A binary star system consists of two stars of masses m1 and m2. The stars, which gravitationally attract each other, revolve around the center of mass of the system. The star with mass mi has a centripetal acceleration of magnitude a1. Find a2, the magnitude of the centripetal acceleration of the star with mass m2 Express the acceleration in terms of quantities given in the problem introduction.

a₂ = G(m₁ + m₂)/(R - √[G(m₁ + m₂)/a₁])²

Step-by-step explanation:

Since the gravitation force of attraction of the center of mass on m₁ equals the centripetal force, then

G(m₁ + m₂)m₁/R₁ = m₁a₁ where (m₁ + m₂) = total mass of the system and R₁ = distance of m₁ from the center of mass

So, G(m₁ + m₂)m₁/R₁² = m₁a₁

G(m₁ + m₂)m₁/m₁a₁ = R₁²

R₁² = G(m₁ + m₂)/a₁

R₁ = √[G(m₁ + m₂)/a₁]

The distance of m₂ from the center of mass is thus R₂ = R - R₁ = R - √[G(m₁ + m₂)/a₁] where R is the distance between the two stars.

Also,  the gravitation force of attraction of the center of mass on m₂ equals the centripetal force, then

G(m₁ + m₂)m₂/R₂ = m₂a₂ where (m₁ + m₂) = total mass of the system and R₂ = distance of m₂ from the center of mass

So, G(m₁ + m₂)m₂/R₂² = m₂a₂

G(m₁ + m₂)m₂/m₂R₂² = a₂

a₂ = G(m₁ + m₂)/R₂²

Substituting R₂ into the equation, we have

a₂ = G(m₁ + m₂)/(R - √[G(m₁ + m₂)/a₁])²

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.