Knowing that you have the mass of each sun, M1=M2=M which is 2x10^30 kg, and you have the distance between the two stars a=1 AU, you can use Kepler's third law to solve for the period in seconds:
p^2=((4pi)^2(a)^3)/(Gm)
where p is measured in seconds, a is measured in metres, that is, 1 AU = 1.49x10^11 m, and the total mass m is the sum of M1 and M2, and finally G is the gravitational constant 6.67x10^-11:
p=(4pi(1.49x10^11)^3/2)/sqrt((6.67x10^-11)(2x(2x10^30)))
The final answer should be:
4.43x10^7 seconds or 1.4 years
Two equal-mass stars orbit their centre of mass. The distance between the stars (measured from the centre of mass of each) is 1 AU. What is the period of revolution in years? How much energy would be required to unbind these stars if they each have the same mass as the sun? The mass of the sun is 2×10^30 kg.
Can someone help please!
I got 3.5x10^39 joules? for the second part to the question.
Thanks in advance.
2 answers
Bill,
Would you not use:
M = 2x10^30
a = 0.5 AU
Giving:
p = sqrt(((4*pi^2)(0.5*1.49x10^11)^3)/((6.67x10^-11)(2x10^30)))
= 1.106x10^7 seconds = 0.35 years
Note: Bill's verison of Kepler's 3rd Law is incorrect.
Where Bill said: (4pi)^2, it should be: (2pi)^2 = 4*(pi^2)
Would you not use:
M = 2x10^30
a = 0.5 AU
Giving:
p = sqrt(((4*pi^2)(0.5*1.49x10^11)^3)/((6.67x10^-11)(2x10^30)))
= 1.106x10^7 seconds = 0.35 years
Note: Bill's verison of Kepler's 3rd Law is incorrect.
Where Bill said: (4pi)^2, it should be: (2pi)^2 = 4*(pi^2)
1 answer