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Newton-Hooke Equasion

  1. Jan 15, 2004 #1

    Hooke's Law:
    [tex]F_h (x) = -kx[/tex]
    k - spring force constant

    Newtons Law: (Gravitation)
    [tex]F_g (r) = -G \frac{M^2}{r^2}[/tex]

    [tex]F_g (r) = F_h (r)[/tex]

    Newton-Hooke Equasion:
    [tex]-G \frac{M^2}{r^2} = -kr[/tex]

    [tex]k = G \frac{M^2}{r^3}[/tex]

    [tex]G = k \frac{r^3}{M^2}[/tex]

    Does Newtons Law obey Hooke's Law?

    Last edited: Jan 22, 2004
  2. jcsd
  3. Jan 16, 2004 #2


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    "i.e. Do two stars of equal mass eventually rebound after being pulled apart?"

    Explain what you mean by "rebound", and by what mechanism do you mean "pulled apart"?
  4. Jan 21, 2004 #3


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  5. Jan 22, 2004 #4

    [tex]U_g = U_h[/tex]

    [tex]- G \frac{M^2}{r} = - k \frac{r^2}{2}[/tex]

    [tex]k = G \frac{2 M^2}{r^3}[/tex]

    [tex]G = k \frac{r^3}{2 M^2}[/tex]

    [tex]E = K + U[/tex]

    [tex]E_g = E_h[/tex]

    Newton-Hooke Energy Theorem:
    [tex]E_g = \frac{Mv^2}{2} - G \frac{M^2}{r} = \frac{Mv^2}{2} - k \frac{r^2}{2}[/tex]

    [tex]E_g = m \left( \frac{v^2}{2} - G \frac{M}{r} \right) = \frac{1}{2} \left( mv^2 - kr^2 \right)[/tex]

    [tex]E_g = m \left( 2 \left( \frac{ \pi r}{T} \right)^2 - G \frac{M}{r} \right) = \frac{r^2}{2} \left( m \left( \frac{2 \pi}{T} \right)^2 - k \right)[/tex]

    [tex]E_h = \frac{r^2}{2} \left( m \left( \frac{2 \pi}{T} \right)^2 - k \right)[/tex]

    [tex]k = 2 \left( 2m \left( \frac{ \pi}{T} \right)^2 - \frac{E_h}{r^2} \right)[/tex]

    Does Newtons Law obey Hooke's Law?

    Are the above theorems true for all non-relativistic gravitational systems?

    What is the value for Hooke's Constant (k)?

    Last edited: Jan 22, 2004
  6. Jan 22, 2004 #5


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    Ok, thanks. I just didn't understand the previous point since it was all formulae with no verbiage.
  7. Jan 22, 2004 #6


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    The answer would be no. As can be seen from the equations, Hooke's law, which applies to elastic stretching and squeezing, increases in force with extension, whilst gravitational force does the opposite. Thus, while an object in a gravitational field has an escape velocity, you can't escape a spring.

    This, as you can see, is nonsense, as G is a constant whilst M and r vary from situation to situation.

    Your equation only represents a system where a gravitationally attracted body is kept from falling in by a gigantic spring of natural length 2r.
  8. Jan 29, 2004 #7
    Welcome after to me. What are non-relativistic gravitatioanl systems

  9. Jan 31, 2004 #8
    Hooke's Law...

    [tex]G = k \frac{r^3}{M^2}[/tex]
    What are the Standard International units for G as described by this formula as opposed to Newton's G SI units?

    Newton's G Law increases in force with contraction.

    Does Hooke's Law increase in force with contraction, or decrease?

    Why is it presumed that Hooke's Constant (k) is a universal constant described by this formula as opposed to a static constant applied to a given system?

    What are the values of Hooke's Constant(k) for a Mercury-Sol system as opposed to a Jupiter-Sol system?

    Q: What are non-relativistic gravitatioanl systems?

    A non-relativistic gravitational system is a gravity dominated system in which the mass velocities involved are only a small fraction of luminous velocity.

    [tex]v_m \ll c[/tex]

    All Newtonian Theorems fail at relativistic velocities.

    Last edited: Jan 31, 2004
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