A gravitation problem with two masses

[toothpaste666]

Homework Statement

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.]

Homework Equations

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

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?

 

[SammyS]

Homework Statement

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.]

Homework Equations

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

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?

Did you intend for M_C 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 M_F = M + M_C .


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

 

[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

 

[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?

 

[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:

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

 

[SammyS]

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:

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

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 M_c in terms of M.

 

[toothpaste666]

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

 

[SammyS]

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

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?

 

[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?

 

[SammyS]

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?

Mass is be proportional to volume in this case.

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

 

[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?

 

[haruspex]

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?

That's it.

 

[toothpaste666]

but how do i relate that back to mass?

 

[haruspex]

but how do i relate that back to mass?

Through density, which is the same for both.

 

[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?

 

[haruspex]

how do we know the density is equal?

See the OP:

A uniform sphere ...

 

[toothpaste666]

Thank you.