0 Members and 1 Guest are viewing this topic.

I was able to calculate the force between a thin ring and a particle somewhere on the line perpendicular to the plane of the ring as G*r*m1*m2/(r^2+d^2)^1.5 where r is the radius of the ring and d is the distance from the particle to the center of the ring.I'm having trouble extending this to a thin circle, a disc, oriented the same way. Can anyone help me out?Cheers :-)

Thanks for the reply! Would you mind explaining the terms used in your equations? I'm having a hard time trying to figure it out still.

So I need to integrate G*r*m1*m2/(r^2+d^2)^1.5 in terms of r, from 0 to the radius of the circle. But to do that I need to represent the ring mass, m2, as determined by r?

So if M is the disk mass, and R is the disc radius...m2/M=(pi(r+dr)^2-pir^2)/piR^2=(2rdr+dr*dr)/R^2So m2=M(2rdr+dr*dr)/R^2Can I disregard the dr*dr?So now I have int G*r*m1*M*2rdr/(R^2*(r^2+d^2)^1.5) from 0 to RWhich works out to 2G*M*m1*d/R^2 * (1/d-1/(R^2+d^2)^.5)Is this correct? My biggest concern is what to do with that dr*dr

I was able to calculate the force between a thin ring and a particle somewhere on the line perpendicular to the plane of the ring as G*r*m1*m2/(r^2+d^2)^1.5 where r is the radius of the ring and d is the distance from the particle to the center of the ring.

Quote from: David LaPierre on 04/03/2013 20:15:18I was able to calculate the force between a thin ring and a particle somewhere on the line perpendicular to the plane of the ring as G*r*m1*m2/(r^2+d^2)^1.5 where r is the radius of the ring and d is the distance from the particle to the center of the ring.Are you sure it's not, instead, G*d*m1*m2/(r^2+d^2)^1.5 ?

Ah! I love the smell of physics in the morning. Smells like ... victory!

Quote from: Pmb on 05/03/2013 22:53:08Ah! I love the smell of physics in the morning. Smells like ... victory! Probably it's because I have flu, but I can't grasp what you mean []

Is it correct to say that the ratio of the ring mass to the disc mass is equal to the ratio of the ring's surface area to the disc's surface area? (surface area only in regard to the face nearest the particle)

So, dm/M = (pi*(r+dr)^2 - pi*r^2) / (pi*R^2)Isolating dm and canceling pi givesdm = M * ((r+dr)^2 - r^2) / R^2Ordm = M * (2rdr + (dr)^2) / R^2If that is all good, what do I do with the squared dr? Those sorts of terms never cease to confuse me..*edit*It think it can just be disregarded. Conceptually, it seems as if integrating a function with a term like dr is almost like adding an infinite number of infinitely small objects. Infinitely small because one of their dimensions gets indefinitely close to 0. Whereas with the squared dr term, it falls short by 2 'dimensions,' and is therefore practically meaningless.That explanation will probably make a lot of people cringe haha. I'd love a more precise explanation if the concept is at all accurate.

That's actually what I was thinking about. I wanted to see how GMm/r^2 compared to whatever resulted from this :-)

This can be integrated to get you (assuming I didn't drop any factors)

Yes, you're absolutely right. I neglected the test mass.