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

Complete Question:

How massive is the brown dwarf

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

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.