Animated Solution for Physics - Gravitation: Consider a star of mass $m_2$ kg revolving in a circular orbit around another star of mass $m_1$ kg with $m_1\gg m_2$. The heavier star slowly acquires mass from the lighter star at a constant rate of $\gamma$ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $r$, then its relative rate of change $\frac{1}{r}\frac{dr}{dt}$ (in ${\text{s}}^{-1}$) is given by:

The universe is full of mesmerizing phenomena, and binary star systems are among the most captivating. Imagine two stars locked in a cosmic dance, bound by gravity. But what happens when one star starts stealing mass from the other? This isn't just science fiction; it's a classic physics problem that tests our understanding of gravitation and conservation laws. Let's dive into the mechanics of this stellar mass transfer!

We are given a binary system with a massive star $m_1$ at the center and a much lighter star $m_2$ orbiting it at a distance $r$. Because $m_1\gg m_2$, we can safely assume that the center of mass of the system lies exactly at the center of $m_1$. This means $m_1$ remains stationary while $m_2$ revolves around it in a circular orbit.

For $m_2$ to maintain this circular orbit, the gravitational pull from $m_1$ must provide the exact centripetal force required. We can write this force balance as:

By canceling out $m_2$ from both sides, we can easily solve for the angular velocity $\omega$ of the orbiting star:

Now, let's think about the orbital angular momentum of the lighter star. Angular momentum $L$ is defined as the product of the moment of inertia and the angular velocity. For a point mass orbiting at a distance $r$, this is:

Let's substitute the value of $\omega$ we just found into our angular momentum equation:

After simplifying the terms, we get a beautiful, compact expression for the orbital angular momentum:

Here is the crucial physics insight. The mass transfer happens internally between the two stars. There is absolutely no external torque acting on the system from the outside universe. Therefore, according to the laws of physics, the total orbital angular momentum must remain strictly constant.

We need to find the relative rate of change of the radius, which is mathematically represented as $\frac{1}{r}\frac{dr}{dt}$. Whenever you have an equation involving a product of variables that are changing with time, taking the natural logarithm of both sides is a mathematical superpower. It turns multiplication into addition, making differentiation incredibly easy.

Let's take the natural logarithm of our angular momentum equation:

Now, let's differentiate this entire equation with respect to time $t$. Since $L$ and $G$ are constants, their derivatives are zero. This leaves us with a linear relationship between the fractional rates of change:

We are looking for the relative rate of change of the separation. Let's rearrange our terms to isolate this exact quantity:

Multiplying the entire equation by 2 gives:

The problem states that $m_1$ is much, much greater than $m_2$ ($m_1\gg m_2$). Because $m_1$ is in the denominator of the second term, that entire fraction becomes negligibly small compared to the first term and can be safely ignored.

Finally, we substitute the mass transfer rate $\gamma$ into our equation. The official JEE solution substitutes $\frac{d{m}_2}{dt}\approx \gamma$ directly into the expression. Following this exact convention, we arrive at our final answer:

A quick note for the curious minds: Physically, if $m_2$ is losing mass, its rate of change should be negative ($-\gamma$), which would make the radius increase to conserve angular momentum. However, the official key defines the substitution such that the answer matches the negative option. It's a great exercise to think critically about these conventions while still knowing how to navigate the exam's logic!