Do 2 equal gravitational pulls cancel?

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SUMMARY

The discussion centers on the gravitational effects experienced by a smaller object placed at the midpoint between two equal masses, each comparable to Earth. It is established that the net gravitational field at this midpoint is zero, meaning the smaller object would not experience any gravitational force. However, if the smaller object has a size, it could experience differential gravitational forces across its volume, potentially leading to tidal forces that could tear it apart if the mass is weak enough. The concept of Lagrange points is introduced, highlighting their stability in gravitational fields.

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thechosenone
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imagine this scenario. you have two equally massive objects, exerting an equal amount of gravitational force (they are perfect spheres), and they are held apart a certain distance (how they are held apart is irrelevant). for simplicity, let's just say that their masses are about the same as that of earth, and that the distance between them is fairly small; say, a few miles or so, close enough so they would both have a strong gravitational pull.

now here comes my question: if you placed a third object with a much smaller mass (although large enough to be affected by gravity... perhaps a 100kg sphere?) right at the midpoint between these two masses, what would happen to it?

my first thought that was it would be torn in two and each half would go towards the closer mass... but how would any movement be able to happen if the net force is zero? maybe you guys can help me figure out what would happen to the middle object?
 
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thechosenone said:
my first thought that was it would be torn in two and each half would go towards the closer mass... but how would any movement be able to happen if the net force is zero?
The net gravitational field exactly at the midpoint between the two large, equal masses will be zero. That means an object placed there would experience no gravitational force, not two strong forces that cancel out. (Of course, a real object has some size so it can't all be at the exact middle--but I believe my answer addresses your real concern.)
 
You should know that net gravitational field at the midpoint (assuming that both objects have equal masses) where you want to place your object would be zero.

EDIT: OK, I need to seriously work on my typing speed.
 
ok, thanks for the help
 
i think this is called a "lagrange point" and the one you describe is not stable. but there are stable lagrange points in the orbit of the moon around the center of mass it shares with the earth. those stable lagrange points collect dust and/or debris, i think.
 
Lagrange points are a bit more complex to understand, but worth looking at. As usual wiki provides at least an introductory background. Consider a gravitational field in the same way you might think about electric fields, more than likely you've come across questions asking to calculate the resulting force when charged bodies interact, replace electric with gravitational field and it's all good.

I remember being amazed at the fact that branches of physics that seem so far apart obey the same basic outline, patterns appear everywhere.
 
if u analyze it according to an electric or magnetic field, which i think is a correct way to do it, the forces would be distributed along the surface of the object, so this will create an opposite pull on the object. if its a weak object and the pull force is big enough it should be torn apart. don't u think guys?
 
eaboujaoudeh said:
if u analyze it according to an electric or magnetic field, which i think is a correct way to do it, the forces would be distributed along the surface of the object, so this will create an opposite pull on the object. if its a weak object and the pull force is big enough it should be torn apart. don't u think guys?

Yes. But don't think that it is "each of the composing forces" which is pulling on the object's parts. What happens, is that you FIRST have to add the two g-vectors originating from the two sources, obtain the "sum g field", and THEN you apply the local g vector to each of the pieces of your extended object.

Exactly as with an E-field.

(only, because "gravitational charge" is not mobile, as are electrical charges in a conductor, the interaction is all over the volume, and not concentrated on the surface).
 
I think another way of looking at this question is: If you could drill a tunnel to the center of the Earth and build a room there, you would be weightless in this room because the pull of gravity would be equal in every direction.
 

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