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Gravitation

Section: Physics

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Question : 5 of 21

Marks: +1, -0

Two spherical stars

A

B

\rho_A

\rho_B

, respectively.

A

B

have the same radius, and their masses

M_A

M_B

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

\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

v_A

v_B

are the escape velocities from

A

B

after the interaction process, the ratio

\frac{v_B}{v_A}=\sqrt{\;\frac{10 n}{15^{1/3}}}

n

[JEE Adv 2022 P1]

Your Answer:

Solution:

Given

R_A = R_B = R

M_B = 2 M_A

Calculation of escape velocity for A:

Radius of remaining star

= \;\frac{R_A}{2}

.

Mass of remaining star

= \rho_A \; \frac{4}{3} \pi \; \frac{R_A^3}{8} = \frac{M_A}{8}

\; \frac{- GM_{A/B}}{R_{A/2}} + \frac{1}{2} m v_A^2 = 0

\Rightarrow v_A = \sqrt{\; \frac{2 GM_{A/B}}{R_{A/2}}} = \sqrt{\; \frac{GM_A}{2 R}}

Calculation of escape velocity for B

Mass collectedover

B = \;\frac{7}{8} M_A

Let the radius of

B

becomes

r

.

\therefore \; \frac{4}{3} \pi ( r^3 - R_B^3 ) \rho_A = \frac{7}{8} \rho_A \; \frac{4}{3} \pi R_A^3 \Rightarrow \pi^3 = \frac{7}{8} R_A^3 + R_B^3 = \frac{(15)^{1/3} R}{2}

\frac{V_B^2}{2} = \frac{23 GM_A}{8 \times 15^{1/3} \;\frac{R}{2}} = \frac{23 GM_A}{4 \times 15^{1/3} R}

\therefore V_B = \sqrt{\; \frac{23 GM_A}{2 \times 15^{1/3}} R}

\therefore \frac{V_B}{V_A} = \sqrt{\; \frac{23}{15^{1/3}}} = \sqrt{\; \frac{10 \times 2.30}{15^{1/3}}}

n = 2.30

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