Find the force of attraction for a particle outside sphere

[Vitani11]

Homework Statement

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.

Homework Equations

g=GMm/R²

The Attempt at a Solution

Will upload picture and solution.

 

[Vitani11]

 

[Mayan 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

 

[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)

 

[Vitani11]

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

 

[Mayan Fung]

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)

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

 

[Vitani11]

Okay then. Thank you, that helped

 

[Mayan Fung]

Okay then. Thank you, that helped

No problem