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Escape Velocity and planet’s gravitational field

  1. Apr 27, 2009 #1
    1. The problem statement, all variables and given/known data
    The escape velocity is the velocity of an object at the surface of a planet that would
    allow it to be removed completely from the planet’s gravitational field.
    Calculate the escape velocity for an object on the surface of Pluto.

    mass of Pluto: 1.3 *1022
    radius of Pluto: 1.2 *106
    G= 6.7 *10-11
    2. Relevant equations
    EK= 1/2mv2
    Potential Energy= GM1M2/R


    3. The attempt at a solution
    I think i have this correct but i dont have the answers to check!!! So i need a professional opinion =]
    So at the surface would i be right in saying the kinetic energy required to leave the planet will be equal to the Potential the body has at the surface of the planet??
    So, 1/2mv2= GM1M2/R
    (the mass of the object will cancel)
    v=[square root]{2GM/R}
    v=[square root]{2*6.7 *10-11*1.3 *1022/1.2 *106
    v= 1205 ms-1
    Is this the right answer?? not entirely sure!
     
  2. jcsd
  3. Apr 27, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Yes. You need to overcome the potential at the surface. 1/2mV2 = Gmp/rp

    Where rp and mp are the radius and mass of the dwarf planet Pluto
     
  4. Apr 27, 2009 #3
    ahhh sweet cheers!! =]
     
  5. Apr 27, 2009 #4
    Hi ,
    I suggest in question like this in a test:
    Don't write right away GMm/R=Mv^2/2
    say that Potential energy = -GMm/R notice the negative sign.
    and in infinity the energy==0
    than Initial kinetic energy + (which will get us minus here) potential energy=0(getting to infinity).

    Just An advice from some 1 that suffered this thing on his flesh :D
     
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