Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA and rB, respectively. If TA and TB are the time period of star A and star B, respectively, then:

Since binary mass system performs circular motion about is common centre of mass, so m A ω A 2 r A = G m B m A ( r A + r B ) 2 = G m B m A r 2 m A ω A 2 × m B ( m A + m B ) r = G m B m A r 2 ω A = G ( m A + m B ) r 3 Similarly we can show that ω B = G ( m A + m B ) r 3 Hence their angular velocity will be same, time period will be same, i.e. TA = TB

Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA and rB, respectively. If TA and TB are the time period of star A and star B, respectively, then:

TA = TB

TA > TB (if mA > mB)

TA > TB (if rA > rB)

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Physics Gravitation Physics Class 11th Gravitation

Consider a binary star system of star A and star B with masses m_A and m_B revolving in a circular orbit of radii r_A and r_B, respectively. If T_A and T_B are the time period of star A and star B, respectively, then:

Option 1 -

T_A = T_B

Option 2 -

T_A > T_B (if m_A > m_B)

Option 3 -

$\frac{T_{A}}{T_{B}} = {(\frac{r_{A}}{r_{B}})}^{\frac{3}{2}}$

Option 4 -

T_A > T_B (if r_A > r_B)

3 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

Since binary mass system performs circular motion about is common centre of mass, so

$m_{A} ω_{A}^{2} r_{A} = \frac{G m_{B} m_{A}}{{(r_{A} + r_{B})}^{2}} = \frac{G m_{B} m_{A}}{r^{2}}$

$m_{A} ω_{A}^{2} \times \frac{m_{B}}{(m_{A} + m_{B})} r = \frac{G m_{B} m_{A}}{r^{2}}$

$ω_{A} = \sqrt[]{\frac{G (m_{A} + m_{B})}{r^{3}}}$

Similarly we can show that

$ω_{B} = \sqrt[]{\frac{G (m_{A} + m_{B})}{r^{3}}}$

Hence their angular velocity will be same, time period will be same, i.e. TA = TB

Since binary mass system performs circular motion about is common centre of mass, so <math> <mrow> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <msubsup> <mrow> <mi>ω</mi> </mrow> <mrow> <mi>A</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msub> <mrow> <mi>r</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>G</mi> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> </mrow> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>r</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>r</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>G</mi> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> </mrow> <mrow> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> <math> <mrow> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <msubsup> <mrow> <mi>ω</mi> </mrow> <mrow> <mi>A</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>×</mo> <mfrac> <mrow> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> <mrow> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mi>r</mi> <mo>=</mo> <mfrac> <mrow> <mi>G</mi> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> </mrow> <mrow> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> <img > <math> <mrow> <msub> <mrow> <mi>ω</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <mo>=</mo> <mroot> <mrow> <mfrac> <mrow> <mi>G</mi> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math> Similarly we can show that <math> <mrow> <msub> <mrow> <mi>ω</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> <mo>=</mo> <mroot> <mrow> <mfrac> <mrow> <mi>G</mi> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>m</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math> Hence their angular velocity will be same, time period will be same, i.e. TA = TB

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Consider a binary star system of star A and star B with masses m_A and m_B revolving in a circular orbit of radii r_A and r_B, respectively. If T_A and T_B are the time period of star A and star B, respectively, then:

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