Question

The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 Km/s. The second star, Beta, has an orbital speed of 12.0Km/s. The orbital period is 137 d.(A)What is the mass of the star Alpha?(B)What is the mass of the star Beta?(C)One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole. The orbital period of A0620-0090 is 7.75hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of the orbit of the orange star.(D)Find the radius of the orbit of the black hole.(E)Find the orbital speed of the orange star.(F)Find the orbital speed of the black hole.

Answer 1

Complete Question

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Answer:

a) the mass of the star Alpha is $M_{\alpha}$= $7.80*10^{29} kg$

b) the mass of the star Beta is $M_{\beta}$ = $2.34*10^{30}kg$

c)  the radius of the orbit of the orange star $R_{0}$= $1.9*10^{9}m$

d) the radius of the orbit of the black hole $R_{B} = 34*10^{8}m$

e) the orbital speed of the orange star $V_{0} = 4.4*10^{2}km/s$

f)  the orbital speed of the black hole $V_{B} =77 km/s$

Explanation:

The generally formula for orbital speed is given as $v =\frac{2\pi R}{T}$

    where v is the orbital speed

                R  is the radius of the star

               T  is  the orbital period

From the question we are given that

  alpha star has an orbital speed of  $V_{\alpha} = 36km/s = 36000m/s$

  beta star has an orbital speed of $V_{\beta} = 12km/s = 12000m/s$

   the orbital period is  $T = 137d = 137(86400)sec$   1 day is equal 86400 seconds

  Making R the subject of formula  we have   for the radius of the alpha star as

 

                 $R_{\alpha} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(36000m/s)}{2\pi}$

                       $=6.78*10^{10}m$

   for the radius of the Beta star as

                 $R_{\beta} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(12000m/s)}{2\pi}$

                                 $= 2.26*10^{10}m$

Looking at the value obtained for $R_{\alpha} and R_{\beta}$

Generally the moment about the center of the mass are equal then

         $M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$

Thus $3M_{\alpha} = M_{\beta} -------(1)$

Generally the formula for the orbital period is given as

                  $T =\frac{2\pi(R_{\alpha+R_{\beta}})^{3/2 }}{\sqrt{G(M_{\alpha}+M_{\beta})} }$

Then

        $M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Where G is the gravitational constant given as $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$

        $M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(6.78*10^{10}m +2.26*10^{10}m)^{2}}{(137d*86400s/d)^{2}(6.67*10^{-11}m^{3}kg^{-1}s^{-2})}$

       $M_{\alpha} + M_{\beta} = 3.12*10^{30}kg -------(2)$

Thus solving (1) and (2) equations

    Mass of alpha star is  $M_{\alpha}=7.80*10^{29}kg$

and the Mass of Beta is $M_{\beta} =2.34*10^{30}kg$

Considering the equation

                     $M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Making $R_{\alpha} + R_{\beta}$ the subject

                        $R_{\alpha}+ R_{\beta} =\sqrt[3]{\frac{(M_{\alpha+M_{\beta}})T^{2}G}{4\pi^{2}} } -------(3)$

and considering this equation  $M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$ from above

               we have that $R_{\beta} =\frac{M_{\alpha}R_{\alpha}}{M_{\beta}}$

  Considering Question C

Let the Orange star be denoted by (0) and

Let the black-hole be denoted by (B)

And we are told from the question that

    Mass of orange star $M_{0} = 0.67M_{sun}$ and

    Mass of black hole  $M_{B} =3.8M_{sun}$

And mass of sun is $M_{sun} = 1.99*10^{30}kg$

Then $R_{B} = [\frac{0.67M_{sun}}{3.8M_{sun}} ]R_{0} =0.176R_{0}--------(4)$

We are also given that the period is $T =7.75 days = 7.75 (86400s)$

Considering equation 3

   

$R_{0} + 0.176R_{0} = \sqrt[3]{\frac{(0.67+3.8)(1.99*10^{30}kg)(7.75(86400s))^{2} 6.67*10^{-11}Nm^{2}/s^{2}}{4\pi ^{2}} }$

     Thus for V616 Monocerotis, $R_{0} =1.9*10^{9}m$

Considering  equation 4

The black-hole is

     $R_{B}= 0.176R_{0}= 0.176*1.9*10^9 =34*10^8m$

From the formula for velocity of  $V_{0} = \frac{2\pi R_{0}}{T} = 4.4*10^{9}km/s$

         the velocity of $V_{B} = \frac{2\pi R_{B}}{T} =77km/s$