A massive planet of radius R has a diametrical hole as shown in figure, such that hole does not effect the spherecityof the planet. The planet has uniform density. Two equal mass and are simultaneously released,from positions shown. If the collisions are elastic then the total distance traveled by mass after being released atthe time of collision is : [ mass of planet, mass of and neglect the gravitational interaction

1. Gravitational field inside a uniform sphere

For any point at a distance $r$ from the centre of a uniform sphere (radius $R$, mass $M$), only the mass inside that radius contributes to the gravitational pull. The effective mass inside radius $r$ is
$M_r = M \left(\frac{r^3}{R^3}\right)$ Hence the gravitational acceleration is
$g(r) = G\frac{M_r}{r^2} \,=\, G\frac{M r}{R^3}$ which is directly proportional to $r$.

2. Link with Simple Harmonic Motion

Because $g(r) \propto r$, the force on a small mass $m$ is
$F = -m g(r) = -m\Big(G\frac{M}{R^3}\Big) r$ This is the same form as the spring force $F=-k r$, so the motion is SHM with angular frequency
$\omega = \sqrt{\frac{G M}{R^3}}$ The time-dependent position for a mass released from the surface ($x=R$) with zero initial speed is
$x(t) = R \cos(\omega t)$

3. When do the two masses collide?

One marble starts at $+R$ and the other at $-R$. Their positions are $x_1(t) = \phantom{-}R \cos(\omega t)$ $x_2(t) = -R \cos(\omega t)$ They meet when $x_1 = x_2 \;\Rightarrow\; x_1 = 0 \;\Rightarrow\; \cos(\omega t)=0$. The first time this happens is at
$t_c = \frac{\pi}{2\omega}$

4. Distance travelled up to the collision

Path length for one marble: it starts at $x=R$, ends at $x=0$, so distance $\text{distance} = |R - 0| = R$ Collision is elastic and the masses are equal, so after the impact they merely exchange velocities and carry on; the distance up to the first contact is still $R$.

What is happening?

Imagine a huge solid ball (the planet) with a straight, smooth tunnel dug right through the centre from one side to the other.
Now place two identical marbles at the two openings of this tunnel and let them go at the same time.
Because the planet pulls everything towards its centre, each marble begins to slide inward.
Inside a uniform sphere, that pull (gravity) gets weaker the closer you get to the centre, in just the right way to make the motion a perfect back-and-forth swing, just like a spring.
So each marble acts like it is on a giant, invisible spring: it starts at the edge (maximum stretch), heads to the centre (where the spring is relaxed), then would go out the other side.

Where do they bump into each other?

Both marbles start at opposite edges, so they meet exactly in the middle at the very same time.

How far has each marble actually slid when they meet?

They started at the edge (distance $R$ from centre) and travelled straight to the centre (distance $0$).
That means each marble covered a path‐length of exactly $R$ metres before bumping.

So, when the problem asks "total distance travelled by a marble until the collision", the answer for one marble is simply $R$. (If it had asked for both marbles together, it would be $2R$.)

Step-by-step calculation

1. Gravitational field inside planet
   $g(r) = G\frac{M r}{R^3}$

2. Equation of motion (restoring force $\propto -r$)
   $m\ddot r = -m\Big(G\frac{M}{R^3}\Big) r \;\Rightarrow\; \ddot r + \omega^2 r = 0$
   where $\omega = \sqrt{\frac{G M}{R^3}}$

3. SHM solution for an object released from $r=R$ with zero initial speed
   $r(t) = R \cos(\omega t)$

4. Collision time: first time when $r=0$
   $R \cos(\omega t_c)=0 \;\Rightarrow\; \cos(\omega t_c)=0 \;\Rightarrow\; \omega t_c = \frac{\pi}{2}$
   $t_c = \frac{\pi}{2\omega}$

5. Distance travelled by one mass up to collision
   Straight-line motion from $r=R$ to $r=0$
   $\Delta s = |R - 0| = R$

[ \boxed{\text{Distance} = R} ] (If the question had asked for the combined distance of both masses, it would be $2R$.)