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I'm just struggling a little with this question:

A uniform sphere, of radius $R$, contains a spherical cavity of radius $R/4$, whose centre is $3R/8$ from the surface. The diameter passing through the centres of the sphere and cavity meets the surface at points $X$ and $Y$. Find the ratio of the gravitational field at $X$ and $Y$.

My attempt at the solution goes something like this:

Using the superposition principle, the gravitation field due to the whole mass is equal to the sum of the gravitational fields due to the remaining mass and the removed mass.

The gravitational field due to a uniform solid sphere is zero at its centre. Therefore, the gravitational field due to the removed mass is zero at its centre.

The gravitational field due to the solid sphere is equal to the gravitational field due to the remaining mass. Now we know g acts towards the centre of the sphere. As such, both the gravitational field of the combination of the sphere and removed mass and the gravitational field of the sphere only act in the same direction, so we can use the scalar form of the equation.

Therefore the gravitational field is given by $g=GMr/R^2$.

Then insert $r=-R$ and $R$ for the gravitational field at $X$ and $Y$.

But this doesn't seem to be correct as it is just the same as if the removed mass wasn't there..... have I gone wrong in my logic somewhere?

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2 Answers

I think you are assuming the cavity is centered on the body which the question says it isn't.

You are almost correct in your approach. Imagine the main body was filled with mass and the cavity was a separate body with negative mass - then sum the result

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As you correctly write, the gravitational field of the sphere with a cavity is the difference between the gravitational field of the mass $M$ full sphere and the gravitational field of the mass $M/4^3=M/64$ little sphere located at the right places (different centers, as you said but you neglected this subtlety). I am choosing $M$ so that the mass of the actual hollow sphere is $M-M/64 = 63M/64$.

Both terms may be expressed simply by $GM/a$ where $a$ is the distance from the center because the distance from the center is never smaller than the radius of the respective sphere (either the solid one or the "negative one").

You wrote that the center of the cavity is $3R/8$ from the nearest point of the surface, $X$, which means it's $5R/8$ from the opposite point on the surface, $Y$. So the gravitational acceleration in $X$ and $Y$, respectively, is: $$ a_x = \frac{64 GM}{ 64 R^2} - \frac{1 GM\cdot 8^2}{64 R^2 \cdot 3^2},\quad a_y = \frac{64 GM}{ 64 R^2} - \frac{1 GM\cdot 8^2}{64 R^2 \cdot 5^2},\quad $$ So we have $$\frac{a_x}{a_y} = \frac{64-(8/3)^2}{64-(8/5)^2} = \frac{25}{27} \sim 0.925 925 925\dots $$ The gravitational field at the point closer to the cavity is weaker, of course.

Sorry if there is a numerical error somewhere; there's a big chance of it.

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Thanks for the reply! Why are the 3/8 and 5/8 terms squared in your expressions for a_x and a_y? –  Fky78 Apr 30 '12 at 18:49
    
Dear Fky78, thanks for your interest. They're squared because the gravitational force obeys the inverse square law. Those factors in the terms come from $1/{\rm distance}^2 = 8^2 / R^2\cdot 3^2$ or $3$ replaced with $5$, respectively because ${\rm distance}=3R/8$ or with $5$. –  Luboš Motl Apr 30 '12 at 18:51
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Thank you! It all makes sense now! –  Fky78 Apr 30 '12 at 18:52
    
Just to say that Lubos's answer is mostly correct but the radius used in the second one should be 13/8 R rather than 5/8. This will then give a value of 169/189 for the relative strength at X to Y. –  user23981 May 4 '13 at 14:50
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