91Ó°ÊÓ

The given data can be listed below as,

• The masses of the two stars are,$M_1$ and$M_2$.
• Theradius of the first star is, $R_1$.
• Theradius of the second star is, $R_2$.
• The orbital speed of first star alpha is,$v_{\mathrm{Î\pm }}=36.0\text{ k³¾}/\text{s}$
• The orbital speed of second star beta is,$v_{\mathrm{β}}=12.0\text{ k³¾}/\text{s}$
• The orbital period is $137d$.

The attraction between all masses in the universe, the gravitational pull of the Earth's mass on objects, mainly near the surface.

a)

The radius$R_1$ and $R_2$are measured with respect to the center of mass, center of mass not accelerated, so it is expressed as,

$\begin{array}{c}{M}_1{R}_1=M_2{R}_2\\ \frac{R_1}{R_2}=\frac{M_2}{M_1}\end{array}$

Hence the ratio of two stars orbital radius is, $\frac{R_1}{R_2}=\frac{M_2}{M_1}$.

b)

The relation of force with mass and acceleration is expressed as,

$F_g=M{a}_{rad}$ ...(i)

Here$M$is the mass of moving particle andlocalid="1655726158095" $a_{rad}$is the radial acceleration and is always directed toward the center of the circle.

The relation of the radial acceleration in terms of the mass and the radius is expressed as,

$a_{rad}=\frac{v^2}{R}$ ...(ii)

Here$v$is the speed of moving particle in a circular orbit and$R$is radius of orbit.

And the relation of a period of the orbit is in terms of the radius of orbit and the speed of moving particle in a circular orbit is expressed as,

$\begin{array}{l}T=\frac{2\mathrm{Ï€}R}{v}\\ v=\frac{2\mathrm{Ï€}R}{T}\end{array}$

Here$T$is the period, the time for one revolution and$R$is the radius of circular orbits.

The relation of gravitational force is expressed as,

Substitute the value of$v$in the equation (ii).

Substitute the value of$a_{rad}$in the equation (i).

$\begin{array}{c}{F}_g=M×\frac{4{\mathrm{π}}^{2}R}{T^2}\\ =\frac{4{\mathrm{π}}^{2}MR}{T^2}\end{array}$

Hence the gravitational force for the first stars is expressed as,

$F_{g1}=\frac{4{\mathrm{Ï€}}^2{M}_1{R}_1}{T_1^2}$ ...(iii)

Here is the time period of star first.

And the gravitational force for the first stars is expressed as,

$F_{g2}=\frac{4{\mathrm{Ï€}}^2{M}_2{R}_2}{T_2^2}$ ...(iv)

Here$T_2$is the time period of star second.

The forces of two stars are equal in magnitude, so it is express as,

$F_{g1}=F_{g2}$ ...(v)

Substitute the value of $F_{g1}$and $F_{g2}$in the equation (v).

$\begin{array}{c}\frac{4{\mathrm{Ï€}}^2{M}_1{R}_1}{T_1^2}=\frac{4{\mathrm{Ï€}}^2{M}_2{R}_2}{T_2^2}\\ \frac{M_1{R}_1}{T_1^2}=\frac{M_2{R}_2}{T_2^2}\end{array}$

From the result of part (a) $M_1{R}_1=M_2{R}_2$. So it is expressed as,

$\begin{array}{c}\frac{1}{T_1^2}=\frac{1}{T_2^2}\\ T_1=T_2\end{array}$

Hence thetwo stars have the same orbital period $T_1=T_2$.

The time period is same, thenlet’stake $T_1=T_2=T$.

The relation of gravitational force is expressed as,

$F_g=G\frac{m_1{m}_2}{r^2}$

Here $F_g$is the gravitational force,$G$is the gravitational constant ,${\mathrm{m}}_1$is the mass of star one,$m_2$is the mass of star second and$r$is the distance between stars.

In given above figure, the two stars separated by a distance$(R_1+R_2)$and the mass of first star is$M_1$, the mass of second star is$M_2$. Then, the gravitational force is expressed as,

$F_g=G\frac{M_1{M}_2}{{(R_1+R_2)}^2}$ ...(vi)

The forces on each star are equal in magnitude, so the equation (iii) and (iv) are equated with equation (vi) and substitute the value of$T_1$and $T_2$. Then it is expressed as,

$\frac{4{\mathrm{Ï€}}^2{M}_1{R}_1}{T^2}=G\frac{M_1{M}_2}{{(R_1+R_2)}^2}$

$M_2{T}^2=\frac{4{\mathrm{Ï€}}^2{R}_1{(R_1+R_2)}^2}{G}$ ...(vii)

And

$\frac{4{\mathrm{Ï€}}^2{M}_2{R}_2}{T^2}=G\frac{M_1{M}_2}{{(R_1+R_2)}^2}$

...(viii)

$M_1{T}^2=\frac{4{\mathrm{Ï€}}^2{R}_2{(R_1+R_2)}^2}{G}$

The addition of the equation (vii) and (viii), is expressed as,

Hence the time period of two star is, $\frac{2\mathrm{Ï€}{(R_1+R_2)}^{\frac{3}{2}}}{\sqrt{G(M_1+M_2)}}$.

c)

The relation of a period of the orbit is in terms of the radius of orbit and the speed of moving particle in a circular orbit is expressed as,

$\begin{array}{l}T=\frac{2\mathrm{Ï€}R}{v}\\ R=\frac{vT}{2\mathrm{Ï€}}\end{array}$

The radius of first star alpha is expressed as,

$R_{\mathrm{Î\pm }}=\frac{v_{\mathrm{Î\pm }}T}{2\mathrm{Ï€}}$

Here$R_{\mathrm{Î\pm }}$is the radius of star alpha.

Substitute $36.0×{10}^3\text{″¾}/\text{s}$for $v_{\mathrm{Î\pm }}$, and $137d$for $T$in the above equation.

$\begin{array}{c}{R}_{\mathrm{Î\pm }}=\frac{36.0×{10}^3\text{″¾}/\text{s}×137\text{ d}×(86400\text{ s}/\text{d})}{2\mathrm{Ï€}}\\ =6.78×{10}^{10}\text{″¾}\end{array}$

And the radius of second star beta is expressed as,

$R_{\mathrm{β}}=\frac{v_{\mathrm{β}}T}{2\mathrm{π}}$

Here$R_{\mathrm{β}}$is the radius of star beta.

Substitute $12.0×{10}^3\text{″¾}/\text{s}$for $v_{\mathrm{β}}$, and $137\text{ d}$for $T$in the above equation.

$\begin{array}{c}{R}_{\mathrm{β}}=\frac{12.0×{10}^3\text{″¾}/\text{s}×137\text{ d}×(86400\text{ s}/\text{d})}{2\mathrm{Ï€}}\\ =2.26×{10}^{10}\text{″¾}\end{array}$

From the time period equation, the sum of two masses is expressed as,

$(M_{\mathrm{Î\pm }}+M_{\mathrm{β}})=\frac{4{\mathrm{Ï€}}^2{(R_{\mathrm{Î\pm }}+R_{\mathrm{β}})}^3}{T^{2}G}$

Here$M_{\mathrm{Î\pm }}$is the mass of star alpha and$M_{\mathrm{β}}$is the mass of star beta.

Substitute$137\text{ d}$for$T$,$6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$for$G$,$6.78×{10}^{10}\text{″¾}$for$R_{\mathrm{Î\pm }}$, and$2.26×{10}^{10}\text{″¾}$for$R_{\mathrm{β}}$in the above equation.

$(M_{\mathrm{Î\pm }}+M_{\mathrm{β}})=\frac{4{\mathrm{Ï€}}^2{(6.78×{10}^{10}\text{″¾}+2.26×{10}^{10}\text{″¾})}^3}{{(137\text{ d}×(86400\text{ s}/\text{d}))}^2×6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}}$

$(M_{\mathrm{Î\pm }}+M_{\mathrm{β}})=3.12×{10}^{30}\text{ k²µ}$ ...(ix)

From part (a) result, it is expressed as,

$\begin{array}{c}{M}_{\mathrm{Î\pm }}{R}_{\mathrm{Î\pm }}=M_{\mathrm{β}}{R}_{\mathrm{β}}\\ M_{\mathrm{β}}=\frac{M_{\mathrm{Î\pm }}{R}_{\mathrm{Î\pm }}}{R_{\mathrm{β}}}\end{array}$

Substitute$6.78×{10}^{10}\text{″¾}$for$R_{\mathrm{Î\pm }}$, and $2.26×{10}^{10}\text{″¾}$for$R_{\mathrm{β}}$in the above equation.

$\begin{array}{l}{M}_{\mathrm{β}}=M_{\mathrm{Î\pm }}×\frac{6.78×{10}^{10}\text{″¾}}{2.26×{10}^{10}\text{″¾}}\\ M_{\mathrm{β}}=3M_{\mathrm{Î\pm }}\end{array}$

Substitute the value of $M_{\mathrm{β}}$in equation (ix).

$\begin{array}{c}4M_{\mathrm{Î\pm }}=3.12×{10}^{30}\text{ k²µ}\\ M_{\mathrm{Î\pm }}=7.8×{10}^{29}\text{ k²µ}\end{array}$

And

$M_{\mathrm{β}}=3M_{\mathrm{Î\pm }}$

Substitute the value of$M_{\mathrm{Î\pm }}$ in the above equation.

$\begin{array}{c}{M}_{\mathrm{β}}=3×7.8×{10}^{29}\text{ k²µ}\\ \text{=2.34}×{10}^{30}\text{ k²µ}\end{array}$

Hence the masses of two stars are, $7.8×{10}^{29}\text{ k²µ}$and $\text{2.34}×{10}^{30}\text{ k²µ}$.

d)

Assume$\mathrm{Î\pm }$refer to star and$\mathrm{β}$refer to black hole. Use relation derives in part (a) and (b). Then the radius of star and black hole is expressed as,

$\begin{array}{c}{M}_{\mathrm{Î\pm }}{R}_{\mathrm{Î\pm }}=M_{\mathrm{β}}{R}_{\mathrm{β}}\\ R_{\mathrm{β}}=R_{\mathrm{Î\pm }}\frac{M_{\mathrm{Î\pm }}}{M_{\mathrm{β}}}\end{array}$

The mass of star 0.67 time mass of the sun and mass of black hole 3.8 time mass of the sun, then it is expressed as,

$\begin{array}{c}{R}_{\mathrm{β}}=\frac{0.67}{3.8}{R}_{\mathrm{Î\pm }}\\ R_{\mathrm{β}}=0.176R_{\mathrm{Î\pm }}\end{array}$

From the time period equation, the sum of two masses is expressed as,

$R_{\mathrm{Î\pm }}+R_{\mathrm{β}}=\sqrt[3]{\frac{(M_{\mathrm{Î\pm }}+M_{\mathrm{β}})T^{2}G}{4{\mathrm{Ï€}}^2}}$

Substitute the value of $M_{\mathrm{Î\pm }}$,$M_{\mathrm{β}}$and$R_{\mathrm{β}}$in the above equation.

$1.176R_{\mathrm{Î\pm }}=\sqrt[3]{\frac{4.47M_s{T}^{2}G}{4{\mathrm{Ï€}}^2}}$

Here$M_S$is the mass of sun.

Substitute $1.9891×{10}^{30}\text{ k²µ}$ for $M_s$, $7.75$for $T$, and $6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$for $G$in the above equation.

$\begin{array}{c}1.176R_{\mathrm{Î\pm }}=\sqrt[3]{\frac{4.47×1.9891×{10}^{30}\text{ k²µ}×{\left(7.75\text{ h°ù}×\frac{3600\text{ s}}{1\text{ h°ù}}\right)}^2×6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}}{4{\mathrm{Ï€}}^2}}\\ =2.27×{10}^9\text{″¾}\\ R_{\mathrm{Î\pm }}\text{=1.9}×{\text{10}}^9\text{″¾}\end{array}$

And

$\begin{array}{c}{R}_{\mathrm{β}}=0.176×\text{1.9}×{\text{10}}^9\text{″¾}\\ =3.3×{\text{10}}^8\text{″¾}\end{array}$

Hence the radius of star and black hole are$\text{1.9}×{\text{10}}^9\text{″¾}$ and $3.3×{\text{10}}^8\text{″¾}$.

The relation of a period of the orbit is in terms of the radius of orbit and the speed of moving particle in a circular orbit is expressed as,

$\begin{array}{l}T=\frac{2\mathrm{Ï€}R}{v}\\ v=\frac{2\mathrm{Ï€}R}{T}\end{array}$

The speed of star is expressed as,

$v_{\mathrm{Î\pm }}=\frac{2\mathrm{Ï€}{R}_{\mathrm{Î\pm }}}{T}$

Substitute$\text{1.9}×{\text{10}}^9\text{″¾}$for$R_{\mathrm{Î\pm }}$and$7.75hr$for $T$, in the above equation.

$\begin{array}{c}{v}_{\mathrm{Î\pm }}=\frac{2\mathrm{Ï€}×\text{1.9}×{\text{10}}^9\text{″¾}×\frac{1\text{ k³¾}}{1000\text{″¾}}}{7.75\text{ h°ù}×\frac{3600\text{ s}}{1\text{ h°ù}}}\\ =4.28×{10}^2\text{ k³¾}/\text{s}\end{array}$

The speed of black hole is expressed as,

$v_{\mathrm{β}}=\frac{2\mathrm{π}{R}_{\mathrm{β}}}{T}$

Substitute$3.3×{\text{10}}^8\text{″¾}$for$R_{\mathrm{β}}$and$7.75hr$for$T$, in the above equation.

$\begin{array}{c}{v}_{\mathrm{β}}=\frac{2\mathrm{Ï€}×3.3×{\text{10}}^8\text{″¾}×\frac{1\text{ k³¾}}{1000\text{″¾}}}{7.75\text{ h°ù}×\frac{3600\text{ s}}{1\text{ h°ù}}}\\ =74.3\text{ k³¾}/\text{s}\end{array}$

Hence the speed of star and black hole are$4.28×{10}^2\text{ k³¾}/\text{s}$ and $74.3\text{ k³¾}/\text{s}$.