Find the force of attraction for a particle outside sphere

1. Oct 25, 2016

Vitani11

1. The problem statement, all variables and given/known data
A sphere of radius R contains two spherical cavities. Each cavity has a radius of R/2 and touches both the outside surface of the sphere and its center as shown. The mass of a similar sphere without the cavities is M. Find the force of attraction on a small particle of mass m located on the x - axis a distance d > R from the center of the sphere.

2. Relevant equations
g=GMm/R2

3. The attempt at a solution

2. Oct 25, 2016

Vitani11

3. Oct 25, 2016

Chan Pok Fung

I am sorry that I don't quite understand your work. You defined g = GMm/R^2, g is a quantity with unit of force, but in your work, g seems to be acceleration.

Here I have an idea:
$$\frac{GMm}{d^2} - 2\frac{GMm}{8r^2}cosθ = ma =F$$
The first term is the force by a fully filled sphere and the second term is the force by two small spheres. As the radius is halved, mass is 1/8. r is the distance between the small sphere and the particle.
After that, you can write
$$r^2 = d^2 + (\frac R 2)^2$$
$$cosθ = \frac {d}{\sqrt{(\frac R 2)^2+d^2}}$$

You have enough information for the solution. I hope it helps

4. Oct 25, 2016

Vitani11

Okay. I see how the mass is 1/8 when the radius becomes 1/2 by writing out the equation for density. But why can you just put an 8 in the denominator? I know it makes sense conceptually, but I mean mathematically (without introducing density into the equation)

5. Oct 25, 2016

Vitani11

PS. the drawing should have the r vectors pointing towards the center of each small sphere from d

6. Oct 25, 2016

Chan Pok Fung

Um.. I mean
$$\frac{Gm\frac M 8 }{r^2}$$
You can directly know it is 1/8 by the argument learnt in math class. For similar objects ,
$$\frac {V_1}{V_2} = (\frac {l_1}{l_2})^3$$

7. Oct 25, 2016

Vitani11

Okay then. Thank you, that helped

8. Oct 25, 2016

No problem