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Gravitational Law - Very Difficult

  1. May 5, 2009 #1
    An object is fired vertically upwards from the surface of a planetary body; it moves under the action of Newton’s Gravitational Law, without
    resistance, so the equation is z'' = -gR^2 / (z + R)^2 . Find the relation between v = z' and z and use this model, and the relation that you have
    found, to obtain a numerical estimate for the escape speed on the surface of the Earth.

    Its the wording of the question I don't get? I assume I integrate z'' twice to find the velocity and distance respectively, but i already have a z on the RHS so i cant integrate, but the escape speed to escape the surface of the earth? is this like when g (9.81) begins to decrease as it moves from the planetry body? I havn't got a clue at all. Any help appreciated
  2. jcsd
  3. May 5, 2009 #2


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    The escape velocity is the velocity at which the kinetic energy is equal to the gravitational potential energy of the planet.

    Using the chain rule you can write [tex]\frac{d^2 z}{dt^2}=\frac{d v}{dt}=\frac{dv}{dz}\frac{dz}{dt}=v\frac{dv}{dz}[/tex] After rearranging your equation you can now integrate.

    How did you get the equation for z''?
  4. May 5, 2009 #3
    it was given in the question, the equation for z'' that is
  5. May 5, 2009 #4


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    Odd that the mass of the planet isn't a part of the equation.
  6. May 5, 2009 #5
    Does it help to know I'm doing a Maths degree and not a Physics one? We were told none of our questions would require knownledge of classical mechanics/physics, but just differential equations etc
  7. May 5, 2009 #6


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    Regardless of what they said this is somewhat of a physics problem, but it matters not. Just use the equation you were given.
  8. May 5, 2009 #7
    Hi there,

    Your first equation: [tex]\frac{d^2z}{dt^2} = ...[/tex] is not Newton's gravitational law. It is the result of the Newton's gravitational law, considering a planetary body, like the Earth, the Sun, or the Moon. Newton's real gravitational law is define as the force that attracts two body of masses, and is express with : [tex]F_g = G\frac{m_1 \cdot m_2}{r^2}[/tex] where [tex]G[/tex] is Universal gravitational constant, [tex]m_i[/tex] are the mass of each body, and [tex]r[/tex] is the distance between the two body's center of mass.

    Hope this helps you understand it more.
  9. May 5, 2009 #8
    yeah i remember all this from A-level physics, but i dont know any masses or distances so i assume theyre kept as constants in the ODE
  10. May 6, 2009 #9
    There is no mention to the mass of the planet because it's inside g, the formula is correct (physically speaking). If this was merely a maths problem, you'd have to solve the ODE with the initial conditions z(0)=0, z'(0)=v and find out for which v z(infty)->infty. But this is too much work for a physicist... We know that there is a conserved quantity, energy:

    [tex]E=z'^2 - {gR^2\over z+R}[/tex]

    and that for E=0, the particle will escape, so... we set z=0 in that equation and solve for z'
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