Fatal attraction

Since the gravitational force acting as a returning force is proportional to the displacement, the fragments undergo SHM. Let's say the mass of the star is $M$ , and the mass of a fragment is $m'$ . Then the answer should be half the period of the oscillation, or

$\pi \sqrt{\frac{m'}{Gm'M}}=\frac {\pi}{\sqrt{GM}}=314.2 \textrm{s}$

Let us place the star such that its CoM is at origin.

Consider the net force on the $ith$ particle due to other particles. The net force is given by:

$\displaystyle \vec{F_{\text{net}}}=\sum_j Gm_jm_i(\vec{r_i}-\vec{r_j})=Gm_i\sum_j m_j\vec{r_i}-Gm_i\sum_jm_j \vec{r_j}$

The second term is zero as it represents the product of position of CoM and total mass of star. Also, $\sum_j m_j=M$ . Hence,

$\displaystyle \Rightarrow \vec{F_{\text{net}}}=-GMm_i\vec{r_i}$

The minus sign is due to fact that force is attractive and is always directed towards origin. Rewriting the above equation,

$\displaystyle \Rightarrow \frac{d^2r}{dt^2}=-GMr$

where I have replaced $r_i$ with $r$ .

The above differential equation represent SHM motion with the time period: $\displaystyle T=\frac{2\pi}{\sqrt{GM}}=\frac{2\pi}{10^{-2}}$ .

It takes time $\dfrac{T}{2}$ to reach back the origin. Hence $\boxed{\tau=314.519\,\, \text{seconds}}$

Say the mass is situated at $\vec{r} = 0$ , $m_{i}$ denotes the mass and $\vec{r_{i}}$ denotes the position vector of $i^{\text{th}}$ fragment, let N be the total number of fragments.

Consider the $n^{\text{th}}$ fragment ,

$\vec{F_{net}} = \displaystyle \sum_{k=1}^{N} Gm_{n} m_{k}(\vec{r_{k}} - \vec{r_{n}})$

Note that I have not excluded $k = n$ , as it yields $0$ .

Hence , $\vec{F_{net}} = Gm_{n} \displaystyle \sum_{k=1}^{N} m_{k} \vec{r_{k}} - Gm_{n}\sum_{k=1}^{N} m_{k} \vec{r_{n}}$

Note that $\displaystyle \sum_{k=1}^{N} m_{k} \vec{r_{k}} = M \vec{r_{cm}} = \vec{0}$ , and $\displaystyle \sum_{k=1}^{N} m_{k} \vec{r_{n}} = M \vec{r_{n}}$

Hence, $\vec{F_{net}} = -Gm_{n}M\vec{r_{n}}$

$\Rightarrow \vec{a_{n}} = -GM \vec{r_{n}}$ , clearly SHM with time period $T= 2 \pi \sqrt{\frac{1}{GM}}$

Required time = $\frac{T}{2} = \boxed{\pi \sqrt{\frac{1}{GM}}}$

assuming the star to be a point mass,we can assume the explosion to be in the form of an expanding shell.let at any instant the radius of the shell be "x".using the procedure of integration we use for proving the shell theorem,we find the field at a distance of"x1".we find it to be GM(x1).putting x1=x, we have the field due to the star fragments on a point on the shell to be GMx.now acceleration of a small object on the shell is [GMx(dm)]/dm=GMx.therefore,a=GMx,where GM=k.this is an SHM equation.thus half its time period is pi/sq. root GM.this is equal to pi(E+2).=314(approx.)

Suppose that the planet splits up into $N$ particles, which have masses $m_1,m_2,\ldots,m_N$ and position vectors $\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N$ . The total mass of the system is $M = m_1 + m_2 +\cdots + m_N$ , and the position vector of the centre of mass of the system is $\mathbf{R}$ , where $M\mathbf{R} \,=\, m_1\mathbf{r}_1 + m_2\mathbf{r}_2 + \cdots + m_N\mathbf{r}_N$ . Suppose that $\mathbf{r}_1 = \mathbf{r}_2 = \cdots = \mathbf{r}_N = \mathbf{0}$ when $t=0$ , at the time when the star explodes. Newton's Law for the $j$ th particle says that (note that $\mathbf{F}_{jj} = \mathbf{0}$ for any $j$ ): $m_j\ddot{\mathbf{r}}_j \; = \; \sum_{i \neq j} \mathbf{F}_{ij} \; = \; \sum_{i=1}^N \mathbf{F}_{ij} \; = \; Gm_jM(\mathbf{R} - \mathbf{r}_j)$ and so $\ddot{\mathbf{r}}_j + MG\mathbf{r}_j \; = \; MG\mathbf{R}$ Moreover $\ddot{\mathbf{R}} \; = \; \sum_{j=1}^N m_j\ddot{\mathbf{r}}_j \; = \; \sum_{i,j=1}^N \mathbf{F}_{ij} \; = \; \mathbf{0}$ so that $\mathbf{R}$ is a linear function of time, and $\mathbf{R} \; = \; t\dot{\mathbf{R}}(0) \; = \; \frac{t}{M}\mathbf{p}$ where $\mathbf{p}$ is the initial momentum of the system. Solving the differential equations ( $\mathbf{r}_j = \mathbf{R}$ is a particular integral), we note that $\mathbf{r}_j \; = \; \mathbf{R} + \mathbf{a}_j \cos\omega t + \mathbf{b}_j \sin\omega t$ for some vectors $\mathbf{a}_1,\mathbf{b}_1,\ldots,\mathbf{a}_N,\mathbf{b}_N$ , where $\omega^2 = MG$ . Matching the initial conditions we deduce that $\mathbf{r}_j \; = \; \frac{t}{M}\mathbf{p} + \frac{1}{\omega}\sin \omega t \dot{\mathbf{r}}(0)$ and hence the fragments coalesce at time $\tau = \frac{\pi}{\omega}$ , at the point $\frac{\tau}{M}\mathbf{p}$ . With the given parameters, $\tau = 314.2$ seconds.

Note that there is no need for the star to be stationary initially.

Notice that the weird Force in the problem satisfies F = -kx, so the motion is simple harmonic.

The problem says that all the particles will come back after a while, even if the explosion is anisotropic. So let's assume that when the star explodes it splits into two masses, one with almost all the mass and the other with negligible mass. The massive particle will essentially not move, and the not massive one will go away and then come back, with time exactly half the Period.

As the motion is simple harmonic, the period is just T = 2pi/(sqrt(GM)), and half of that turns out to be 100pi.