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Kepler's First Law

  1. Jan 4, 2012 #1
    I can understand the mathematics behind the deduction of "expanded" (using Newton's laws) Kepler's First law that shows the orbits can be only parabolic, hyperbolic, linear or ellipses, but a friend (who doesn't know derivatives and differential equations) asked me why, taking in account only the gravity of the sun, the earth will not "fall", I would like to know if it's possible explain that using only concepts like momentum/energy conservation, Newton's laws, etc, without using any complicated mathematical formulas.

    And I have another question, the r=k/(1+e*cos(theta+theta0)) equation only shows the positions of a orbiting object at steady state? I mean, if there is a rocket near to the earth with "engine on" it will not follow this position (because the resultant force will not be only the gravity), but, one moment after turning the engine off (assuming the only force acting will be the gravity) it will start following this position equation, or it will take some time to happen?
     
    Last edited: Jan 4, 2012
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  3. Jan 4, 2012 #2

    D H

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    The Earth is constantly falling toward the Sun. It just keeps missing the Sun thanks to the non-zero component of the velocity vector normal to the radial vector. The easiest way to think of orbits is that the orbiting object is perpetually falling toward the central object but is at the same time perpetually missing that central object because of the orthogonal component of the velocity vector.

    To keep it simple, look at circular orbits only. For something to be moving in uniform circular motion there must be a force [itex]F=mv^2/r=m\omega^2r=m(2\pi/T)^2r[/itex] directed toward the center of the circle. In the case of an orbiting body, this central force is gravity. Newton's law of gravitation says that the gravitational force exerted on this body is GMm/r2. Equating this gravitational force with the centripetal force yields [itex]m(2\pi/T)^2r = GMm/r^2[/itex], or [itex]T^{\,2} = ((4\pi^2)/(GM))\;r^3[/itex]. Kepler's third law!

    It will start following that equation as soon as the engine is turned off.
     
    Last edited: Jan 4, 2012
  4. Jan 4, 2012 #3
    Just to make it clear, when I say "fall" I mean make a spiral like movement approaching to the sun.

    Thanks for your explanation, you gave me a light how to explain to him.

    Ok, I will try to make some matlab simulations to visualize that, thank you.
     
    Last edited: Jan 4, 2012
  5. Jan 4, 2012 #4

    D H

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    That would violate the conservation laws. Gravitation in Newtonian mechanics is a conservative force.

    In general relativity, orbiting objects emit gravitational waves as they orbit. These gravitational waves drain energy from the system, and hence the objects spiral in toward one another. This has even been observed: IH Stairs et al. Measurement of Relativistic Orbital Decay in the PSR B1534+12 Binary System, 1998 ApJ 505 352 doi:10.1086/306151.

    The Sun-Earth system loses so little energy (theoretically, about 200 watts) due to gravitational waves that this is a non-effect. What is observable is that the Earth is instead slowly spiraling out from the Sun. Slowly. Very slowly. The Sun slowly loses mass due to nuclear fusion in its core and due to solar wind. This loss of mass means the Earth must spiral out thanks to conservation of energy.
     
  6. Jan 4, 2012 #5
    Could you check if my explanation is plausible?
    If the Earth wasn't moving in a "closed loop" (on a "perfect world", taking in account just the Sun's gravity) each period of translation would represent some work done by the gravity, but there is no "energy input" on the system to realize/consumes this work, that means the only possible orbit trajectories are closed paths (where the work done by a conservative force is zero)?
     
  7. Jan 4, 2012 #6

    D H

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    That's not quite right. For a conservative central force of the form [itex]F=krn[/itex], closed paths only result for n=1 (e.g., a spring) and n=-2 (e.g., gravity). Yet the force is conservative because it can be expressed as the gradient of a potential function.
     
  8. Jan 4, 2012 #7
    Symon says: a force is conservative when it's only function of position and its curl is zero, using both facts and the Stokes theorem it can be proven that the closed path integral of the force is always zero, so if the earth moves along a closed path, the work done by the Sun's gravity is zero (as expected). But it doesn't prove that the only path that brings a zero work is the closed path.

    Am I missing something?

    I am trying to connect this two facts:
    "1-Spiral movement would violate the conservation laws.
    2-Gravitation is a conservative force"
     
    Last edited: Jan 4, 2012
  9. Jan 5, 2012 #8

    D H

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    A force is conservative when it be can be expressed as the gradient of some potential function. Conservation of energy is the key.

    Orbits in the two body problem are closed because gravitation is an inverse square force. A conservative central force is not a sufficient condition for closed orbits. A conservative inward central force is a sufficient condition for the existence of bounded behavior.
     
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