Binary Stars energy loss calculation

AI Thread Summary
The discussion focuses on calculating energy loss in a binary star system with two stars of equal mass in orbit. The key equations include the force on each star, the period of revolution, and the total energy of the system. Participants explore the relationship between energy loss and the fractional decrease in the orbital period, leading to the conclusion that the fractional loss of energy per unit time can be expressed with a proportionality constant of 2/3. The estimated fractional energy loss per year is calculated as 1.7x10^-9 J yr^-1, and the lifetime of the binary star system is approximated to be 6x10^8 years. The calculations and derivations provided clarify the energy dynamics of the binary stars.
Pseudopro
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I can't seem to get any of the answers for this - please help!

Two stars of equal mass M orbit a common centre. The radius of the orbit of each star is R. Assume that each of the stars has a mass equal to 1.5 solar masses (solar mass=2x10^30kg) and that the initial separation of the stars is 2.0x10^9m.

Note: the figure shows two stars on opposite sides of a circular orbit and a distance 2R separating them.

Force on each star=GM^2/4R^2
Period of revolution of each star T^2=16pi^2 R^3/GM
Period=7.8
Total energy of two stars E=-GM^2/4R
The two-star system loses energy as a result of emitting gravitational radiation so that the stars will move closer to each other.

1)Explain why the fractional loss of energy per unit time may be calculated from the expression delta E/E/delta t=3/2 delta T/T/delta t where delta T/T/delta t is the fractional decrease in period per unit time.
2)The orbital period decreases at a rate of delta T/delta t=72 micro s yr^-1. Estimate the fractional energy loss per year. (Ans: 1.7x10^-19 J yr^-1)
3)The two stars will collapse into each other when delta E~E. Estimate the lifetime, in years, of this binary star system. (Ans: 6x10^8 yr)

Basically, I attempted by saying E is proportionate to R^-1 so delta E/delta t is proportionate to R^-2, T proportion to R^3/2 so delta T... proportion to 3/2T^1/2... I replaced these 'proportionates' into the equation and it worked out. But when I place the numbers for 2), the answer is totally different. I don't understand how 3) would work.

Long question I know. Thank you so much for your patience. Thank you very much if you can help me!
 
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I suspect that the proportionality constant should be -2/3, not 3/2.

If you solve the expression for the period for R as a function of T, then substitute that into the energy expression, you end up with E being proportional to -(1/T)2/3.

So write E = -k (1/T)2/3. Find dE/dT, and carry on.
 
Pseudopro said:
1)Explain why the fractional loss of energy per unit time may be calculated from the expression delta E/E/delta t=3/2 delta T/T/delta t where delta T/T/delta t is the fractional decrease in period per unit time.
2)The orbital period decreases at a rate of delta T/delta t=72 micro s yr^-1. Estimate the fractional energy loss per year. (Ans: 1.7x10^-19 J yr^-1)
3)The two stars will collapse into each other when delta E~E. Estimate the lifetime, in years, of this binary star system. (Ans: 6x10^8 yr)

Basically, I attempted by saying E is proportionate to R^-1 so delta E/delta t is proportionate to R^-2, T proportion to R^3/2 so delta T... proportion to 3/2T^1/2... I replaced these 'proportionates' into the equation and it worked out. But when I place the numbers for 2), the answer is totally different. I don't understand how 3) would work.

Long question I know. Thank you so much for your patience. Thank you very much if you can help me!
For 1:

\frac{d}{dt}E=-\frac{d}{dt}\left(\frac{G\,M^2}{4\,R}\right)\quad\to\quad\frac{d}{dt}E=\left(\frac{G\,M^2}{4\,R^2}\right)\,\frac{dR}{dt}

\frac{(\Delta E)/E}{\Delta t}=\frac{\left(\frac{G\,M^2}{4\,R^2}\right)}{\left(\frac{G\,M^2}{4\,R}\right)}\,\frac{\Delta R}{\Delta t}\quad\to\quad\frac{(\Delta E)/E}{\Delta t}=\frac{(\Delta R)/R}{\Delta t}

Note: You can also get this by taking d/dt of   \ln(-E)=\ln\left(\frac{G\,M^2}{4\,R}\right)\,.

T^2=\left(\frac{16\pi^2}{GM}\right)\,R^3

2T\,\frac{dT}{dt}=3\left(\frac{16\pi^2}{GM}\right)\,R^2\frac{dR}{dt}

Divide the left side by 2T2 and the right side by 2×16π2 R3/GM .

\frac{(\Delta T)/T}{\Delta t}=\frac{3}{2}\,\frac{(\Delta R)/R}{\Delta t}

So, it looks as if the proportionality constant should be 2/3, i.e. (ΔE/E)/Δt = (2/3) (ΔT/T)/Δt, which is what you get if you follow gneil's suggestion.
gneil said:
...
So write E = -k (1/T)2/3. Find dE/dT, and carry on.
 
Pseudopro said:
...

Period=7.8 hours
...

1)Explain why the fractional loss of energy per unit time may be calculated from the expression delta E/E/delta t=2/3 delta T/T/delta t where delta T/T/delta t is the fractional decrease in period per unit time.
2)The orbital period decreases at a rate of delta T/delta t=72 micro s yr -1. Estimate the fractional energy loss per year. (Ans: 1.7x10^-19 J yr^-1)
...
After "playing around" with the numbers, I made some additions/corrections in red.

delta T/delta t=72 μs/yr, and T = (7.8 hours)×(3600 seconds/hour)

So, the fractional decrease in the period is: (72x10-6)/(7.8)/(3600)yr-1.

Multiply this by 2/3 to get the fractional energy loss per year.
 
SammyS said:
After "playing around" with the numbers, I made some additions/corrections in red.

delta T/delta t=72 μs/yr, and T = (7.8 hours)×(3600 seconds/hour)

So, the fractional decrease in the period is: (72x10-6)/(7.8)/(3600)yr-1.

Multiply this by 2/3 to get the fractional energy loss per year.

Thank you for your help (SammyS and gneill), the expression derivation makes sense. SammyS, would the answer be 1.7x10-9 as opposed to 1.7x10-19?
 
Yes, I got 1.7x10-9yr-1.

That seems consistent with:
Estimate the lifetime, in years, of this binary star system. (Ans: 6x10^8 yr)
 
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