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Katyanochek1 [597]

3 years ago

7

Suppose you observe a binary system containing a main-sequence star and a brown dwarf. The orbital period of the system is 1 yea

r, and the average separation of the system is 1 AUAU . You then measure the Doppler shifts of the spectral lines from the main-sequence star and the brown dwarf, finding that the orbital speed of the brown dwarf in the system is 22 times greater than that of the main-sequence star.

1 answer:

Agata [3.3K]3 years ago

6 0

Complete Question:

How massive is the brown dwarf

Answer:

Mass of the brown dwarf, M₂ = 4.132 * 10²⁷ kg

Explanation:

Let M₁ = Mass of the main-sequence star

M₂ = Mass of the brown dwarf

v₁ = speed of the main-sequence star

v₂ = speed of the brown dwarf

The average separation of the system r₁ = r₂ = 1 AU

Orbital speed of the brown dwarf is 22 times that of the main - sequence star

That is, v₂ = 22 v₁

Since the sun is a main sequence star

Mass of the main - sequence star = 2 * 10³⁰ kg

Centripetal force, F = Mv²/r

M₁v₁²/r₁² = M₂v₂²/r²

M₁v₁² = M₂v₂²

M₂/M₁ = v₁²/v₂²

M₂/M₁ = (1/22)²

M₂/M₁ = 0.002066

M₂ = 0.002066M₁

M₂ = 0.002066 * 2 * 10³⁰

M₂ = 4.132 * 10²⁷ kg

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