Gravitational Force between two bodies when Distance is zero

[nvijayprakash]

What will be the Gravitaional force between two bodies when the distance between them is zero
e.g., If we place a ring around a sphere, the center of gravity of the two bodies coincide and thus the distance of separation is zero.

F = G m₁m₂\d²
d = 0 ,F = Infinity

 

[A.T.]

What will be the Gravitaional force between two bodies when the distance between them is zero
e.g., If we place a ring around a sphere, the center of gravity of the two bodies coincide and thus the distance of separation is zero.

F = G m₁m₂\d²
d = 0 ,F = Infinity

This formula is for point masses with d > 0 and non intersecting spherical masses

 

[Doc Al]

You can't just stick the distance between the centers of mass into that formula. It applies to each element of mass; you'll have to integrate. In this case the net gravitational force will be zero, not infinity.

 

[arpan251089]

Well, the answer is (As far as i think)

This equation applies to "point like" bodies
if two (POINT LIKE) bodies have zero distance between them that means they are located at the same location.
So they will behave as single object and this unified body (which is again POINT LIKE) will not exert any gravitational force on itself!

 

[arpan251089]

Well when we place the ring and sphere as you said that time

(i) Ring is outside the sphere so sphere can be treated as point like object with its mass concentrated at the centre (Let us say it S)
(ii) Now break the ring into elementary parts each elementary part will be at a distance of r (radius of ring) from S.
(iii) now each elementary part will exert a force on S. So the net force on S will be Zero
(iv) But the elementary parts of ring will not have gravitational force in all directions. So if the ring is not rigid it will try to shrink[otherwise restoring forces in the ring will balance the gravitational force on each elementary ring and it will remain intact]

 

[denisv]

x/0 is undefined, not infinity.

 

[D H]

You are missing the point, denisv. Newton's law of gravitation applies only to point masses. To compute the force between two non-point masses you will have to apply Newton's laws to all pairs of points in the two bodies. In general, this is a rather nasty double volume integral:

\mathbf F = - \int_{V_1} \int_{V_2}<br /> \frac{G\rho(\mathbf x_1)\rho(\mathbf x_2)}<br /> {||\mathbf x_1-\mathbf x_2||^3}(\mathbf x_1-\mathbf x_2)<br /> d\mathbf x_2 d\mathbf x_1<br />

In this particular case, the force is zero. Everything cancels.

 

[denisv]

What you say in no way changes that x/0 is undefined, which is what I was pointing out.

 

[A.T.]

What you say in no way changes that x/0 is undefined, which is what I was pointing out.

No one said that you are wrong, just that you miss the point. What would the difference between "infinity" and "undefined" by anyway for physical questions?

 

[DaveC426913]

This equation applies to "point like" bodies
if two (POINT LIKE) bodies have zero distance between them that means they are located at the same location.
So they will behave as single object and this unified body (which is again POINT LIKE) will not exert any gravitational force on itself!

Objects within a unified body (such as the Earth) do indeed exert gravitational force on each other i.e. every atom in the Earth affects every other atom in the Earth.