Gravity of a Ring on a Particle

[student34]

Homework Statement

Consider a ring-shaped body in a fixed position with mass M. A particle with mass m is placed at a distance x from the center of the ring and perpendicular to its plane. Calculate the gravitational potential energy U of the system (the picture has a small sphere traveling towards the center of a ring perpendicular to the ring's plane).

Homework Equations

Fg = (G*m*M)/r^2

Fg*r = U = (G*m*M)/r, where r is the radius, and G = 6.67*10^(-11) (gravitational constant)

The Attempt at a Solution

Let r be the hypotenuse of a right triangular distance to any part of the ring and a be the distance from the center of the ring to join the hypotenuse. x will make the 90° angle with a.

So, I thought that I would multiply the perpendicular component of force to the perpendicular distance x to get a function for gravitational potential energy,
F(perpendicular component) = (G*m*M)/(x^2) = (G*m*M)/(r^2 - a^2), where x^2 = r^2 - a^2.
Then, U = (G*m*M)/(r^2 - a^2)*(r^2 - a^2)^(1/2) = (G*m*M)/(r^2 - a^2)^(1/2).
But apparently this is wrong.

The right answer is U = (G*m*M)/(r^2 + a^2)^(1/2), the same as my answer except for the +.

I just don't understand why my answer is not right.

 

[paisiello2]

Provide picture please.

 

[Bashkir]

Are you positive that that is the correct answer?

Each section of the ring has some infinitesimal mass that is proportional to some infinitesimal length. Mathematically,

$$dM = \frac{M}{2πa}dl\\ dM = \frac{M}{2πa}adø$$

If this substitution is not obvious to you, all I did was say that the infinitesimal length, dl, is actually an arc length, that, with a little bit of trig, can be given as L = aø. where ø is in radians, and is the angle that is swept out by the ring, and a is the radius. This obviously is 2pi for the entire ring, but we will come back to that. We can plug this infinitesimal mass into our potential energy problem and should see that for each tiny bit of mass, we should get a tiny bit of potentially energy.

Can you see where you need to go from here? I don't know what your physical level is, or what mathematical background you come from.

 

[student34]

Are you positive that that is the correct answer?

Each section of the ring has some infinitesimal mass that is proportional to some infinitesimal length. Mathematically,

$$dM = \frac{M}{2πa}dl\\ dM = \frac{M}{2πa}adø$$

If this substitution is not obvious to you, all I did was say that the infinitesimal length, dl, is actually an arc length, that, with a little bit of trig, can be given as L = aø. where ø is in radians, and is the angle that is swept out by the ring, and a is the radius. This obviously is 2pi for the entire ring, but we will come back to that. We can plug this infinitesimal mass into our potential energy problem and should see that for each tiny bit of mass, we should get a tiny bit of potentially energy.

Can you see where you need to go from here? I don't know what your physical level is, or what mathematical background you come from.

Oh thank-you so much! :)

 

[BvU]

Dear 34,

You want to consider where your relevant equations apply. You confuse me (and likely yourself as well) by using the term radius to describe r.

Then in your solution you use x (which is smaller than the distance between m and M -- meaning you would get a larger magnitude for the force). Why ?

There is another issue: your expression goes to infinity when x -> 0 which should go against all intuition.

[edit]So where did you end up with your answer ? And: is your r the same r as the r in the answer ?

 

[student34]

Dear 34,

You want to consider where your relevant equations apply. You confuse me (and likely yourself as well) by using the term radius to describe r.

Then in your solution you use x (which is smaller than the distance between m and M -- meaning you would get a larger magnitude for the force). Why ?

There is another issue: your expression goes to infinity when x -> 0 which should go against all intuition.

[edit]So where did you end up with your answer ? And: is your r the same r as the r in the answer ?

Yeah, I was out of control with this one; I get it now though. Thanks anyways!