Gravitational force between a particle and sphere

[UrbanXrisis]

the question is http://home.earthlink.net/~urban-xrisis/phy002.jpg

I thought that...
$$F=G\frac{m_1m}{R_1^2}-G\frac{m_2m}{R_2^2}$$
a.
$$F=G\frac{m_1m}{a^2}-G\frac{m_2m}{(R_2-a)^2}$$
b.
$$F=G\frac{m_1m}{b^2}-G\frac{m_2m}{(R_2-b)^2}$$
c.
$$F=G\frac{m_1m}{c^2}+G\frac{m_2m}{c^2}$$

is this correct?

 

[whozum]

The only correct one is c.

For a and b, note that the mass is inside the shell, consider the acceleration from the mass ofthe shell closest to it in comparison to the mass farthest of it. Is there a net effect?

 

[UrbanXrisis]

when the mass is in the middle of the shell, it has a net force of zero, but I don't know how to calculate the force when it is not in the center of the shell

 

[whozum]

It's very similar to the reasoning behind the gravitational effects of a mass inside a solid sphere. Tell me what the gravitational effect for a mass inside a sphere is, and think of the reasoning why it is so. Take a guess as to how this applies for a shell.

 

[UrbanXrisis]

for gravitational effect inside a shere...

$$F=G\frac{mMr}{R^3}$$

where r<R

I'm not sure how this relates to a shell

 

[whozum]

Ugh, I had this whole thing typed out and something went wrong and lost it.

Anyway, for a particle (at radius $r$) inside a solid sphere of radius $R$, the net force of gravitaiton on the particle results only from the mass contained inside the sphere with radius $r$.

 

[UrbanXrisis]

what about the gravity inside a shell? it's not the same as a sphere is it?

 

[whozum]

Same principle applies, except there's no mass in the smaller sphere

 

[UrbanXrisis]

$$F=G\frac{m_1mr}{R_1^3}-G\frac{m_2mr}{R_2^3}$$
a.
$$F=G\frac{m_1ma}{R_1^3}-G\frac{m_2ma}{R_2^3}$$
b.
$$F=G\frac{m_1m}{(R_1+b)^2}-G\frac{m_2mb}{R_2^3}$$

 

[whozum]

Reconsider what I said about a particle inside a hollow shell

 

[whozum]

And for B, check your radius, R_1 + b doesn't make sense.

 

[UrbanXrisis]

are you saying it should be:
a.
$$F=G\frac{m_1ma}{R_1^3}$$
b.
$$F=G\frac{m_1m}{b^2}$$

 

[whozum]

Yeah, do you understand why?

 

[UrbanXrisis]

but the formula: $$F=G\frac{mMr}{R^3}$$

it doesn't include the mass of the inner sphere...so it can't equal zero

m- is the point mass
M- mass of sphere/shell
R- radius of sphere/shell
r- little radius

also, what about the force pulling it towards the shell? as in since it's in the shell, would the shell exert a force towards the shell's surface? hence the negative force? (that's why I subtracted)

 

[whozum]

The force pulling it towards the shell cancels in the horizontal direction due to symmetry, in the vertical direction, the forces cancel since there's a certain amount of mass on top of the particle at a small radius away, but there's also a lot more mass at a farther radius. These two end up cancelling also, so everything cancels.

That formula is for a uniform sphere. The shell isn't a uniform sphere.

It is also analogous to

$$F = G\frac{mM_{contained}}{R^2}$$

 

[UrbanXrisis]

$M_{contained}$ being the sphere that is below the point mass? which in my case would = 0?

 

[whozum]

Yes, with respect to the hollow shell, it contains no mass inside the sphere with radius b, therefore exerts no net force.

 

[Werg22]

Mass is proportional to radius. So M2/radius 2=Mcountained/b. So the force that the big mass exerts is

if M=Particle mass
M2=Big sphere mass

M2*b*M*/radius^3

That is why you get this equation.

 

[Werg22]

The big sphere density is:

$$\frac{3M_{2}} {4 r^3 {\pi}}$$

and since the volume of the sphere with radius b is

$$\frac{4 b^3 {\pi}} {3}$$

So the mass is

$$M_{countained} = \frac{M_{2}b^3} {r^3}$$

since

$$a = \frac{M_{countained}G} {b^2}$$

$$a = \frac{G M_{2} b^3} {r^3 b^2}$$

Simplifying

$$a= \frac{G M_{2} b} {r^3}$$