1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Let's test our basic knowledge of physics

  1. Apr 6, 2009 #1

    Newton derived F=GMm/R^2 by puting one kg of sphere on the surface of earth. Now consider if the same sphere is bigger and bigger till it reach the size of earth. OR simply put imaginary earth on earth

    Now apply F= GMm/R2 where g = GM/d2. M= m= mass of earth = mass of imaginary earth, centre to centre distance between two masses = diameter of earth.
    According to universal law of gravitation; there is force of F=GM^2/d^2 between aforementioned masses.

    As gravities of both earths cancel out and both masses exert a force which is equal but opposit in direction on each other therefore it is wrong to say that there is force of F=GM^2/d^2 between aforementioned masses.

    This means that weight of mass of 1 kg of sphere is start decreasing by increasing its size on the surface of ground.


    Suppose there are two sphere of masses m1 = m2 = 1 kg, diameter of m1 = m2 = 0.5 m. Both masses are in space and the centre to center distance between them is 1 metre = r , then
    F= Gm1m2/r2 = G = 6.67*10-11 Newton.

    But technically, both masses exert a force which is equal but opposit in direction on each other (gravaties of both masses g= Gm2/r2= Gm1/r2) and hence cancel out.

    if not, how come F= Gm1m2/r2 = G = 6.67*10-11 Newton exist and how both are gravitating and falling masses at the same time.


    Let “P” is a point or an origin of two circles of radius r1=1 meter and r2= 2 meter. Consider these two circle are spheres (empty from inside) or two bangles in space. OR a spherical boiled egg with removed shell such that the center of the sperical white and yolk coinicide each other at one point say “P”.

    Now apply Newton’s law of gravitation i.e. F=GMm/R2 to aforementioned two masses and neglect all other local attractions.

    As center to center distance b/t masses is zero, therefore F= infinity but in reality it's not.

    Now how much force is required to separate aforementioned masses, infinity or less? If less then what about the Newton’s law?
  2. jcsd
  3. Apr 6, 2009 #2


    User Avatar
    Gold Member

    You seem to be ignoring density.
  4. Apr 7, 2009 #3


    User Avatar

    Staff: Mentor

    No, that's not it, he's just ignoring how forces interact:
    Having an equal and opposite force pair is not cancelling out. The equation is correct.

    An example: if you throw a baseball, the baseball and your hand exert equal and opposite forces on each other. Does that mean they cancel out and the ball drops straight down to the ground as soon as it leaves your hand? Of course not!

    For question two, you are applying the equation incorrectly. The gravity equation does indeed assume you are dealing with point masses, so if you are not, in fact, dealing with point masses, you must adjust your usage of the equation. How to deal with this issue is, in fact, part of Newton's own gravity theory. http://en.wikipedia.org/wiki/Shell_theorem

    You may want to consider that the issues you are raising (when real, anyway...) are so spectacularly obvious that not even the very first physicist to start to generate theories of motion and gravity overlooked them -- and as a result, perhaps you should adjust your posture/tone to one of accepting and questioning your own ignorance instead of questioning the theories.
  5. Apr 7, 2009 #4
    Very well stated.

  6. Apr 7, 2009 #5


    User Avatar
    Gold Member

    That's what I meant by density, but I stated it pretty badly. Thanks for the tune-up.
  7. Apr 7, 2009 #6
    the size is as big as earth ,but the mass is 1kg,it's a constant.
  8. Apr 7, 2009 #7

    so simple F1=GM^2/d^2= GM^2/d^2=F2 so 1=1 and in q2 just ignore the spherical boiled egg portion
  9. Apr 8, 2009 #8


    User Avatar

    Staff: Mentor

    O didn't ignore it, I pointed you to a link that discusses how to solve it!
  10. Apr 8, 2009 #9
    please consider the center of mass
  11. Apr 13, 2009 #10
    Actually I am quit busy and didn't go through his link therefore I advised for instance just to ignore a portion of q2. i will be back once i go through the link regarding spherical gravity.

    its weird but real

  12. Apr 19, 2009 #11
    q1: Both gees (accelerations) are equal but opposite in direction
    q2: Your link is irrelevant
  13. Apr 19, 2009 #12

    Doc Al

    User Avatar

    Staff: Mentor

    What are you talking about?
  14. Apr 19, 2009 #13


    User Avatar
    Science Advisor
    Gold Member

    The force on the first earth is only from the second earth and it is attractive. The force on the second earth is only from the first earth and it is attractive as well. If one were placed equidistant between the two bodies, then you would experience a net force of zero only if the second earth had a mass equal to the first earth (but you stated it had a mass of 1 kg), but you would still experience the two forces pulling you asunder. Either way, the two bodies still experience a force, the first earth towards the second earthand the second earthto the first earth.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook