Pull Your Fair Share

Relevant wiki: Gravitation

If the density of the sphere is $\rho$ , using polar coordinates we can write the attractive force from the blue hemisphere as $Gm\rho R \int_0^1 \, dr \int_0^\pi \,d\theta \int_{-\frac12\pi}^{\frac12\pi}\, d\phi \frac{r^2\sin\theta (2 - r\sin\theta\cos\phi)}{(4 - 4r\sin\theta\cos\phi + r^2)^{\frac32}}$ while the total attractive force from the whole sphere is $Gm\rho \times \tfrac43\pi R^3 (2R)^{-2} \; = \; \tfrac13\pi Gm\rho R$ which is of course the same as we obtain from the above integral, replacing the $\phi$ limits by $-\pi$ and $\pi$ .

Evaluating this multiple integral numerically, we see that the blue sphere is responsible for $\boxed{67.376}$ % of the total gravitational attraction.

This solution will use methods similar to those used to derive the shell theorems. G and m will not be omitted, and the overall sign will also be omitted since they don't affect the answer (can adjust at the end). First, the potential for the blue region will be dealt with first. Since potential is not a vector, it will be easier to deal with since the potentials can all be added together, whereas integrating the force would require dealing with components. First we find the potential a disk of density $\sigma$ $\int_{0}^{R'}\ \frac{2\pi \sigma ydy}{\sqrt{r^2+y^2}}$ The general problem is 3-d dimensional, so later it will turn out that $\sigma$ = $\rho$ dx where x is the distance of the center of the disk away from the point mass. Evaluating the integral yields $2\pi\sigma(\sqrt{R'^2+r^2}-r)$ Now the potential for a thin disk of the sphere yields $dV=2\pi (\sqrt{R^2\sin^2(\theta)+x^2}-x)\rho dx$ Drawing a right triangle and knowing that cos( $\theta$ =(r-x)/R, sin( $\theta$ ) can be calculated. Also I'm going to get rid of $\rho$ to be lazy. Plugging in the value for sin( $\theta$ ): $dV=2\pi (\sqrt{(R^2-r^2-x^2+2rx)+x^2}-x) dx = 2\pi (\sqrt{2rx+R^2-r^2}-x)dx$ Now integrating it from r-R to r: (You can verify that integrating from r-R to r+R yields 4 $\pi\rho$ R^3/(3r) which is M/r, and U=-GMm/r.) We will be evaluating things about the blue half of the sphere for most of the problem. $V=\int_{r-R}^{r}\ 2\pi (\sqrt{2rx+R^2-r^2}-x)dx$ Using u-sub for u=2rx+r^2-r^2, du=2rdx for the first part, and some more work, the integral becomes $\frac{2\pi}{3r}(2rx+R^2-r^2)^{\frac32}-\pi x^2|_{r-R}^{r}$ $\frac{2\pi}{3r}(r^2+R^2)^{\frac32}-\pi r^2-\frac{2\pi}{3r}(2r^2-2Rr+R^2-r^2)^{\frac32}+\pi(r-R)^2$ Note how it's r-R instead of R-r, since raising something squared then taking the square root makes it positive. We know r>R, so we leave it as r-R. $\frac{2\pi}{3r}((r^2+R^2)^{\frac32}-(r-R)^3)-\pi(2Rr-R^2)$ Now we need to find the force. By definition of potential energy, we can find it by differentiating the above by r. There may be errors with the overall sign, but again that will be adjusted at the end. (F=-dV/dr, not +dV/dr. The order of integration above may also affect the overall sign, but whatever). Pretty messy quotient rule: $\frac{2\pi}{3}\frac{\frac32(r^2+R^2)^{1/2}\cdot2r^2-(r^2+R^2)^{\frac32}-3(r-R)^2r+(r-R)^3}{r^2}-2\pi R$ After a bit of simplifying we end up with a not particularly nice equation. $2\pi\frac{r^2\sqrt{r^2+R^2}-\frac13(r^2+R^2)^{\frac32}-(r-R)^2r+\frac13(r-R)^3-Rr^2}{r^2}$ Finally we plug in the situation given in the problem r=2R. After some simplifying, it ends up as $2\pi R^3(\sqrt{5}-\frac{5\sqrt{5}}{12}-\frac{1}{2}+\frac{1}{12}-1)$ This, multiplied by density and G will be the gravitational acceleration at that point particle. For the entire sphere, the force (excluding density, G, and m) $\frac43\pi\frac{R^3}{r^2}=\frac{\pi R}{3}$ Finally finding the ratio $\frac{2\pi R(\sqrt{5}-\frac{5\sqrt{5}}{12}-\frac12+\frac{1}{12}-1)}{\frac{\pi R}{3}}$ And getting (sign got flipped) $-6(\sqrt{5}-\frac{3}{2}-\frac{5\sqrt{5}-1}{12})=\frac{17-7\sqrt{5}}{2} \approx$ $\boxed{67.3762}$ %

Relevant wiki: Gravitation

Here is an analysis which gives an *approximate * answer to the problem.

Since we are looking for a ratio we can cut ourselves some slack, take $R=1$ and ignore $\pi$ , G, $\rho$ and other numerical factors!

Measure x downwards from the red point, towards the centre of the sphere. Slice the sphere up like an onion, with the layers perpendicular to the x-axis. Now we have a stack of disks with their centres of gravity on the x axis, their distance from the red point being $x$ , and their area being proportional to the square of their radius. By an easy application of Pythagoras this is $1^2 - (2-x)^2=4x-3-x^2$ . Now we imagine that all the mass of each disk of thickness $dx$ is concentrated at its centre (see note).

Using the inverse square law for gravity, we can sum the effects of the disks by integration to find our ratio -

$\frac{\int_{1}^{2}\frac{4x-3-x^2}{x^2}dx}{\int_{1}^{3}\frac{4x-3-x^2}{x^2}dx}=\frac{\log{16}-\frac{5}{2}}{\log{81}-4}=0.6911= \boxed{69.11\%}$

note.

The gravitational field of the disk and an equally massive point at its centre of gravity are only approximately the same (see Mark Henning's comment below). Treating them as equal is what turns this solution from an exact analysis into an approximation. The further the red point is from the sphere, the better this approximation becomes. I'm surprised to be within 5% of the correct answer when the point is only $2R$ from the centre of the sphere!

Well this is an approximate method just like that of Peter Macgregor's, the correct solution is already provided by Mark Hennings.
Let's cut the sphere in two parts and let the whole mass of each sphere (i.e. M/2) be concentrated at its centre of mass i.e. 3R/8 from base, now you have three point size particles, and gravitational field could be easily calculated and the ratio of required quantity comes out to be 68.11% (that's damn close) Caution-this was just a quick objective approach, in practise we can never change the mass distribution in space to calculate field, it alters the field at every point.
:)

I did a numerical simulation in a spreadsheet (one of my favorite methods).

I created a semicircle in a 10 by 20 array, labeled .05 to .95 along the short X axis and -.95 to .95 along the longer Y axis (steps of .10 in both directions). Within the semicircle, each cell represents an annulus of mass within the sphere. The gravitational effect of each annulus is proportional to the mass, which is proportional to the radius of the annulus, thus the X label of the cell.it is inversely proportional to the distance from the cell's (X,Y) label to the point (0,2). This is then adjusted by the cosine of the angle between the center line and the line between (X,Y) and (0,2), to eliminate the outward direction of the gravity.

Summing over the 2 halves of the semicircle yields 5.480902 for blue and 11.340153 for green, with a ratio of 67.4164%.

Using Cartesian coordinates, otherwise similar to Mark Hennings' solution.

$\text{Assuming}\left[-1\leq x\leq 1\land -1\leq y\leq 1\land -1\leq z\leq 1,\int_0^1 \left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(\int_{-\sqrt{-x^2-y^2+1}}^{\sqrt{-x^2-y^2+1}} \frac{2-x}{\left((2-x)^2+y^2+z^2\right)^{3/2}} \, dz\right) \, dy\right) \, dx\right] \Rightarrow -\frac{1}{6} \left(7 \sqrt{5}-17\right) \pi$

$\text{Assuming}\left[-1\leq x\leq 1\land -1\leq y\leq 1\land -1\leq z\leq 1,\int_{-1}^1 \left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(\int_{-\sqrt{-x^2-y^2+1}}^{\sqrt{-x^2-y^2+1}} \frac{2-x}{\left((2-x)^2+y^2+z^2\right)^{3/2}} \, dz\right) \, dy\right) \, dx\right] \Rightarrow \frac{\pi }{3}$

$-\frac{\left(7 \sqrt{5}-17\right) \pi }{\frac{6 \pi }{3}} \Rightarrow \frac{1}{2} \left(17-7 \sqrt{5}\right) \approx 0.673762078750736$