Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Incorporating Inverse Square Law In Gravity

  1. Oct 21, 2012 #1
    First of all, I'm 13 so I might not comprehend the complex vocabulary or symbols others might use. Second, I just joined!

    Okay, let's get to it.

    I think I know what the inverse square law is: if a number goes up by x, then the other number is the square of x but in the negative side. Right?

    I also think that the force of gravity on a 1kg object on the surface of the Earth (from the average radius) is 9,818,373.084 N . Now what if the distance gets increased by 10 metres? In the equation for the force: F=Gm1m2/r2, do you just add 10 to the r? So it would equal 9,787,623.422 N?

    1st equation)
    F = 9,818,373.084 N
    G = 6.673 X 10-11
    m1= 1kg
    m2= 5.97219 X 1024 (Earth)
    r2 = 40,589,641 (63712)

    2nd equation)
    F = 9,787,623.422 N
    G = 6.673 X 10-11
    m1 = 1kg
    m2 = 5.97219 X 1024 (Earth)
    r2 = 40,717,161 (63812)

    If so, where does the inverse square law go?

    And is this correct?
     
  2. jcsd
  3. Oct 21, 2012 #2

    Mute

    User Avatar
    Homework Helper

    I'm not sure what you mean by "in the negative side", so let me try to explain it for you in words.

    Suppose I said the magnitude of some force between two objects increases linearly with separation. That means that if I double the distance between the two objects, the force between the two objects doubles. That is, the factor by which the force increased is the same as the factor by which the separation increased.

    Now suppose I said that the magnitude of some force between two objects decreases inversely with the separation. This means that if I double the distance between the two objects, the force between the two objects is cut in half. The factor by which the force changed is 1 over the factor by which the separation changed.

    Now suppose I said that the magnitude of some force between two objects increases as the square of the separation. Then if I double the separation between the objects, the force between them is quadrupoled (four times): the factor by which the force increases is the square of the factor by which the distance increased.

    If we put these three observations together, you can conclude that if the force between two objects follows and inverse square relationship, then if we double the separation between two objects, the force between them is reduced by 1/4: the factor by which the force is decreased is 1 over the factor by which the separation increased.

    Does this make it clearer?

    The calculations look correct. The inverse square law is the 1/r2 part of the equation for the force: F=Gm1m2/r2.

    If the force of gravity were inversely proportional to the separation, then the equation would involve 1/r instead of 1/r2.
     
  4. Oct 21, 2012 #3
    It is a good idea to use (and keep track of) units.
    The weight of a 1 kg object on the surface of the Earth is around 9.8 N.
    You are off by 6 orders of magnitude. The reason is using the radius in km whereas G is in standard SI units (N*m^2/kg^2).

    For the meaning of the inverse square law, Mute has explained it already.
     
  5. Oct 22, 2012 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The same km<->m-issue occurs at the height difference: It is 10m (6371.000 -> 6371.010), not 10km (6371->6381).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Incorporating Inverse Square Law In Gravity
  1. Inverse square law (Replies: 5)

  2. Inverse Square Law (Replies: 9)

  3. Inverse square law (Replies: 8)

Loading...