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The first step is to find the partial derivative of $$P = \frac{3 M}{(2 a e)} \left[B \left(1 - \frac{r^2}{(2 a^2 e^2)} + \frac{z^2}{(a^2 e^2}\right) + \frac{r^2}{(2 a^2 e^2)} \sin(B) \ \cos(B) - \frac{z^2}{(2 a^2 e^2)} \tan(B) \right] \ \ $$ with respect to r.

The result occupies several pages, so I will not post the result here, but it can be seen in this Mathematica worksheet.

The next step is to substitute the value of z (The vertical coordinate) as a function of r (the horizontal coordinate).

$$z(r) = \sqrt{(a^2 - a^2 e^2)*(1 - r^2/a^2)}$$

which is readily obtained from the definition of an ellipse.

We can now plot the horizontal gravitational force as a function of r:

enter image description here

In the above plot it can be seen that although the curve of $F_r$ is not trivial, it is in fact a straight line within the confines of $r<a$, i.e. within the radius of the oblate sphere.

A little experimentations shows that the straight line portion of the graph for $F_r$ is approximated fairly well by the following:

$$F_r = \frac{r}{a} \left( \frac{M}{a^2} \exp{(e^2)}^{\left(\frac{2 e^2}{3}\right)} + \frac{1}{\sqrt{1 - e^2}} \right)$$

The important thing to note is that everything in the equation is a constant except for r and the equation is proportional to $r M$ and not the expected $M/r^2$. It means the surface gravity vector of an oblate sphere that points towards the central axis increases as r gets larger!

Of course the above equation is an approximation, but it is not too difficult to show that for any value of a for the radius of the galaxy and any value of $e$ for the eccentricity of the galaxy, the $r M$ relationship always holds. For the picky, the exact equations are contsined in the Mathematica document linked above.

For completeness, the force component $F_z$ that acts down towards the disk parallel to the rotation z axis, can be approximated by:

$$F_r = \sqrt{a^2-r^2} \left(\frac{M}{a^3} (\exp{e^2})^{\left(\frac{9 e^2}{10}\right)} + \frac{1}{a \sqrt{1 - e^2}} \right)$$

With these two approximations we can readily plot the force vectors and visualise the resultant total force vector at the surface.

enter image description here

The first step is to find the partial derivative of $$P = \frac{3 M}{(2 a e)} \left[B \left(1 - \frac{r^2}{(2 a^2 e^2)} + \frac{z^2}{(a^2 e^2}\right) + \frac{r^2}{(2 a^2 e^2)} \sin(B) \ \cos(B) - \frac{z^2}{(2 a^2 e^2)} \tan(B) \right] \ \ $$ with respect to r.

The result occupies several pages, so I will not post the result here, but it can be seen in this Mathematica worksheet.

The next step is to substitute the value of z (The vertical coordinate) as a function of r (the horizontal coordinate).

$$z(r) = \sqrt{(a^2 - a^2 e^2)*(1 - r^2/a^2)}$$

which is readily obtained from the definition of an ellipse.

We can now plot the horizontal gravitational force as a function of r:

enter image description here

In the above plot it can be seen that although the curve of $F_r$ is not trivial, it is in fact a straight line within the confines of $r<a$, i.e. within the radius of the oblate sphere.

A little experimentations shows that the straight line portion of the graph for $F_r$ is approximated fairly well by the following:

$$F_r = \frac{r}{a} \left( \frac{M}{a^2} \exp{(e^2)}^{\left(\frac{2 e^2}{3}\right)} + \frac{1}{\sqrt{1 - e^2}} \right)$$

The important thing to note is that everything in the equation is a constant except for r and the equation is proportional to $r M$ and not the expected $M/r^2$. It means the surface gravity vector of an oblate sphere that points towards the central axis increases as r gets larger!

Of course the above equation is an approximation, but it is not too difficult to show that for any value of a for the radius of the galaxy and any value of $e$ for the eccentricity of the galaxy, the $r M$ relationship always holds.

For completeness, the force component $F_z$ that acts down towards the disk parallel to the rotation z axis, can be approximated by:

$$F_r = \sqrt{a^2-r^2} \left(\frac{M}{a^3} (\exp{e^2})^{\left(\frac{9 e^2}{10}\right)} + \frac{1}{a \sqrt{1 - e^2}} \right)$$

With these two approximations we can readily plot the force vectors and visualise the resultant total force vector at the surface.

enter image description here

The first step is to find the partial derivative of $$P = \frac{3 M}{(2 a e)} \left[B \left(1 - \frac{r^2}{(2 a^2 e^2)} + \frac{z^2}{(a^2 e^2}\right) + \frac{r^2}{(2 a^2 e^2)} \sin(B) \ \cos(B) - \frac{z^2}{(2 a^2 e^2)} \tan(B) \right] \ \ $$ with respect to r.

The result occupies several pages, so I will not post the result here, but it can be seen in this Mathematica worksheet.

The next step is to substitute the value of z (The vertical coordinate) as a function of r (the horizontal coordinate).

$$z(r) = \sqrt{(a^2 - a^2 e^2)*(1 - r^2/a^2)}$$

which is readily obtained from the definition of an ellipse.

We can now plot the horizontal gravitational force as a function of r:

enter image description here

In the above plot it can be seen that although the curve of $F_r$ is not trivial, it is in fact a straight line within the confines of $r<a$, i.e. within the radius of the oblate sphere.

A little experimentations shows that the straight line portion of the graph for $F_r$ is approximated fairly well by the following:

$$F_r = \frac{r}{a} \left( \frac{M}{a^2} \exp{(e^2)}^{\left(\frac{2 e^2}{3}\right)} + \frac{1}{\sqrt{1 - e^2}} \right)$$

The important thing to note is that everything in the equation is a constant except for r and the equation is proportional to $r M$ and not the expected $M/r^2$. It means the surface gravity vector of an oblate sphere that points towards the central axis increases as r gets larger!

Of course the above equation is an approximation, but it is not too difficult to show that for any value of a for the radius of the galaxy and any value of $e$ for the eccentricity of the galaxy, the $r M$ relationship always holds. For the picky, the exact equations are contsined in the Mathematica document linked above.

For completeness, the force component $F_z$ that acts down towards the disk parallel to the rotation z axis, can be approximated by:

$$F_r = \sqrt{a^2-r^2} \left(\frac{M}{a^3} (\exp{e^2})^{\left(\frac{9 e^2}{10}\right)} + \frac{1}{a \sqrt{1 - e^2}} \right)$$

With these two approximations we can readily plot the force vectors and visualise the resultant total force vector at the surface.

enter image description here

Removed references to dark matter
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KDP
  • 10.1k
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The first step is to find the partial derivative of $$P = \frac{3 M}{(2 a e)} \left[B \left(1 - \frac{r^2}{(2 a^2 e^2)} + \frac{z^2}{(a^2 e^2}\right) + \frac{r^2}{(2 a^2 e^2)} \sin(B) \ \cos(B) - \frac{z^2}{(2 a^2 e^2)} \tan(B) \right] \ \ $$ with respect to r.

The result occupies several pages, so I will not post the result here, but it can be seen in this Mathematica worksheet.

The next step is to substitute the value of z (The vertical coordinate) as a function of r (the horizontal coordinate).

$$z(r) = \sqrt{(a^2 - a^2 e^2)*(1 - r^2/a^2)}$$

which is readily obtained from the definition of an ellipse.

We can now plot the horizontal gravitational force as a function of r:

enter image description here

In the above plot it can be seen that although the curve of $F_r$ is not trivial, it is in fact a straight line within the confines of $r<a$, i.e. within the radius of the oblate sphere.

A little experimentations shows that the straight line portion of the graph for $F_r$ is approximated fairly well by the following:

$$F_r = \frac{r}{a} \left( \frac{M}{a^2} \exp{(e^2)}^{\left(\frac{2 e^2}{3}\right)} + \frac{1}{\sqrt{1 - e^2}} \right)$$

The important thing to note is that everything in the equation is a constant except for r and the equation is proportional to $r M$ and not the expected $M/r^2$. This is the crux of the matter. It means the surface gravity vector of an oblate sphere that points towards the central axis increases as r gets larger! Thirty years ago, when it was found that the rotation curves of galaxies maintain either constant or increasing velocity with radius, it was thought the explanation must be a large sphere or halo of invisible matter (dark matter) and this would yield the required $rM$ relationship. The idea is that as you move outward, the enclosed volume of matter increases as per $r^3$ while the gravitational attraction is proportional to $M/r^2$ which together yield $rM$. This analysis shows we do not need a halo of invisible matter, because the natural equation for the surface gravity of a flat oblate sphere is also $rM$.

Of course the above equation is an approximation, but it is not too difficaultdifficult to show that for any value of a for the radius of the galaxy and any value of $e$ for the eccentricity of the galaxy, the $r M$ relationship always holds.

In other RIP dark matter, for any galaxies that have a flattened elliptical profile,.

For completeness, the force component $F_z$ that acts down towards the disk parallel to the rotation z axis, can be approximated by:

$$F_r = \sqrt{a^2-r^2} \left(\frac{M}{a^3} (\exp{e^2})^{\left(\frac{9 e^2}{10}\right)} + \frac{1}{a \sqrt{1 - e^2}} \right)$$

With these two approximations we can readily plot the force vectors and visualise the resultant total force vector at the surface.

enter image description here

The first step is to find the partial derivative of $$P = \frac{3 M}{(2 a e)} \left[B \left(1 - \frac{r^2}{(2 a^2 e^2)} + \frac{z^2}{(a^2 e^2}\right) + \frac{r^2}{(2 a^2 e^2)} \sin(B) \ \cos(B) - \frac{z^2}{(2 a^2 e^2)} \tan(B) \right] \ \ $$ with respect to r.

The result occupies several pages, so I will not post the result here, but it can be seen in this Mathematica worksheet.

The next step is to substitute the value of z (The vertical coordinate) as a function of r (the horizontal coordinate).

$$z(r) = \sqrt{(a^2 - a^2 e^2)*(1 - r^2/a^2)}$$

which is readily obtained from the definition of an ellipse.

We can now plot the horizontal gravitational force as a function of r:

enter image description here

In the above plot it can be seen that although the curve of $F_r$ is not trivial, it is in fact a straight line within the confines of $r<a$, i.e. within the radius of the oblate sphere.

A little experimentations shows that the straight line portion of the graph for $F_r$ is approximated fairly well by the following:

$$F_r = \frac{r}{a} \left( \frac{M}{a^2} \exp{(e^2)}^{\left(\frac{2 e^2}{3}\right)} + \frac{1}{\sqrt{1 - e^2}} \right)$$

The important thing to note is that everything in the equation is a constant except for r and the equation is proportional to $r M$ and not the expected $M/r^2$. This is the crux of the matter. It means the surface gravity vector of an oblate sphere that points towards the central axis increases as r gets larger! Thirty years ago, when it was found that the rotation curves of galaxies maintain either constant or increasing velocity with radius, it was thought the explanation must be a large sphere or halo of invisible matter (dark matter) and this would yield the required $rM$ relationship. The idea is that as you move outward, the enclosed volume of matter increases as per $r^3$ while the gravitational attraction is proportional to $M/r^2$ which together yield $rM$. This analysis shows we do not need a halo of invisible matter, because the natural equation for the surface gravity of a flat oblate sphere is also $rM$.

Of course the above equation is an approximation, but it is not too difficault to show that for any value of a for the radius of the galaxy and any value of $e$ for the eccentricity of the galaxy, the $r M$ relationship always holds.

In other RIP dark matter, for any galaxies that have a flattened elliptical profile,.

For completeness, the force component $F_z$ that acts down towards the disk parallel to the rotation z axis, can be approximated by:

$$F_r = \sqrt{a^2-r^2} \left(\frac{M}{a^3} (\exp{e^2})^{\left(\frac{9 e^2}{10}\right)} + \frac{1}{a \sqrt{1 - e^2}} \right)$$

With these two approximations we can readily plot the force vectors and visualise the resultant total force vector at the surface.

enter image description here

The first step is to find the partial derivative of $$P = \frac{3 M}{(2 a e)} \left[B \left(1 - \frac{r^2}{(2 a^2 e^2)} + \frac{z^2}{(a^2 e^2}\right) + \frac{r^2}{(2 a^2 e^2)} \sin(B) \ \cos(B) - \frac{z^2}{(2 a^2 e^2)} \tan(B) \right] \ \ $$ with respect to r.

The result occupies several pages, so I will not post the result here, but it can be seen in this Mathematica worksheet.

The next step is to substitute the value of z (The vertical coordinate) as a function of r (the horizontal coordinate).

$$z(r) = \sqrt{(a^2 - a^2 e^2)*(1 - r^2/a^2)}$$

which is readily obtained from the definition of an ellipse.

We can now plot the horizontal gravitational force as a function of r:

enter image description here

In the above plot it can be seen that although the curve of $F_r$ is not trivial, it is in fact a straight line within the confines of $r<a$, i.e. within the radius of the oblate sphere.

A little experimentations shows that the straight line portion of the graph for $F_r$ is approximated fairly well by the following:

$$F_r = \frac{r}{a} \left( \frac{M}{a^2} \exp{(e^2)}^{\left(\frac{2 e^2}{3}\right)} + \frac{1}{\sqrt{1 - e^2}} \right)$$

The important thing to note is that everything in the equation is a constant except for r and the equation is proportional to $r M$ and not the expected $M/r^2$. It means the surface gravity vector of an oblate sphere that points towards the central axis increases as r gets larger!

Of course the above equation is an approximation, but it is not too difficult to show that for any value of a for the radius of the galaxy and any value of $e$ for the eccentricity of the galaxy, the $r M$ relationship always holds.

For completeness, the force component $F_z$ that acts down towards the disk parallel to the rotation z axis, can be approximated by:

$$F_r = \sqrt{a^2-r^2} \left(\frac{M}{a^3} (\exp{e^2})^{\left(\frac{9 e^2}{10}\right)} + \frac{1}{a \sqrt{1 - e^2}} \right)$$

With these two approximations we can readily plot the force vectors and visualise the resultant total force vector at the surface.

enter image description here

Source Link
KDP
  • 10.1k
  • 1
  • 11
  • 61

The first step is to find the partial derivative of $$P = \frac{3 M}{(2 a e)} \left[B \left(1 - \frac{r^2}{(2 a^2 e^2)} + \frac{z^2}{(a^2 e^2}\right) + \frac{r^2}{(2 a^2 e^2)} \sin(B) \ \cos(B) - \frac{z^2}{(2 a^2 e^2)} \tan(B) \right] \ \ $$ with respect to r.

The result occupies several pages, so I will not post the result here, but it can be seen in this Mathematica worksheet.

The next step is to substitute the value of z (The vertical coordinate) as a function of r (the horizontal coordinate).

$$z(r) = \sqrt{(a^2 - a^2 e^2)*(1 - r^2/a^2)}$$

which is readily obtained from the definition of an ellipse.

We can now plot the horizontal gravitational force as a function of r:

enter image description here

In the above plot it can be seen that although the curve of $F_r$ is not trivial, it is in fact a straight line within the confines of $r<a$, i.e. within the radius of the oblate sphere.

A little experimentations shows that the straight line portion of the graph for $F_r$ is approximated fairly well by the following:

$$F_r = \frac{r}{a} \left( \frac{M}{a^2} \exp{(e^2)}^{\left(\frac{2 e^2}{3}\right)} + \frac{1}{\sqrt{1 - e^2}} \right)$$

The important thing to note is that everything in the equation is a constant except for r and the equation is proportional to $r M$ and not the expected $M/r^2$. This is the crux of the matter. It means the surface gravity vector of an oblate sphere that points towards the central axis increases as r gets larger! Thirty years ago, when it was found that the rotation curves of galaxies maintain either constant or increasing velocity with radius, it was thought the explanation must be a large sphere or halo of invisible matter (dark matter) and this would yield the required $rM$ relationship. The idea is that as you move outward, the enclosed volume of matter increases as per $r^3$ while the gravitational attraction is proportional to $M/r^2$ which together yield $rM$. This analysis shows we do not need a halo of invisible matter, because the natural equation for the surface gravity of a flat oblate sphere is also $rM$.

Of course the above equation is an approximation, but it is not too difficault to show that for any value of a for the radius of the galaxy and any value of $e$ for the eccentricity of the galaxy, the $r M$ relationship always holds.

In other RIP dark matter, for any galaxies that have a flattened elliptical profile,.

For completeness, the force component $F_z$ that acts down towards the disk parallel to the rotation z axis, can be approximated by:

$$F_r = \sqrt{a^2-r^2} \left(\frac{M}{a^3} (\exp{e^2})^{\left(\frac{9 e^2}{10}\right)} + \frac{1}{a \sqrt{1 - e^2}} \right)$$

With these two approximations we can readily plot the force vectors and visualise the resultant total force vector at the surface.

enter image description here

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