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Variation of gravitational force with calculus

  1. Jun 15, 2014 #1
    Hi all,

    This is Newton's universal law of gravitation:

    F = GMm/r2, where r is the distance between the centre of the two bodies.

    Therefore, considering two objects in mutual gravitational acceleration, with only linear motion and acceleration, they shall be moving in closer and closer. Since the force is inversely proportional to the distance between the two bodies centre, it will increase as r decreases. Therefore as they move gravitational force gets stronger and stronger.

    The question is, how to compute the instantaneous acceleration in those conditions using calculus?

    Also, considering the two objects again, how would I compute the velocity needed for mass m to enter in orbit around mass M?

    Thanks for any answers
     
    Last edited: Jun 15, 2014
  2. jcsd
  3. Jun 15, 2014 #2

    Matterwave

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    The instantaneous acceleration is pretty easy to find simply by applying Newton's second law. What's Newton's second law tell you? This should not require any calculus.

    Computing the orbital velocity requires knowledge of which orbit you are talking about. Different orbits require different velocities.

    In the case of elliptical orbits, there is no 1 velocity that is the "orbit velocity" because by Kepler's 2nd law, the orbiting object moves fastest when it is closest to the gravitating body.

    Circular orbits have 1 velocity, and they depend on the distance R from which you are orbiting. The problem is easier still when one object (say M) is much more massive than the other object so we can make an approximation that M just stays stationary (just an approximation).

    Which conditions are you considering?
     
  4. Jun 16, 2014 #3
    So my acceleration at any point would be the magnitude of the force at the distance r divided by the mass of object 2?
    The conditions I am considering are a circular orbit of distance r, where mass M around which mass m orbits can be approximated to remain stationary.

    Thanks.
     
  5. Jun 16, 2014 #4

    Drakkith

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    I believe in that case you simply find the magnitude of the force and calculate the acceleration. Since you have a circular orbit, the distance between the two objects will not change and you will have no variation in either the force or the acceleration.
     
  6. Jun 16, 2014 #5

    Matterwave

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    You should be careful to specify which object you are talking about. The acceleration of object 2 will be the force you wrote down divided by the mass of object 2.

    For a circular orbit, you can use the above fact AND the fact that for circular motion the centripetal acceleration is v^2/r. Using these two facts, you can set the two accelerations equal to each other and solve for the velocity. This should also only require algebra. Although deriving a=v^2/r takes a little bit of vector manipulation. You should be able to find that in any introductory physics textbook.
     
  7. Jun 17, 2014 #6

    UltrafastPED

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