Two spherical stars A and B have densities rho_A and rho_B, respectively. A and B have the same radi

Q: Why is the new mass of star A exactly $M_A/8$?

Mass is volume times density. Since the density is constant and the star remains spherical, mass is proportional to $R^3$. Halving the radius ($R/2$) reduces the volume, and thus the mass, by a factor of $(1/2)^3 = 1/8$.

Q: How do we find the volume of the shell deposited on star B?

The shell is the region between the original radius $R$ and the new outer radius $r$. Its volume is the difference between the volume of the large sphere (radius $r$) and the inner sphere (radius $R$), which is $\frac{4}{3}\pi(r^3 - R^3)$.

Q: Why do we use the density of star A for the shell on star B?

The problem explicitly states that the mass lost by A is deposited as a shell on B, and the density of this shell is $\rho_A$. This is a crucial constraint to find the new radius of B.

Q: Does the original density of star B ($\rho_B$) matter?

Interestingly, no! We are given the initial mass relationship $M_B = 2M_A$. We only need the total mass of B and its final radius to find its escape velocity. The internal density distribution of B doesn't affect the escape velocity from its surface, as long as it's spherically symmetric.

Q: What is the formula for escape velocity and why?

The escape velocity $v = \sqrt{2GM/R}$ is derived from energy conservation. A particle needs enough initial kinetic energy to overcome the gravitational potential energy at the surface, meaning its total energy must be at least zero: $\frac{1}{2}mv^2 - \frac{GMm}{R} = 0$.

Question

Two spherical stars A and B have densities $\rho_A$ and $\rho_B$, respectively. A and B have the same radius, and their masses $M_A$ and $M_B$ are related by $M_B = 2M_A$. Due to an interaction process, star A loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains $\rho_A$. The entire mass lost by A is deposited as a thick spherical shell on B with the density of the shell being $\rho_A$. If $v_A$ and $v_B$ are the escape velocities from A and B after the interaction process, the ratio $\frac{v_B}{v_A} = \sqrt{\frac{10n}{15^{1/3}}}$. The value of n is _______.

Sigma Percentile · Accepted Answer

The universe is full of dynamic processes, and this problem takes us on a thrilling journey of cosmic mass transfer! We are presented with two stars, A and B, and we need to figure out how their escape velocities change after a dramatic interaction. Let's break down this beautiful application of gravitation and density.

### The Initial Cosmic Setup

Imagine two spherical stars, A and B, suspended in space. Initially, they share the exact same radius, $R$. However, they are not identical twins. Star B is twice as massive as star A, meaning $M_B = 2M_A$.

This difference in mass, despite having the same volume, implies that star B is significantly denser than star A. But for our calculations, we will focus primarily on the density of star A, denoted as $ho_A$, because it plays a crucial role in the upcoming mass transfer.

### The Shrinking of Star A

Suddenly, an interaction occurs! Star A loses a significant portion of its mass. The problem states that its new radius is exactly half of its original radius, so $R'_A = \frac{R}{2}$.

Crucially, star A retains its spherical shape and its original density, $ho_A$. This is the key to finding its new mass. Since mass is the product of volume and density, and the density is constant, the mass is directly proportional to the cube of the radius.

$$M \propto R^3$$

By halving the radius, the volume becomes $\left(\frac{1}{2}ight)^3 = \frac{1}{8}$ of the original volume. Therefore, the new mass of star A is simply one-eighth of its original mass:

$$M'_A = \frac{M_A}{8}$$

With its new mass and radius, we can immediately calculate the new escape velocity from star A. The escape velocity formula is $v = \sqrt{\frac{2GM}{R}}$. Substituting our new values:

$$v_A = \sqrt{\frac{2G \left(\frac{M_A}{8}ight)}{\frac{R}{2}}} = \sqrt{\frac{GM_A}{2R}}$$

### The Accretion onto Star B

Now, what happens to the mass that star A lost? It doesn't just vanish; it is entirely deposited onto star B!

First, let's calculate exactly how much mass was transferred. Star A started with mass $M_A$ and ended up with $\frac{M_A}{8}$. The difference is the transferred mass:

$$\Delta M = M_A - \frac{M_A}{8} = \frac{7}{8}M_A$$

This mass forms a thick spherical shell around star B. The problem gives us a vital piece of information: the density of this new shell is $ho_A$. Let's denote the new outer radius of star B as $r$. The volume of this shell is the difference between the volume of the new large sphere and the original sphere of star B:

$$V_{	ext{shell}} = \frac{4}{3}\pi (r^3 - R^3)$$

The mass of this shell is its volume multiplied by its density, $ho_A$. We know this mass must equal the transferred mass, $\frac{7}{8}M_A$. Also, remember that the original mass of star A can be written as $M_A = \frac{4}{3}\pi R^3 ho_A$. Equating these gives us a beautiful relationship:

$$\frac{4}{3}\pi (r^3 - R^3) ho_A = \frac{7}{8} \left( \frac{4}{3}\pi R^3 ho_A ight)$$

Notice how the $\frac{4}{3}\pi$ and $ho_A$ terms cancel out perfectly! We are left with a simple equation for the radii:

$$r^3 - R^3 = \frac{7}{8}R^3$$

Solving for $r^3$, we get $r^3 = \frac{15}{8}R^3$. Taking the cube root reveals the new radius of star B:

$$r = \frac{15^{1/3}}{2}R$$

Before we can find the escape velocity for star B, we need its new total mass. It's simply its original mass plus the mass of the accreted shell:

$$M'_B = 2M_A + \frac{7}{8}M_A = \frac{23}{8}M_A$$

### The Final Escape Velocity Ratio

We now have everything we need to find the new escape velocity from star B. Using the standard formula again with our newly found mass and radius:

$$v_B = \sqrt{\frac{2G \left(\frac{23}{8}M_Aight)}{\frac{15^{1/3}}{2}R}} = \sqrt{\frac{23GM_A}{2 \cdot 15^{1/3} R}}$$

The final step is to find the ratio of the two escape velocities, $\frac{v_B}{v_A}$. Let's divide the two expressions we've derived:

$$\frac{v_B}{v_A} = \frac{\sqrt{\frac{23GM_A}{2 \cdot 15^{1/3} R}}}{\sqrt{\frac{GM_A}{2R}}}$$

Watch as the $G$, $M_A$, and $2R$ terms elegantly cancel out, leaving us with a clean numerical ratio:

$$\frac{v_B}{v_A} = \sqrt{\frac{23}{15^{1/3}}}$$

The problem states that this ratio is equal to $\sqrt{\frac{10n}{15^{1/3}}}$. By comparing our result with this given expression, it is immediately clear that:

$$10n = 23$$

**Therefore, the value of n is 2.3.**

This problem is a fantastic demonstration of how fundamental principles like density, volume, and energy conservation intertwine in astrophysical scenarios!

Enter Numerical Value:

Visualized Solution

$R_{A} = R_{B} = R, M_{B} = 2 M_{A}$

$M_{A}^{'} = \frac{M _{A}}{8}$

$v_{A} = \frac{G M _{A}}{2 R}$

$Δ M = \frac{7}{8} M_{A}$

$V_{shell} \cdot ρ_{A} = \frac{7}{8} M_{A}$

$r = \frac{1 5 ^{1/3}}{2} R$

$M_{B}^{'} = \frac{23}{8} M_{A}$

$v_{B} = \frac{23 G M _{A}}{2 \cdot 1 5 ^{1/3} R}$

$\frac{v _{B}}{v _{A}} = \frac{23}{1 5 ^{1/3}}$

$n = 2.3$

The Sigma Insight: Gravitational Potential and Potential Energy

The Initial Cosmic Setup

The Shrinking of Star A

The Accretion onto Star B

The Final Escape Velocity Ratio

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RA​=RB​=R,MB​=2MA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MA′​=8MA​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vA​=2RGMA​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔM=87​MA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vshell​⋅ρA​=87​MA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=2151/3​R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MB′​=823​MA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vB​=2⋅151/3R23GMA​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vA​vB​​=151/323​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=2.3