know calculus.

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asked by Dwayne
  1. Then set the initial KEnergy equal to the potential energy at the surface of EArth.

    1/2 mv^2= GMe*m/re

    solve for escape velocity, v

    posted by bobpursley
  2. v^2 = 2Mg/r + (u^2 -2Mg/d )

    Is that right

    posted by Dwayne
  3. Why did you make it so complicated?

    1/2 m v^2=GMe m/re

    v^2= 2GMe/re
    take the square root of each side.

    posted by bobpursley

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