Suppose that a binary star system consists of two stars of equal mass.......?

Question

they are observed to be separated by 360 million km and to take 5.0 Earth years to orbit about a point midway between them. what is the mass of each?

Answer 1

Staying with astronomical terms, 360 million km = 2.406 AU.

Ma + Mb = 2M = a3 / P^2 = (2.406)^3 / (5.0^2)
M = 0.2787 solar masses

One solar mass = 1.989 * 10^30 kg, so M = 5.543*10^29 kg.

Answer 2

The Gravitational force on each star is given by:

F = Gm^2/d^2..........(1)

Where d is the distance between the centres of the two stars.

The balancing force on each star due to circular motion is given by:

F = m(r.w)^2/r = m.r.w^2..........(2)

Where w is the angular velocity in radians per second and r is half the distance between the centres of the stars (d/2).

So, equation (1) = equation (2)

Gm^2/d^2 = m.r.w^2...........(3)

Rearranging to make m the subject of the equation, and substituting d/2 for r, we get

m = d^3w^2/2G................(4)

Substituting for d, w and G, we get.....

m = (3.6x10^11)^3x(2xpi/5x365x24x3600)^2/2x6.642x10^-11

m = 2.231x10^30kg approx

Answer 3

the centripetal force is provided by the gravitational force..........
so
mv^2/r/2=Gmm/r^2....
as
they orbit a point midway between the dis......
frm this....
2mv^2=Gm^2/r
2v^2r=Gm
m=2v^2r/G
...
G is const...(6.67*10^-11)
r=360,000,000
this is the hard part......
i think v=[r*360/(5*365*24*60*60]/2
[v=r*angular speed]
[angular speed=degrees moved/time]
[/2 as mid way point]
[*24*60*60 as v r convering to SI units]
substitute.......

my final calc..
is
1.82*10^24kg
check my calculations alone
hope tat's right!!!!!!

Answer 4

F = -G MM/R² = -MRω²
M = R³ω²/G
G = 6.642*10^-11 m³/kg-s²
M = ((3.6*10^14 m)^3)((2π/5 yr)((365.259 da/yr)(24 hr/da)(3600 s/hr))²/6.642*10^-11 m³/kg-s²
M = (46.656*10^42 m)(1.5727*10^15)/6.642*10^-11 m³/kg-s²
M = 1.1047*10^69 kg