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The purpose of this thread is to present an analysis of the attraction of a particle to a solid disk of material of density rho. It will be shown that the result is finite and very small - and incidentally that the accleration is nearly uniform along the vertical axis regardless of one's height above the disk.

This problem (the gravitation of a uniform disk) is often called an Alderson disk

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Above is the cross section of a circular disk, of radius a, and thickness t. The left side of the disk has a cordinate x=-a. The right side of the disk has a coordinate x=+a. The top of the disk has a y coordinate 0. The bottom of the disk has a y coordiante of -t.

We wish to find the gravitational force on a particle located above the center of the disk, at x=0, y=b, where b << a.

We will assume that the disk is uniform and continuous.

From the geometry of the problem, it is obvious that the force on the particle will be directed downward.

Thus we need to consider only the 'y' component of the gravitational force. This will be equal to the intergal over the volume of the disk of

force/mass = G*dM*y/R^3

Here G is Newton's gravitational constant, dM is the mass of a small 'piece' of the disk, y is the vertical component of the distance from the small piece of the disk to the test mass, and R is the total distance from the small piece of the disk to the test mass.

Let x be the x and y be the coordinates of a point on the two dimensional surface above. This point represents the ring generated by revolving the 2 dimensional figure above around the y axis.

We can then write the above intergal as

[tex]

accel = \iint_{x=0,y=-t}^{x=a,y=0} \frac{G*(\rho*2*pi*x*dx*dy)*(b-y)}{((b-y)^2+x^2)^{\frac{3}{2}}}

[/tex]

here [tex]\rho[/tex] is the density of the material of the disk, G is the gravitational constant, x and y are the integration variables, and a,b, and t are the radius of the disk, the height of the particle above the disk, and the thickness of the disk as previously described.

2*pi*x*dx*dy is the volume of the annular ring of constant vertical distance (b-y) from the particle at y=b, and constant total distance R=sqrt((b-y)^2+x^2) from the particle. Multiplying this by rho gives us dM.

We will perform the integration over x first.

[clarification]

The indefinite integral (performed via computer) is easily verified by differentiation

[tex]

-2\,{\frac {G\rho\,\pi\, \left( b-y \right) }{\sqrt {{b}^{2}-2\,by+{y}

^{2}+{x}^{2}}}}

[/tex]

Substiituting the limits x=0..infinity gives the result

[tex]2 \pi G \rho dy[/tex]

[end clarification]

Next we will perform the integration over y. We let y vary from -t to zero. The result of the integration is a very simple formula

[tex]

2 \pi G \rho t

[/tex]

Note that this is not a function of b (the height above the disk) at all. The gravitational acceleration along the axis of the disk is truly constant only for an infinite disk, for a finite size disk it eventaully starts to drop off, but only when the height becomes much greater than the width of the disk. The formulas can be worked out for a finite radius disk, but the formula for an infinite disk is much simpler.

Note also that the result is finite even in the limit as b->0.

To put this result into numerical context, we will calculate the acceleration at the center of an infinite disk of aluminum, 2 feet thick.

The density of aluminium is 2700 kg/ m^3. Plugging this into the above formula using the Google calculator we get 7*10-7 meters/second squared. This is slightly under one tenth of one millionth of a gravity. Gravity is a very weak force.

google calculator link

This problem (the gravitation of a uniform disk) is often called an Alderson disk

---------------------

---------------------

---------------------

Above is the cross section of a circular disk, of radius a, and thickness t. The left side of the disk has a cordinate x=-a. The right side of the disk has a coordinate x=+a. The top of the disk has a y coordinate 0. The bottom of the disk has a y coordiante of -t.

We wish to find the gravitational force on a particle located above the center of the disk, at x=0, y=b, where b << a.

We will assume that the disk is uniform and continuous.

From the geometry of the problem, it is obvious that the force on the particle will be directed downward.

Thus we need to consider only the 'y' component of the gravitational force. This will be equal to the intergal over the volume of the disk of

force/mass = G*dM*y/R^3

Here G is Newton's gravitational constant, dM is the mass of a small 'piece' of the disk, y is the vertical component of the distance from the small piece of the disk to the test mass, and R is the total distance from the small piece of the disk to the test mass.

Let x be the x and y be the coordinates of a point on the two dimensional surface above. This point represents the ring generated by revolving the 2 dimensional figure above around the y axis.

We can then write the above intergal as

[tex]

accel = \iint_{x=0,y=-t}^{x=a,y=0} \frac{G*(\rho*2*pi*x*dx*dy)*(b-y)}{((b-y)^2+x^2)^{\frac{3}{2}}}

[/tex]

here [tex]\rho[/tex] is the density of the material of the disk, G is the gravitational constant, x and y are the integration variables, and a,b, and t are the radius of the disk, the height of the particle above the disk, and the thickness of the disk as previously described.

2*pi*x*dx*dy is the volume of the annular ring of constant vertical distance (b-y) from the particle at y=b, and constant total distance R=sqrt((b-y)^2+x^2) from the particle. Multiplying this by rho gives us dM.

We will perform the integration over x first.

[clarification]

The indefinite integral (performed via computer) is easily verified by differentiation

[tex]

-2\,{\frac {G\rho\,\pi\, \left( b-y \right) }{\sqrt {{b}^{2}-2\,by+{y}

^{2}+{x}^{2}}}}

[/tex]

Substiituting the limits x=0..infinity gives the result

[tex]2 \pi G \rho dy[/tex]

[end clarification]

Next we will perform the integration over y. We let y vary from -t to zero. The result of the integration is a very simple formula

[tex]

2 \pi G \rho t

[/tex]

Note that this is not a function of b (the height above the disk) at all. The gravitational acceleration along the axis of the disk is truly constant only for an infinite disk, for a finite size disk it eventaully starts to drop off, but only when the height becomes much greater than the width of the disk. The formulas can be worked out for a finite radius disk, but the formula for an infinite disk is much simpler.

Note also that the result is finite even in the limit as b->0.

To put this result into numerical context, we will calculate the acceleration at the center of an infinite disk of aluminum, 2 feet thick.

The density of aluminium is 2700 kg/ m^3. Plugging this into the above formula using the Google calculator we get 7*10-7 meters/second squared. This is slightly under one tenth of one millionth of a gravity. Gravity is a very weak force.

google calculator link

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