# A gravitation problem with two masses

1. Apr 11, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

A uniform sphere has mass M and radius r. A spherical cavity (no mass) of radius r/2 is then carved within this sphere (the cavity's surface passes through the sphere's center and just touches the sphere's outer surface). The centers of the original sphere and the cavity lie on a straight line, which defines the x axis.

With what gravitational force will the hollowed-out sphere attract a point mass m which lies on the x axis a distance d from the sphere's center? [Hint: Subtract the effect of the "small" sphere (the cavity) from that of the larger entire sphere.]

2. Relevant equations

$F = G\frac{m_1m_2}{r^2}$

3. The attempt at a solution

Let mass $M_f$ = the mass of the sphere without the cavity, $M_c$ = the mass of the cavity, and $R_f$ = the distance between the two masses.

$M_f = M - M_c$
$R_f = d + R_2$
$F = G\frac{(M_f) m}{(R_f)^2}$

is this right?

2. Apr 11, 2014

### SammyS

Staff Emeritus
Did you intend for MC to be negative?

I interpret the Hint to be to calculate the gravitational force that would be exerted on the point mass if the sphere were not hollow, but had the same density. From that, subtract the gravitational force that would be exerted on the point mass by an isolated uniform sphere the size of the cavity, also having the same density.

That would give MF = M + MC .

When you say, "Rf = the distance between the two masses", what two masses are you referring to ?

3. Apr 12, 2014

### toothpaste666

Here is the picture. The two masses are the small particle mass on the right and the sphere on the left with the cavity

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4. Apr 12, 2014

### haruspex

The hint does not say subtract the masses, it says subtract the gravitational effects.
What force would be exerted by the complete sphere?
Suppose we take the "complement" of the cavitated sphere, i.e. throw away the larger sphere and instead have just the smaller sphere (at the position of the cavity). What force would that exert?

5. Apr 13, 2014

### toothpaste666

so if M is the mass of the sphere before the cavity was taken out and Mc is the mass of the cavity then we have:

$G\frac{Mm}{d^2} - G\frac{M_c m}{(d-r_2)^2}$

6. Apr 13, 2014

### SammyS

Staff Emeritus
Yes, if M is the mass of the sphere before the cavity is taken out. That is as the problem is stated. (So, I believe I was incorrect in my previous post.)

You can determine Mc in terms of M.

7. Apr 13, 2014

### toothpaste666

If the mass is evenly distributed wouldn't subtracting the volume be the same as subtracting the mass?

8. Apr 13, 2014

### SammyS

Staff Emeritus
What are you subtracting from what?

What is the mass of a sphere the size of the cavity? How is that related to the mass of the large sphere before the cavity is formed in the sphere?

9. Apr 13, 2014

### toothpaste666

Subtracting the volume of the cavity from the volume of the original sphere i mean. Wouldn't the mass be the same as the volume if its evenly distributed?

10. Apr 13, 2014

### SammyS

Staff Emeritus
Mass is be proportional to volume in this case.

What fraction of the initial solid sphere's mass is removed to make the cavity ?

11. Apr 13, 2014

### toothpaste666

To find the volumes let Vc = Volume of cavity:

$r_2 = \frac{r}{2}$

$V_c = \frac{4pi(\frac{r}{2})^3}{3}$

$V_c = \frac{\frac{4pi(r^3)}{8}}{3}$

$V_c = \frac{\frac{pi(r^3)}{2}}{3}$

$V_c = \frac{pi(r^3)}{6}$

since $V_M =\frac{4pi(r^3)}{3}$

we have

$V_M = 8V_c$

does this get me closer to expressing Mc in terms of M?

12. Apr 14, 2014

### haruspex

That's it.

13. Apr 14, 2014

### toothpaste666

but how do i relate that back to mass?

14. Apr 14, 2014

### haruspex

Through density, which is the same for both.

15. Apr 14, 2014

### toothpaste666

$D = \frac{M}{V_M} = \frac{3M}{4(pi)r^3}$

$M_c = D V_c = D \frac{(pi)r^3}{6} = (\frac{3M}{4(pi)r^3})(\frac{(pi)r^3}{6}) = \frac{M}{8}$

which gives

$G\frac{Mm}{d^2} - G\frac{Mm}{8(d-r_2)^2}$

how do we know the density is equal?

16. Apr 14, 2014

### haruspex

See the OP:

17. Apr 14, 2014

Thank you.