Physics, 22.02.2020 05:19 kyandrewilliams1

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.

Answers: 2

Show answers

Answers

Answer from: PayeA6901

Complete Question

The complete question is shown on the first uploaded image

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

The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an

Answer from: Quest

Fgiryhfdashnmhddvn,

Answer from: Quest

he made 11 calls in one day

Another question on Physics

Physics, 22.06.2019 11:00

Afisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. it takes a time of 2.20 s for the boat to travel from its highest point to its lowest, a total distance of 0.670 m . the fisherman sees that the wave crests are spaced a horizontal distance of 5.80 m apart. (a) how much is the wavelength? (b) find the period of the wave. (c) how fast are the waves traveling? (d) what is the amplitude a of each wave?

Answers: 3

Answer

Physics, 22.06.2019 15:30

To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. a conductor is placed in an external electrostatic field. the external field is uniform before the conductor is placed within it. the conductor is completely isolated from any source of current or charge. part a: which of the following describes the electric field inside this conductor? it is in the same direction as the original external field.it is in the opposite direction from that of the original external field.it has a direction determined entirely by the charge on its surface.it is always zero. part b: the charge density inside the conductor is: 0non-zero; but uniformnon-zero; non-uniforminfinite part c: assume that at some point just outside the surface of the conductor, the electric field has magnitude e and is directed toward the surface of the conductor. what is the charge density η on the surface of the conductor at that point? express your answer in terms of e and ϵ0

Answers: 1

Answer

Physics, 22.06.2019 22:50

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. these cyclotrons typically accelerate h- (the hydride ion, which has one proton and two electrons) to an energy of 5 mev to 20 mev. this ion has a mass very close to that of a proton because the electron mass is negligible—about 1 2000 of the proton’s mass. a typical magnetic field in such cyclotrons is 1.9 t. (a) what is the speed of a 5.0-mev h-? (b) if the h- has energy 5.0 mev and b = 1.9 t, what is the radius of this ion’s circular orbit?

Answers: 1

Answer

Physics, 23.06.2019 00:00

Does the distance travelled by an oscillating mass between times t=0 and t2 equal the displacement of the particle over the same time period if t2> t/2? hint: consider a sine wave starting at t1=0, then consider a closing wave starting at t1=0, for your explanation.

Answers: 1

Answer

You know the right answer?

The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an...

Questions