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Acceleration calculation is mocking me

  1. Jul 24, 2006 #1
    I took a physics class in high school, and considering how little I’ve since exercised the skills I learned in the class, I must say I’m pretty proud of myself for having come this far. But now I’ve hit a wall.

    I enjoy reading and writing science fiction, but to write good science fiction you need a base of science fact, so I’ve been trying to learn how to calculate ballpark travel times between planets for spacecraft using Newtonian physics. I’m having an issue, however, with a concept I either never learned, of have since forgotten completely: Calculating velocity when acceleration is accelerating. Changing velocity of a rock dropped off a roof? No problem. Should be roughly 9.81m/s^2. Changing velocity of a rock that falls from being motionless at a distance of +12,000,000 km? Big problem. That’s a long ways, and some lone firing synapse reminded me that the gravitational force is inversely proportional to the square of the distance.

    So I know how to calculate the gravitational pull for given distance, but no idea how to find the acceleration of acceleration or use it to calculate the velocity of an object at a certain distance when it fell from a much greater distance. If anybody could explain how one goes about calculating things like these, it would be much appreciated.
     
  2. jcsd
  3. Jul 24, 2006 #2

    Kurdt

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    You are indeed correct that the acceleration changes as the distance decreases. The formula for this relation is:

    [tex] a(r) = \frac {GM}{r^2} [/tex]

    The problem with this is that there is no explicit time dependence. This just gives the instantaneous acceleration at some point r.
     
  4. Jul 24, 2006 #3

    HallsofIvy

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    I presume you mean that the acceleration is -9.81 m/s2. Then the velocity after t seconds is -9.81t m/s.

    Gravitational force between two objects of masses m and M, at distance r (between their centers) is [itex]-\frac{GmM}{r^2}[/itex] so, since F= ma, the acceleration on the object of mass m, due to the object of mass M, is given by [itex]F= ma= -\frac{GmM}{r^2}[/itex] so that
    [tex]a= \frac{dv}{dt}= -\frac{GM}{r^2}[/tex]
    Since t does not appear explicitely in that we can use the "chain rule" to swap variables:
    [tex]\frac{dv}{dt}= \frac{dr}{dt}\frac{dv}{dr}= -\frac{GM}{r^2}[/tex]
    But dr/dt is the velocity so
    [tex]v\frac{dv}{dt}= -\frac{GM}{r^2}[/tex]
    which we can write as
    [tex]v dv= -\frac{GM}{r^2}[/tex]
    and integrate:
    [tex]\frac{1}{2}v^2= \frac{GM}{r}+ C[/tex]
    If you look at that closely, you'll see that it is really "conservation of energy":
    [tex]\frac{1}{2}v^2- \frac{GM}{r}= C[/tex]
    The first term is the kinetic energy and the second is the gravitational potential energy at distance r (each missing a factor of m).

    Of course, then,
    [tex]v= \sqrt{\frac{2GM}{r}+ 2C}[/tex]
     
  5. Jul 24, 2006 #4

    Kurdt

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    Just a note on the constant that appears in the equations above. Be careful when determining the constant. For example Halls of Ivy stated the constant is given by:

    [tex]\frac{1}{2}v^2- \frac{GM}{r}= C[/tex]

    For your initial conditions (In the OP) the radius for this constant will be fixed at 12 million km and the velocity will be 0. Of course with this general equation you can have any initial conditions you fancy. Anyhoo just thought I'd point that out.
     
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