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Hunting a black hole. Observations of the light from a certain star indicate that it part of a binary (two-star) system. This visible star has orbital speed v = 270 km/s, orbital period T = 1.70 days, and approximate mass m1​==6Ms​, where Ms​, is the Sun's mass, 1.99×1030 kg Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. 13-47). What integer multiple of Ms​, given the approximate mass m2​ of the dark star?

Respuesta :

The equation has one real solution: α=9.3, which implies m₂ /M s≈9.

What integer multiple of Ms​, given the approximate mass m2​ of the dark star?

The two stars are in circular orbits, not about each other, but about the two-star system's center of mass (denoted as O ), which lies along the line connecting the centers of the two stars. The gravitational force between the stars provides the centripetal force necessary to keep their orbits circular. Thus, for the visible, Newton's second law gives

F=r2Gm₁m₂=r₁m₁v₂

where r is the distance between the centers of the stars. To find the relation between r and r₁, we locate the center of mass relative to m1. Using Equation 9−1, we obtain

r₁=m₁+m₂m₁(0)+m₂r=m₁+m₂m₂r ⇒r=m₂m₁+m₂r₁

On the other hand, since the orbital speed of m₁ is v=2πr₁/T, then r₁=vT/2π and the expression for r can be rewritten as

r=m₂m₁ / m2 × vT/2π

Substituting r and r₁ into the force equation, we obtain

F=(m1+m2)2v2T24π2Gm1m23=T2πm1v

or

6.90×1030kg =3.467Ms

where Ms=1.99×10³⁰kg is the mass of the sun. With m₁=6Ms, we write m₂=αMs and solve the following cubic equation for α:

α³/(6+α)²−3.467=0

What is Black hole?

In space, a black hole is a region where gravity is so strong that even light cannot escape. Because the substance is compressed into such a small area, the gravity is extremely intense. When a star is dying, this may take place. People cannot perceive black holes because no light can escape from them. They are undetectable.

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