Time period of oscillation and gravitation

[Saitama]

If $U_0$ is the potential energy at the equilibrium, and $K_0$ is the kinetic energy at the same position, what does $K_0 = 0$ imply?

If I compare it with a spring block system performing SHM, the kinetic energy at equilibrium is maximum. But then what do I substitute for $K_0$?

 

[voko]

If I compare it with a spring block system performing SHM, the kinetic energy at equilibrium is maximum. But then what do I substitute for $K_0$?

Do you have to?

Motion always depends on some initial conditions, so $K_0$ can be regarded as one of them.

Alternatively, you can differentiate the energy equation, which will eliminate both $U_0$ and $K_0$ (but when it is solved, you will have to specify initial conditions anyway).

 

[Saitama]

Hi voko! I am thinking of writing down the energy of system at any instant and set the derivative with respect to time equal to zero to obtain a relation between $\ddot{\theta}$ and $\theta$.

I rechecked my algebra and there was a mistake in my previous expression for $U$. This time I get,
$$U=U_0-GM_em\left(\frac{a}{(R+a)^2}-\frac{a}{(R-a)^2}+\frac{4b^2}{R^3}\right)\frac{\theta^2}{2}$$
Using the approximation, $1/(R-a)^2=(1/R^2)(1+2a/R)$ and $1/(R+a)^2=(1/R^2)(1-2a/R)$,
$$U=U_0-\frac{GM_em}{2}\left(\frac{4b^2}{R^3}-\frac{4a^2}{R^3}\right)\frac{\theta^2}{2}$$
$$\Rightarrow U=U_0-\frac{2GM_em(b^2-a^2)}{R^3}\theta^2$$
The kinetic energy K is
$$K=ma^2\dot{\theta}^2+mb^2\dot{\theta}^2=m\dot{\theta}^2(a^2+b^2)$$
The total energy E at any instant is $E=K+U$. Substituting K and U and differentiating wrt to time,
$$\frac{dE}{dt}=-\frac{2GM_em(b^2-a^2)}{R^3}(2\theta\dot{\theta})+m(a^2+b^2)(2\dot{\theta}\ddot{\theta})=0$$
Simplifying,
$$\ddot{\theta}=-\frac{2GM_e(a^2-b^2)}{(a^2+b^2)R^3}\theta=-\frac{2GM_e(\eta^2-1)}{(\eta^2+1)R^3}\theta$$
where $\eta=a/b$.
Hence, the time period of small oscillations is
$$T=2\pi\sqrt{\frac{R^3(\eta^2+1)}{2GM_e(\eta^2-1)}}$$
According to the question, this is equal to orbital period which is $2\pi\sqrt{R^3/(GM_e)}$. Equating both the expressions,
$$\frac{\eta^2+1}{2(\eta^2-1)}=1$$
Solving for $\eta$, $\eta=\sqrt{3}$ but this is wrong. :(

Please help me point out the error in my above working. Thanks!

 

[voko]

I cannot spot any error in #33 (except the second equation has one extra 2 in the denominator, but that is corrected in subsequent equations).

Perhaps we need to review our assumptions about the model.

 

[haruspex]

$$U=U_0-\frac{GM_em}{2}\left(\frac{4b^2}{R^3}-\frac{4a^2}{R^3}\right)\frac{\theta^2}{2}$$
$$\Rightarrow U=U_0-\frac{2GM_em(b^2-a^2)}{R^3}\theta^2$$

I think you didn't cancel the 2's quite correctly.

 

[Saitama]

I think you didn't cancel the 2's quite correctly.

Yes, sorry about that but the next equation is correct. I accidentally put another 2 there.

 

[TSny]

You will need to be more accurate in the initial expressions for the distances from the center of the Earth to the masses $m$ as functions of $\theta$. You have to be careful that you are accurate to $\theta ^2$ in $U$. The law of cosines might be helpful. See the figure for the bottom mass $m$.

 

[Saitama]

You will need to be more accurate in the initial expressions for the distances from the center of the Earth to the masses $m$ as functions of $\theta$. You have to be careful that you are accurate to $\theta ^2$ in $U$. The law of cosines might be helpful. See the figure for the bottom mass $m$.

Great, thanks a lot TSny!

I thought that calculating the second derivatives would be a mess because it would involve surds in the denominator but really, it wasn't because the approximations worked fine and there was no need to calculate the derivatives.
Here's what I did:

Number the particles starting from the bottom one and going clockwise. From cosine rule,
$$r_1^2=a^2+R^2-2aR\cos\theta \Rightarrow \frac{1}{r_1}=\cfrac{1}{R\sqrt{1+\cfrac{a^2}{R^2}-\frac{2a}{R}\cos\theta}}$$
Using the approximation,
$$\frac{1}{1-2x\cos\theta+x^2} \approx 1+x\cos\theta+\frac{3\cos(2\theta)+1}{4}x^2$$
$1/r_1$ can be simplified to
$$\frac{1}{r_1}=\frac{1}{R}\left(1+\frac{a}{R} \cos \theta+\frac{3\cos(2\theta)+1}{4}\frac{a^2}{R^2}\right)$$
Similarly,
$$\frac{1}{r_2}=\frac{1}{R}\left(1+\frac{b}{R}\sin\theta+\frac{1-3\cos(2\theta)}{4}\frac{b^2}{R^2}\right)$$
$$\frac{1}{r_3}=\frac{1}{R}\left(1-\frac{a}{R}\cos\theta+\frac{3\cos(2\theta)+1}{4}\frac{a^2}{R^2}\right)$$
$$\frac{1}{r_4}=\frac{1}{R}\left(1-\frac{b}{R}\sin\theta+\frac{1-3\cos(2\theta)}{4}\frac{b^2}{R^2}\right)$$
Potential energy U is:
$$U=-\frac{GM_em}{r_1}-\frac{GM_em}{r_2}-\frac{GM_em}{r_3}-\frac{GM_em}{r_4}+C$$
where C is a constant comprising of potential energy of central mass wrt Earth and of point masses wrt each other and the central mass.
Substituting the expressions,
$$U=-\frac{GM_em}{R}\left(4+\frac{3\cos(2\theta)+1}{2}\frac{a^2}{R^2}+\frac{1-3\cos(2\theta)}{2}\frac{b^2}{R^2}\right)+C$$
$$\Rightarrow U=-\frac{GM_em}{R}\left(4+\frac{a^2}{2R^2}+\frac{b^2}{2R^2}+\frac{3\cos(2\theta)}{2R^2}(a^2-b^2)\right)+C$$
Since $\theta$ is small, $\cos2\theta \approx 1-2\theta^2$.
$$\Rightarrow U=-\frac{GM_em}{R}\left(k'-\frac{3}{R^2}(a^2-b^2)\theta^2\right)+C$$
where k' replaces the constant terms inside the parentheses.
Kinetic energy K is
$$K=m(a^2+b^2)\dot{\theta}^2$$
Hence,
$$E=-\frac{GM_em}{R}\left(k'-\frac{3}{R^2}(a^2-b^2)\theta^2\right)+C+m(a^2+b^2)\dot{\theta}^2$$
$$\frac{dE}{dt}=\frac{3GM_em}{R^3}(a^2-b^2)(2\theta\dot{\theta})+m(a^2+b^2)(2\dot{\theta}\ddot{\theta})=0$$
Simplifying,
$$\ddot{\theta}=-\frac{3GM_em(a^2-b^2)}{R^3(a^2+b^2)}\theta$$
Therefore, the time period of small oscillations is
$$T=2\pi\sqrt{\frac{R^3(a^2+b^2)}{3GM_em(a^2-b^2)}}=2\pi\sqrt{\frac{R^3(\eta^2+1)}{3GM_em(\eta^2-1)}}$$
This is equal to $2\pi\sqrt{R^3/(GM_e)}$. Equating,
$$\frac{\eta^2+1}{3(\eta^2-1)}=1$$
Solving for $\eta$,
$$\eta=\sqrt{2}$$
This is the correct answer. Thanks a lot everyone. :)

Sorry for such a long post, I hope my work is correct. I still have one more question. How do I take those approximations? I had to use Wolfram Alpha for that.

Thank you again!

 

[TSny]

That looks very good to me. (I noticed a couple of typos: there should be a square root on the left side of the second equation and the $m$ cancels in deriving $\ddot{\theta}$.)

For the approximation, show that the Taylor expansion of $\frac{1}{\sqrt{1+x}} \approx 1-\frac{1}{2}x+\frac{3}{8} x^2$.

 

[voko]

Since you are using Wolfram Alpha anyway, taking derivatives should not be a problem:

$$ U''(0) = -GM_e m (-(a R)/(a^2-2 a R+R^2)^{3/2}+(a R)/(a^2+2 a R+R^2)^{3/2}+(6 b^2 R^2)/(b^2+R^2)^{5/2}) $$ (this could have been simplified, but I copied it from WA with minimal editing)

Using WA again, I found that the first derivative of the first two terms at $a = 0$ is zero, and the second derivative is $-12/R^3$; likewise, the first derivative of the third term at $b=0$ is zero, and the second is $12/R^3$. Thus, $$ U''(0) \approx - 6 GM_e m \frac {b^2 - a^2} {R^3} $$ So $$ U(\theta) \approx U_0 - 3 GM_e m \frac {b^2 - a^2} {R^3} \theta^2 $$ Same result as yours, but no magic approximations.

 

[Saitama]

Very sorry for the late reply.

That looks very good to me. (I noticed a couple of typos: there should be a square root on the left side of the second equation and the $m$ cancels in deriving $\ddot{\theta}$.)

For the approximation, show that the Taylor expansion of $\frac{1}{\sqrt{1+x}} \approx 1-\frac{1}{2}x+\frac{3}{8} x^2$.

Thank you TSny! I used Taylor expansion and using the approximations, I ended up with same expression as shown by Wolfram Alpha.