Question

Two stars M1 and M2 of equal mass make up a binary star system. They move in a circular orbit that has its center at the midpoint of the line that separates them. If M1-M2-6.95 sm (solar mass), and the orbital period of each star is 2.20 days, find their orbital speed. (The mass of the sun is 1.99x 1030 kg.) km/s M2

Answer 1

Answer:

V = 365643.04 m/s

Explanation:

mass of the sun = 1.99 x 10^{30} kg

mass of M1 = mass of M2 = 6.95 solar mass = 6.95 x 1.99 x 10^{30} = 13.8305x 10^{30} kg  

orbital period of each star (T) = 2.20 days = 2.20 x 24 x 60 x 60 =190,080 s

gravitational constant (G) = 6.67 x 10^{-11} N m2/kg2

orbital speed (V) = $\sqrt{\frac{G(M1+M2)}{r} }$

we need to find the orbital radius (r) before we can apply the formula above and we can get it from Kepler's third law, $T^{2} = r^{3}$ x k

where

• T = orbital period

• r = orbital radius

• k = $\frac{4n^{2} }{G(M1+M2)}$  (take note that π is shown as $n$)

making r the subject of the formula we now have

$r = (\frac{G(M1+M2).T^{2}}{4n^{2} } )^{\frac{1}{3} }$    (take note that π is shown as $n$)

$r = (\frac{ 6.67 x 10^{-11} ( 13.8305x 10^{30}+ 13.8305x 10^{30} )x190080^{2}}{4x3.142^{2} } )^{\frac{1}{3} }$

r = 1.38 x 10^{10} m

Now that we have the orbital radius (r) we can substitute all required values into the formula for orbital speed

orbital speed (V) = $\sqrt{\frac{G(M1+M2)}{r} }$

$V = \sqrt{\frac{6.67 x 10^{-11} ( 13.8305x 10^{30}+ 13.8305x 10^{30}}{1.38 x 10^{10} } }$

V = 365643.04 m/s