Two spherical stars A and B have densities and , respectively. A and B have the same radius, and their masses and are related by . Due to an interaction process, star A loses some of its mass, [2022]

Question

Two spherical stars A and B have densities $ρ_{A}$ and $ρ_{B}$ , respectively. A and B have the same radius, and their masses $M_{A}$ and $M_{B}$ are related by $M_{B} = 2 M_{A}$ . 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 $ρ_{A}$ . The entire mass lost by A is deposited as a thick spherical shell on B with the density of the shell being $ρ_{A}$ . If $v_{A}$ and $v_{B}$ are the escape velocities from A and B after the interaction process, the ratio $\frac{v_{B}}{v_{A}} = \sqrt{\frac{10 n}{15^{1 / 3}}}$ the value of $n$ is _______. [2022]

Smart Achievers · Accepted Answer

(2.3)

Initially, let $M_{A} = m$ , then $M_{B} = 2 m$

$Now, V_{e} = \sqrt{\frac{2 G M}{R}}$

$\frac{V_{B}}{V_{A}} = \sqrt{\frac{M'_{B}}{M'_{A}} \times \frac{R'_{A}}{R'_{B}}} = \sqrt{\frac{2 m + \frac{7 m}{8}}{\frac{m}{8}} \times \frac{R / 2}{R'_{B}}}$

$Now, ρ \times \frac{4}{3} π ({(R'_{B})}^{3} - R^{3}) = \frac{7}{8} \times ρ \times \frac{4}{3} π R^{3}$

$\Rightarrow {(R'_{B})}^{3} - R^{3} = \frac{7}{8} R^{3} \Rightarrow {(R'_{B})}^{3} = \frac{15}{8} R^{3}$

$\Rightarrow R'_{B} = \frac{15^{1 / 3}}{2} R$

$So, \frac{V_{B}}{V_{A}} = \sqrt{\frac{\frac{23 m}{8}}{\frac{m}{8}} \times \frac{R / 2}{\frac{15^{1 / 3}}{2} R}} = \sqrt{\frac{\frac{23 m}{8}}{\frac{m}{8}} \times \frac{1}{15^{1 / 3}}}$

$= \sqrt{\frac{23}{15^{1 / 3}}}$ $= \sqrt{\frac{10 \times 2.3}{15^{1 / 3}}}$

$∴$ $n = 2.3$

Solution

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Solution

Ans. (2.3) Initially, let MA=m, then MB=2m Now, Ve=2GMR VBVA=M'BM'A×R'AR'B=2m+7m8m8×R/2R'B Now, ρ×43π((R'B)3-R3)=78×ρ×43πR3 ⇒(R'B)3-R3=78R3⇒(R'B)3=158R3 ⇒R'B=151/32R So, VBVA=23m8m8×R/2151/32R=23m8m8×1151/3 =23151/3=10×2.3151/3 ∴ n=2.3

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