Gravity of a Circular Ring at a co-planar external point.

Question

I have a circular ring of unit mass and fixed radius R which lies in the XY plane at point $O$ with coordinates $O:(0,0)$.

I wish to find a formula for the gravitational force at a point $P: (D,0)$ which lies in the same plane as the ring and is at some variable distance D from the ring centre O.

(Note: There are many treatments of the case for a target lying on the axis of the ring. The reference so far found nearest to this co-planar case is Problems 5-12, 5-13, (no solutions given) p.127 in Classical Dynamics of Particles and Systems by Jerry B. Marion.

I expect that the formula should be of the form $F = GM*f(D)$ where $G$ is the gravitational constant, $M$ is the mass and $f$ is some function similar to the Newtonian spherical divergence function $f(D) = \frac{1}{ D^2}$ (where the factor $\frac{1}{4.\pi}$ is absorbed in the value of the constant $G$ ).

So far I have obtained an integral formula by initially modelling the ring as a series of $N$ small point masses of mass $\frac{1}{N}$ separated by angle $\delta\theta$, whose distance from target is $L$ where:

$$L^2 = (D-a)^2+b^2 = D^2-2aD+R^2 = D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)$$

where $a (= R\cos\theta)$ and $b(=R\sin\theta)$ are the $x$ and $y$ coordinates of the point.

Due to symmetry and vector addition of forces there is no net force in the y-direction and so the effective force contribution (along $x$) for a point is given by multiplying by the cosine factor $(D-a)/L$ thus:-

$$ F = \frac{-GM}{N}\frac{1}{4\pi.L^2}\frac{D-a}{L} = \frac{-GM}{ N} \frac{D-a}{L^3} $$

$$ F = \frac{-GM}{ N} \frac{D-R\cos\theta}{\left(D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)\right)^{\frac{3}{2}}} $$

$$ F = \frac{-GM}{ N} \frac{D-R\cos\theta}{D^3 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$

$$ F = \frac{-GM}{ N} \frac{1-(R/D)\cos\theta}{D^2 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$

I then obtained the following integral formula for the force exerted on the target point by the ring:-

$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi}\int_0^{2\pi}\frac{1-Q\cos\theta}{\left(1-2Q\cos\theta+Q^2\right)^{\frac{3}{2}}} \text{d}\theta$$ where $Q = R/D$.

$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi} \frac{1}{(2Q)^{3/2}}\int_0^{2\pi}\frac{1-Q\cos\theta} {\left(\frac{Q^2+ 1}{2Q} - \cos\theta\right)^{\frac{3}{2}}} \text{d}\theta$$

Defining $A = \frac{Q^2+ 1}{2Q}$, Wolfram Alpha gives... $$ \int_0^{2\pi}\frac{ 1 - Q \cos x}{(A -\cos x)^{3/2}} dx $$

$$=\left[\frac{2}{(A^2-1)\sqrt{A - \cos x}}\left(A^2-1\right)Q\sqrt{\frac{A - \cos x}{A-1}} \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-AQ\sin x- (A-1)(AQ-1)\sqrt{\frac{A-\cos x}{A-1}}\operatorname{E}\left(\frac{x}{2}~\big|~\frac{2}{1-A}\right) +\sin x\right]_0^{2\pi}$$

Where $E(x|m)$ is an elliptic integral of the 2nd kind with parameter $m=k^2$, and $F(x|m)$ is an elliptic integral of the 1st kind with parameter $m=k^2$.

Replacing $\cos x$ by $1$ and $\sin x$ by $0$... $$=\frac{2}{(A^2-1)\sqrt{A -1}}*\left[(A^2-1)Q \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-(A-1)(AQ-1) \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)\right]_0^{2\pi}$$

Cancelling $(A^2-1)$... $$=\frac{2}{\sqrt{A -1}}\left[Q\operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right) - \frac{(AQ-1)}{A+1} \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right) \right]_0^{2\pi}$$

Being unfamiliar with Elliptic Integrals, this is as far as I can comfortably go at present.

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After reading the wikipedia article Elliptic Integral, proceeding tentatively, from the definitions of elliptic integrals I think that $E(x|k^2)$ and $F(x|k^2)$ both go to zero when $x$ is zero, thus...

$$=\frac{2Q}{\sqrt{A -1}}\left[\operatorname{F}\left(\pi~\big|~\frac{-2}{A-1}\right)-\frac{(AQ-1)}{AQ+Q} \operatorname{E}\left(\pi~\big|~\frac{-2}{A-1}\right)\right]$$

Next perhaps it would be helpful to reformulate the problem so that the amplitude(?) term in the elliptic integrals changes from $\pi$ to $\pi/2$, thereby making the elliptic integrals "complete" and permitting them to be expressed as power series. This reformulation could be done by modelling the gravitational effect ($Fx$ component only) of two half rings (positive $y$ and negative $y$), independently, and using respectively the angles $\theta_1$ and $\theta_2$ which both range from $0$ to $\pi/2$ but in different directions.

Answer 1

Considering similar problems it is usually simpler to consider the potential rather than force. The latter can be found later as the negative of the potential gradient. Assuming the masses of the test point-like body and the ring be $m$ and $M$, respectively, we have in spherical coordinates with the1 origin at the center of the ring and the polar axis directed perpendicular to the plane of the ring: $$ U({\bf r})=-\frac{GmM}{2\pi}\int_0^{2\pi}\frac{d\theta}{\sqrt{r^2+R^2+2rR\sin\phi\cos\theta}},\tag1 $$ where (following the "maths" convention referred to in the Spherical Coordinates link and for consistency with the Question) $r,\phi,\theta $ are radial distance, polar angle, and azimuthal angle of the point ${\bf r}$, and $R$ is the radius of the circle.

The integral $(1)$ can be dealt in the following way:

$$\begin{align} \int_0^{2\pi}\frac{d\theta}{\sqrt{r^2+R^2+2rR\sin\phi\cos\theta}} &=2\int_0^{\pi}\frac{d\theta}{\sqrt{r^2+R^2+2rR\sin\phi\cos\theta}}\\ &=2\int_0^{\pi}\frac{d\theta}{\sqrt{(r^2+R^2+2rR\sin\phi)-4rR\sin\phi\sin^2\frac\theta2}}\\ &=\frac{4}{\sqrt{r^2+R^2+2rR\sin\phi}} \operatorname{K}\left(\frac{4rR\sin\phi}{r^2+R^2+2rR\sin\phi}\right), \end{align} $$ where we used the convention $$ \operatorname{K}(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m\sin^2\theta}} $$ for the complete elliptic integral of the first kind.

Finally $$ U({\bf r})=-\frac{2GmM}{\pi\sqrt{r^2+R^2+2Rr\sin\phi}}\operatorname{K}\left(\frac{4rR\sin\phi}{r^2+R^2+2rR\sin\phi}\right).\tag2 $$

In the plane of the circle $\phi=\frac\pi2$ and the above equation simplifies to: $$ U({\bf r})=-\frac{2GmM}{\pi(R+r)}\operatorname{K}\left(\frac{4Rr}{(r+R)^2}\right). $$

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To obtain the expression for the acting force recall that: $$ \nabla f={\partial f \over \partial r}\hat{\mathbf r} + {1 \over r}{\partial f \over \partial \phi}\hat{\boldsymbol \phi} + {1 \over r\sin\phi}{\partial f \over \partial \theta}\hat{\boldsymbol \theta}. $$

As the potential $(2)$ does not depend on $\theta$ only two first terms remain.

Tedious but straightforward calculation reveals: $$\begin{align} {\bf F}_r&=\frac{GmM}{\pi}\frac{(R^2-r^2)\operatorname{E}\left(1-\frac {y^2}{x^2}\right)-y^2\operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{rxy^2};\tag3\\ {\bf F}_\phi&=\frac{GmM}{\pi}\frac{(R^2+r^2)\operatorname{E}\left(1-\frac {y^2}{x^2}\right)-y^2\operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{rxy^2}\cot\phi,\tag4\\ \end{align} $$ where $x=\sqrt{R^2+r^2+2Rr\sin\phi},\ y=\sqrt{R^2+r^2-2Rr\sin\phi}$.

Answer 2

In order to express the result in terms of complete elliptic integrals, it is easier to compute the gravitational potential $\phi(D)$ first. Then the (radial) field is given by $F(D) = - \phi'(D)$. Following your approach we find $$ \phi(D) = - \frac{G M}{2 \pi D} \int \limits_0^{2\pi} \frac{\mathrm{d} \theta}{\sqrt{1 - 2 Q \cos(\theta) + Q^2}} = - \frac{G M}{\pi D} \int \limits_0^{\pi} \frac{\mathrm{d} \theta}{\sqrt{1 - 2 Q \cos(\theta) + Q^2}} \, . $$ In the last step we have used the fact that the integral from $0$ to $\pi$ and that from $\pi$ to $2\pi$ have the same value. Now we can write $$ - \cos(\theta) = \cos(\pi - \theta) = 1 - 2 \sin^2\left(\frac{\pi - \theta}{2}\right) $$ and introduce the new integration variable $\alpha = \frac{\pi - \theta}{2}$ to obtain $$ \phi(D) = -\frac{2 G M}{\pi D} \int \limits_0^{\pi/2} \frac{\mathrm{d} \alpha}{\sqrt{1 + 2 Q + Q^2 - 4 Q \sin^2(\alpha)}} = -\frac{2 G M}{\pi D} \frac{1}{1+Q} \int \limits_0^{\pi/2} \frac{\mathrm{d} \alpha}{\sqrt{1 - \frac{4 Q}{(1+Q)^2} \sin^2(\alpha)}} \, . $$ But this integral is just the definition of the complete elliptic integral of the first kind and (using the parameter $m = k^2$ as the argument) $$ \phi(D) = - \frac{2 G M}{\pi D} \frac{1}{1+Q} \operatorname{K}\left(\frac{4 Q}{(1+Q)^2}\right) = - \frac{2 G M}{\pi D} \operatorname{K}(Q^2) = - \frac{2 G M}{\pi D} \operatorname{K}\left(\frac{R^2}{D^2}\right)$$ follows. The final simplification is an application of Gauss's transformation. Taking the derivative we find the field $$ F(D) = - \frac{2 G M}{\pi(D^2 - R^2)} \operatorname{E}\left(\frac{R^2}{D^2}\right) $$ in terms of the complete elliptic integral of the second kind.

Answer 3

Solution from Analysis of Force (not using Potential)

$$ F =\frac{-GM}{ D^2} \frac{1}{\pi} \left[ \frac{1}{\left(1-\frac{R}{D}\right)}\operatorname{K}\left(\frac{-4R/D }{ \left(1 - \frac{R}{D}\right)^2 }\right)+ \frac{1}{(1+\frac{R}{D})}\operatorname{E}\left(\frac{-4R/D}{ \left(1 - \frac{R}{D}\right)^2 }\right) \right]. $$

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As Question Poster I wanted to derive a solution by working purely with force ($F$) (i.e. not involving Gravitational Potential and thereby avoiding any calculus in having to convert between Force and Potential).

I greatly appreciate the answers from /u/USER/ and /u/ComplexYetTrivial/ which were developed using a model of the Potential then converting to Force.

All three solutions are presented in different algebraic forms but they give the same results. The equivalence of the $K()$ components in my equation and that derived from /u/USER/'s answer can be demonstrated using Gauss's Transformation (see Appendix 2 for Details) as follows...

$$ \frac{1}{\left(1-\frac{R}{D}\right)}\operatorname{K}\left(\frac{-4R/D }{ \left(1 - \frac{R}{D}\right)^2 }\right) = \operatorname{K}\left( \left(\frac{-R }{ D }\right)^2 \right) = \operatorname{K}\left( \left(\frac{+R }{ D }\right)^2 \right) = \frac{1}{\left(1+\frac{R}{D}\right)}\operatorname{K}\left(\frac{+4R/D }{ \left(1 + \frac{R}{D}\right)^2 }\right). $$

The equivalence of the $E()$ components in my equation and that derived from /u/USER/'s answer

$$ \left(1-\frac{R}{D}\right) ~\operatorname{E}~\left(\frac{-4R/D }{ \left(1 - \frac{R}{D}\right)^2 }\right) = \left(1+\frac{R}{D}\right) ~\operatorname{E}~\left(\frac{+4R/D }{ \left(1 + \frac{R}{D}\right)^2 }\right). $$

is assumed from the fact that both answers give the same results. But I have no external validation of this at present from analysis or sources.

The equivalence of the solution by /u/ComplexYetTrivial/ with the other two solutions is also demonstrated here (in absence of full derivation)to my present knowledge only by the fact that it gives the same solution. If it is correct then the following identities are indicated (follow-up question):-

$$ (1-x ) ~\operatorname{E}~\left(\frac{-4x }{ \left(1 - x\right)^2 }\right) = 2\operatorname{E}\left(x^2\right)-(1-x^2)\operatorname{K}\left(x^2\right) = (1+x ) ~\operatorname{E}~\left(\frac{+4x }{ \left(1 + x\right)^2 }\right). $$

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As a starting point we have the the following integral formula for the net force purely along the $x$ axis acting on the target point towards the ring centre:-

$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi}\int_0^{2\pi}\frac{1-Q\cos\theta}{\left(1-2Q\cos\theta+Q^2\right)^{\frac{3}{2}}} \text{d}\theta$$ where $Q = R/D$.

$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi} \frac{1}{(2Q)^{3/2}}\int_0^{2\pi}\frac{1-Q\cos\theta} {\left(\frac{Q^2+ 1}{2Q} - \cos\theta\right)^{\frac{3}{2}}} \text{d}\theta$$

Defining $A = \frac{Q^2+ 1}{2Q}$, Wolfram Alpha gives... $$ \int_0^{2\pi}\frac{ 1 - Q \cos x}{(A -\cos x)^{3/2}} dx $$

$$=\left[ \frac{2}{(A^2-1)\sqrt{A - \cos x}} \left( (A^2-1)Q\sqrt{\frac{A - \cos x}{A-1}} \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-AQ\sin x- (A-1)(AQ-1)\sqrt{\frac{A-\cos x}{A-1}}\operatorname{E}\left(\frac{x}{2}~\big|~\frac{2}{1-A}\right) +\sin x \right) \right]_0^{2\pi}$$

Where $E(U|m)$ is an incomplete elliptic integral of the 2nd kind and $F(U|m)$ is an incomplete elliptic integral of the 1st kind. In both cases the parameter $U$ is the upper limit of the integration range and the parameter $m$ corresponds to the term $k^2$ in the elliptic integral. $$ $$ Now the analysis can be simplified if we use Complete Elliptic Integrals. This requires the first parameter ($U$) to have the value $\pi/2$. In this question that can be done by recognising the symmetry either side of the extended line passing through the target and the ring centre. This allows us to replace $$ \int_0^{2\pi}\frac{ 1 - Q \cos x}{(A -\cos x)^{3/2}} dx ~ \text{ by } ~ 2 \int_0^{\pi}\frac{ 1 - Q \cos x}{(A -\cos x)^{3/2}} dx$$ $$ $$ We then have... $$ F = \frac{-GM}{ D^2} \frac{2}{2\pi} \frac{1}{(2Q)^{3/2}}\int_0^{\pi}\frac{1-Q\cos\theta} {\left(A - \cos\theta\right)^{\frac{3}{2}}} \text{d}\theta $$ Next, using the integrand provided by Wolfram Alpha... $$ F =\frac{-GM}{ D^2} \frac{2}{2\pi} \frac{1}{(2Q)^{3/2}} \left[\frac{2}{(A^2-1)\sqrt{A - \cos \theta}}\left(\left(A^2-1\right)Q\sqrt{\frac{A - \cos \theta}{A-1}} \operatorname{F}\left(\frac{\theta}{2}~\big|~\frac{-2}{A-1}\right)-AQ\sin \theta- (A-1)(AQ-1)\sqrt{\frac{A-\cos \theta}{A-1}}\operatorname{E}\left(\frac{\theta}{2}~\big|~\frac{2}{1-A}\right) +\sin \theta \right) \right]_0^{\pi} $$ Let us simplify this expression, using the fact that $\sin(\pi) = \sin(0) = 0$, and cancelling the terms in $\cos\theta$ and some terms in $A$... $$ F =\frac{-GM}{ D^2} \frac{2}{\pi} \frac{1}{(2Q)^{3/2}} \frac{1}{\sqrt{A-1}} \left[ Q \operatorname{F}\left(\frac{\theta}{2}~\big|~\frac{-2}{A-1}\right) - \frac{(A-1)(AQ-1)}{(A-1)(A+1)}\operatorname{E}\left(\frac{\theta}{2}~\big|~\frac{2}{1-A}\right) \right]_0^{\pi} $$

For $\theta=0$ the incomplete EI functions return the value $0$. So within the big square brackets we only retain the terms in $\theta=\pi$. Then for $\theta=\pi$ we can replace the incomplete EI functions E and F by the complete EI functions E and K, giving us...

$$ F =\frac{-GM}{ D^2} \frac{2}{\pi} \frac{1}{(2Q)^{3/2}} \frac{1}{\sqrt{A-1}} \left[Q \operatorname{K}\left(\frac{2}{1-A}\right)+ \frac{(1-AQ)}{(A+1)}\operatorname{E}\left(\frac{2}{1-A}\right) \right] $$

Now substituting for $A$ where $A = \frac{Q^2+ 1}{2Q}$... $$ F =\frac{-GM}{ D^2} \frac{2}{\pi} \frac{1}{(2Q)^{3/2}} \frac{1}{\sqrt{\frac{Q^2+ 1}{2Q}-1}} \left[Q \operatorname{K}\left(\frac{2}{1-\frac{Q^2+ 1}{2Q}}\right)+ \frac{(1-Q*\frac{Q^2+ 1}{2Q})}{(\frac{Q^2+ 1}{2Q}+1)}\operatorname{E}\left(\frac{2}{1-\frac{Q^2+ 1}{2Q}}\right) \right] $$ Simplifying... $$ F =\frac{-GM}{ D^2} \frac{2}{\pi} \frac{1}{(2Q)^{3/2}} \frac{\sqrt{2Q}}{\sqrt{ Q^2+ 1 -2Q}} \left[Q \operatorname{K}\left(\frac{4Q}{ 2Q - Q^2 - 1 }\right)+ \frac{(1-\frac{Q^2+ 1}{2})(2Q)}{( Q^2+ 1+2Q)}\operatorname{E}\left(\frac{4Q}{ 2Q - Q^2 - 1 }\right) \right] $$ Simplifying again... $$ F =\frac{-GM}{ D^2} \frac{2Q}{\pi} \frac{1}{2Q} \frac{ 1 }{(1-Q)} \left[ \operatorname{K}\left(\frac{-4Q}{ 1-2Q + Q^2 }\right)+ \frac{(1-Q^2)}{(1+Q)(1+Q)}\operatorname{E}\left(\frac{-4Q}{ 1-2Q + Q^2 }\right) \right] $$ Simplifying yet again... $$ F =\frac{-GM}{ D^2} \frac{1}{\pi} \frac{ 1 }{(1-Q)} \left[ \operatorname{K}\left(\frac{-4Q}{ (1-Q)^2 }\right)+ \frac{(1+Q)(1-Q)}{(1+Q)(1+Q)}\operatorname{E}\left(\frac{-4Q}{ (1-Q)^2 }\right) \right] $$

And simplifying once more... $$ F =\frac{-GM}{ D^2} \frac{1}{\pi} \left[ \frac{ 1 }{(1-Q)}\operatorname{K}\left(\frac{-4Q}{ (1-Q)^2 }\right)+ \frac{1}{(1+Q)}\operatorname{E}\left(\frac{-4Q}{ (1-Q)^2 }\right) \right] $$ Note that it would be possible to apply Gauss's Transformation to the expression in $\operatorname{K}()$. But as the same cannot be done for the expression $\operatorname{E}()$ I shall leave the equation as is for harmonious appearance.

Finally we can replace $Q$ by $R/D$... $$ F =\frac{-GM}{ D^2} \frac{1}{\pi} \left[ \frac{1}{\left(1-\frac{R}{D}\right)}\operatorname{K}\left(\frac{-4R/D }{ \left(1 - \frac{R}{D}\right)^2 }\right)+ \frac{1}{(1+\frac{R}{D})}\operatorname{E}\left(\frac{-4R/D}{ \left(1 - \frac{R}{D}\right)^2 }\right) \right]. $$

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As a partial check, as $\frac{R}{D}\rightarrow 0$, so... $$ F \rightarrow\frac{-GM}{ D^2} \frac{1}{\pi} \left[ \operatorname{K}\left(0\right)+ \operatorname{E}\left(0\right) \right] ~ \rightarrow ~ \frac{-GM}{ D^2} \frac{1}{\pi} \left[ \frac{\pi}{2} + \frac{\pi}{2} \right] ~ \rightarrow ~ \frac{-GM}{ D^2}. $$

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Note how the final expression for $F$ is rather different to that obtained in the answer by /u/ComplexYetTrivial... $$ F = \frac{-GM}{D^2} \frac{2}{\pi} \frac{1}{\left(1 - \frac{R^2}{D^2}\right)} \operatorname{E}\left(\frac{R^2}{D^2}\right). $$

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In the answer by /u/USER/ the relevant equation (for $F_{\phi=\pi/2}$) can be converted (see Appendix 1) into the following form:-

$$ =\frac{-GM}{D^2} \frac{1}{\pi} \left[ \frac{1}{\left(1+\frac{R}{D}\right)} \operatorname{K}\left(\frac{4R/D}{\left(1+\frac{R}{D}\right)^2}\right) + \frac{ 1}{\left(1-\frac{R}{D}\right)} \operatorname{E}\left(\frac{4R/D}{\left(1+\frac{R}{D}\right)^2}\right) \right] $$

This converted /u/USER/ solution, compared to the solution here, is identical in the pattern of variables, but different in the details of $+$ and $-$ signs.

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Appendix 1 - Converting the relevant /u/USER/ solution

The converted /u/USER/ solution is obtained from the form presented by /u/USER/ as follows...

Using $x=\sqrt{D^2+R^2+2DR\sin\phi},\ y=\sqrt{D^2+R^2-2DR\sin\phi}$; for a target in the plane of the ring $\phi=\pi/2$, $\sin\phi = 1$ and so $x=\sqrt{D^2+R^2+2DR}$,and $y=\sqrt{D^2+R^2-2DR}$; and thus...

$$\left(1-\frac {y^2}{x^2}\right) = \left(\frac{x^2- y^2}{x^2}\right) = \left(\frac{(D^2+2DR+R^2)- (D^2-2DR+R^2)}{ D^2+2DR+R^2 }\right) = \left(\frac{4RD}{ D^2+2DR+R^2 }\right) = \left(\frac{4R}{D~\left(1+\frac{2R}{D}+\frac{R^2}{D^2}\right)}\right) = \left(\frac{4R/D}{\left(1+\frac{R}{D}\right)^2}\right) $$ 1 $$ \operatorname{F}=\frac{-GM}{\pi}\frac{(D^2-R^2) \operatorname{E}\left(1-\frac {y^2}{x^2}\right)+ y^2 \operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{Dxy^2} $$ 2 $$ =\frac{-GM}{\pi x} \left[ \frac{~D^2~\left(1-\frac{R^2}{D^2}\right) \operatorname{E}\left(1-\frac {y^2}{x^2}\right)}{D y^2} + \frac{\operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{D} \right] $$ 3 $$ =\frac{-GM}{\pi D \sqrt{1+\frac{2R}{D}+\frac{R^2}{D^2}}} \left[ \frac{D^2~\left(1-\frac{R^2}{D^2}\right) \operatorname{E}\left(1-\frac {y^2}{x^2}\right)} { D^3~(1-\frac{2R}{D}+\frac{R^2}{D^2}) } + \frac{\operatorname{K}\left(1-\frac {y^2}{x^2}\right) } { D } \right] $$ 4 $$ =\frac{-GM}{D^2} \frac{1}{\pi} \frac{1}{\left(1+\frac{R}{D}\right)} \left[ \frac{\left(1-\frac{R^2}{D^2}\right) \operatorname{E}\left(1-\frac {y^2}{x^2}\right)} {\left(1-\frac{R}{D}\right)^2} + \operatorname{K}\left(1-\frac {y^2}{x^2}\right) \right] $$ 5 $$ =\frac{-GM}{D^2} \frac{1}{\pi} \left[ \frac{\left(1-\frac{R^2}{D^2}\right) \operatorname{E}\left(1-\frac {y^2}{x^2}\right)} {\left(1+\frac{R}{D}\right) \left(1-\frac{R}{D}\right) \left(1-\frac{R}{D}\right)} + \frac{\operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{\left(1+\frac{R}{D}\right)} \right] $$ 6 $$ =\frac{-GM}{D^2} \frac{1}{\pi} \left[ \frac{ \operatorname{E}\left(1-\frac {y^2}{x^2}\right)} {\left(1-\frac{R}{D}\right)} + \frac{\operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{\left(1+\frac{R}{D}\right)} \right] $$ 7 $$ =\frac{-GM}{D^2} \frac{1}{\pi} \left[ \frac{ 1}{\left(1-\frac{R}{D}\right)} \operatorname{E}\left(\frac{4R/D}{\left(1+\frac{R}{D}\right)^2}\right) + \frac{1}{\left(1+\frac{R}{D}\right)} \operatorname{K}\left(\frac{4R/D}{\left(1+\frac{R}{D}\right)^2}\right) \right] $$ 8 $$ =\frac{-GM}{D^2} \frac{1}{\pi} \left[ \frac{1}{\left(1+\frac{R}{D}\right)} \operatorname{K}\left(\frac{4R/D}{\left(1+\frac{R}{D}\right)^2}\right) + \frac{ 1}{\left(1-\frac{R}{D}\right)} \operatorname{E}\left(\frac{4R/D}{\left(1+\frac{R}{D}\right)^2}\right) \right] $$

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Appendix 2 - The Use of Gauss' Transformation with Complete Elliptic Integrals of the First Kind.

In the answer given by /u/ComplexYetTrivial/ (to whom all credit is due) Gauss's Transformation is applied. Here are some workings in more detail.

Gauss's Transformation states:-

If: $$(1+x\sin^2\alpha)\sin\beta=(1+x)\sin\alpha \text{, }$$

Then:$$ (1+x)\int_0^{\alpha}\frac{\text{d}\phi}{\sqrt{1-x^2sin^2phi}}=\int_0^\beta\frac{d\phi}{\sqrt{1-\frac{4x}{(1+x)^2}\sin^2\phi}}. $$

Taking $\alpha=\beta=\frac{\pi}{2}$, we have

$$(1+Q\sin^2\frac{\pi}{2})\sin\frac{\pi}{2}=(1+Q)\sin\frac{\pi}{2}$$

which is true, so then: $$ (1+Q)\int_0^{\pi/2}\frac{\text{d}\phi}{\sqrt{1-Q^2sin^2\phi}}=\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-\frac{4Q}{(1+Q)^2}\sin^2\phi}}, $$

Rearranging... $$ \frac{1}{ (1+Q)}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-\frac{4Q}{(1+Q)^2}\sin^2\phi}} = \int_0^{\pi/2}\frac{\text{d}\phi}{\sqrt{1-Q^2sin^2\phi}} = \operatorname{K}(Q^2) . $$

where $\operatorname{K}$ is the complete elliptic integral of the first kind.