Answer :
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) the mass of the star Alpha is [tex]M_{\alpha}[/tex]= [tex]7.80*10^{29} kg[/tex]
b) the mass of the star Beta is [tex]M_{\beta}[/tex] = [tex]2.34*10^{30}kg[/tex]
c) the radius of the orbit of the orange star [tex]R_{0}[/tex]= [tex]1.9*10^{9}m[/tex]
d) the radius of the orbit of the black hole [tex]R_{B} = 34*10^{8}m[/tex]
e) the orbital speed of the orange star [tex]V_{0} = 4.4*10^{2}km/s[/tex]
f) the orbital speed of the black hole [tex]V_{B} =77 km/s[/tex]
Explanation:
The generally formula for orbital speed is given as [tex]v =\frac{2\pi R}{T}[/tex]
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 [tex]V_{\alpha} = 36km/s = 36000m/s[/tex]
beta star has an orbital speed of [tex]V_{\beta} = 12km/s = 12000m/s[/tex]
the orbital period is [tex]T = 137d = 137(86400)sec[/tex] 1 day is equal 86400 seconds
Making R the subject of formula we have for the radius of the alpha star as
[tex]R_{\alpha} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(36000m/s)}{2\pi}[/tex]
[tex]=6.78*10^{10}m[/tex]
for the radius of the Beta star as
[tex]R_{\beta} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(12000m/s)}{2\pi}[/tex]
[tex]= 2.26*10^{10}m[/tex]
Looking at the value obtained for [tex]R_{\alpha} and R_{\beta}[/tex]
Generally the moment about the center of the mass are equal then
[tex]M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}[/tex]
Thus [tex]3M_{\alpha} = M_{\beta} -------(1)[/tex]
Generally the formula for the orbital period is given as
[tex]T =\frac{2\pi(R_{\alpha+R_{\beta}})^{3/2 }}{\sqrt{G(M_{\alpha}+M_{\beta})} }[/tex]
Then
[tex]M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}[/tex]
Where G is the gravitational constant given as [tex]6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}[/tex]
[tex]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})}[/tex]
[tex]M_{\alpha} + M_{\beta} = 3.12*10^{30}kg -------(2)[/tex]
Thus solving (1) and (2) equations
Mass of alpha star is [tex]M_{\alpha}=7.80*10^{29}kg[/tex]
and the Mass of Beta is [tex]M_{\beta} =2.34*10^{30}kg[/tex]
Considering the equation
[tex]M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}[/tex]
Making [tex]R_{\alpha} + R_{\beta}[/tex] the subject
[tex]R_{\alpha}+ R_{\beta} =\sqrt[3]{\frac{(M_{\alpha+M_{\beta}})T^{2}G}{4\pi^{2}} } -------(3)[/tex]
and considering this equation [tex]M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}[/tex] from above
we have that [tex]R_{\beta} =\frac{M_{\alpha}R_{\alpha}}{M_{\beta}}[/tex]
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 [tex]M_{0} = 0.67M_{sun}[/tex] and
Mass of black hole [tex]M_{B} =3.8M_{sun}[/tex]
And mass of sun is [tex]M_{sun} = 1.99*10^{30}kg[/tex]
Then [tex]R_{B} = [\frac{0.67M_{sun}}{3.8M_{sun}} ]R_{0} =0.176R_{0}--------(4)[/tex]
We are also given that the period is [tex]T =7.75 days = 7.75 (86400s)[/tex]
Considering equation 3
[tex]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}} }[/tex]
Thus for V616 Monocerotis, [tex]R_{0} =1.9*10^{9}m[/tex]
Considering equation 4
The black-hole is
[tex]R_{B}= 0.176R_{0}= 0.176*1.9*10^9 =34*10^8m[/tex]
From the formula for velocity of [tex]V_{0} = \frac{2\pi R_{0}}{T} = 4.4*10^{9}km/s[/tex]
the velocity of [tex]V_{B} = \frac{2\pi R_{B}}{T} =77km/s[/tex]
