Potential Energy of a solid sphere and a thin disk

[kudoushinichi88]

A solid sphere with radius r is placed on top of a thin disk with radius R. The contact point is the center of the disk. Both objects are uniform and have the same mass M. Calculate the gravitational potential energy of the system. Take the potential energy to be zero when the sphere and the disk are infinitely far apart.

The gravitational potential equation is of course,

U=\frac{GMm}{r}

I know I have to do some integration here, but I am not sure how to cut up the two objects into small pieces. And I have a feeling that this might require a double integration...

Any hints on how to cut them up?

 

[Delphi51]

I think you can treat the sphere as a point mass because of its symmetry. The disk will need to be integrated over a series of rings.

 

[kudoushinichi88]

I tried the method you suggested and this is the result;

I rename the the radius of the sphere to r_0.

A ring on the circle has an area of

<br /> dA=\pi(r+dr)^2-\pi r^2
=2\pi r dr

the ratio of the mass of the ring to the total mass of the disk should be equal to the ratio of the area of the ring to the total area of the disk. This makes

<br /> \frac{dm}{M}=\frac{dA}{A}=\frac{2\pi r dr}{\pi R^2}

The distance S between the center of the sphere to a ring on the disk is

<br /> S=\sqrt{r^2+r_0^2}

Okay now we are ready to integrate...

<br /> dU=-\frac{GMdm}{S}

Subbing in everything, and integrating,

<br /> \int_{0}^{U}dU=-\int_{0}^{R}\frac{2GM^2}{R^2}\frac{r}{\sqrt{r^2+r_0^2}}dr

which results in

<br /> U=-\frac{GM^2}{R^2}\left(\sqrt{R^2-r_0^2}-r_0\right)

does this result make any sense?

 

[Delphi51]

In the very last step I get a + sign in the square root rather than the minus you have. There is a problem with the minus - when R gets smaller than ro, the radicand goes negative.

I can't think of a good way to test it for sensibility. As both R and ro get very small, U should get very large and the expression does that.

 

[kudoushinichi88]

oh right. that term is supposed to mean the distance between the center of the sphere and the edge of the disk. damn, i keep making these careless mistakes...

So it should be

<br /> U=-\frac{GM^2}{R^2}\left(\sqrt{R^2+r_0^2}-r_0\right)<br />

 

[Delphi51]

Right. I actually cheated and looked up the integral in my old table. But I have since worked it out with a r = ro*tan(θ) substitution followed by a y = cos(θ). Thank you for an interesting revisit with calculus!

 

[kudoushinichi88]

Wow, I didn't do any substitution at all. All I did was realising that by differentiating

(r^2-r_0^2)^{\frac{1}{2}},

you will get

\frac{r}{\sqrt{r^2+r_ 0^2}}

I found out that this is the faster than any substitution method.

Anyway, thank you for your insight to the problem!

Although actually I am still wondering how do we justify that we can take the sphere as a point mass...

 

[Delphi51]

Very smart! I think I would have noticed that 35 years ago.
(you have that minus sign again)

 

[kudoushinichi88]

woops. copy paste error. XP

 

[Delzac]

Just a question of my own. Why can't we just assume everything is point mass and work it out?

 

[oreodunker]

koudos are you from NUS- physics? Jamil?

 

[Delzac]

Please refrain from using this thread as a chat room. There is a private message function. (:

 

[kudoushinichi88]

Just a question of my own. Why can't we just assume everything is point mass and work it out?

Assuming them as point masses would only yield approximate answers. Especially when we are dealing with a disk, and not a spherical mass.

We could imagine that if R is big, then the sphere would be affected by the gravity exerted by the mass on the edges of the disk. Therefore, we can't possibly treat the disk as a point mass.

From the equation that I got, if r_0 is really really big, we arrive at the correct conclusion - the potential energy becomes zero. But If R is really really big, then the equation becomes GM^2/R. I'm not too sure about the validity of this equation though... I am not sure how to explain this physically.

 

[ehild]

Subbing in everything, and integrating,

<br /> \int_{0}^{U}dU=-\int_{0}^{R}\frac{2GM^2}{R^2}\frac{r}{\sqrt{r^2+r_0^2}}dr

which results in

<br /> U=-\frac{GM^2}{R^2}\left(\sqrt{R^2+r_0^2}-r_0\right)

You missed a factor of 2 from the integral , apart from the - sign. (∫x^-½dx = 2x^½).


To find the behaviour of U at very big r₀ and very big R, use the identity :

√(a+b)-√a=b/(√(a+b)+√a)

The asymptotic potential energy function should be the same as that of a point mass in the field of a disk or sphere.


ehild

 

[Delphi51]

You missed a factor of 2 from the integral , apart from the - sign. (∫x-½dx = 2x½).

Another 2 from differentiating r² cancels it out, I think.

 

[ehild]

No, it cancels out with the factor 2 in the integrand itself.

ehild