1. The masses of the two pulsars in the Hulse-Taylor system are both about 1.4 times the mass of the Sun. Their orbits are very elliptical and slightly smaller than the Sun’s diameter: 1 million kilometers. We will make an assumption that the orbits are circular because that simplifies the calculation and still gives us an idea of the energies and timescales involved. Under this assumption, the distance separating the pulsars is twice their orbital radius, or about 2 million km.
Using our expression for energy emission (power) in gravitational waves we get:
Pgw=(2/5)[ ( 6.67e-11 N m2/kg2 ) ( 1.4 * 2e30 kg )/( 2e9 m ) ( 3e8 m/s )2 ]5 *[ ( 1.4 * 2e30 kg )/( 1.4 * 2e30 kg ) ]2 ( 3e8 m/s )5 / ( 6.67e-11 N m2/kg2 ) = 1.9e25 W
If we could divide this into the total energy in the orbit of the two stars we would know the timescale for the stars to spiral into one another. The total energy of a system is the sum of its kinetic and potential energies. But in gravitationally stable systems the potential energy is always twice as large as the kinetic energy (except that potential energy is negative), so the total energy is
Etot = KE - PE = PE/2 - PE = -PE/2 = -GMm/2R
For these two neutron stars orbiting each other about one solar radius apart, this is
Etot = - ( 6.67 × 10-11 N m2/kg2 ) ( 1.4 * 2 × 1030 kg )2 / ( 2 × 1.6 × 109 m ) = -6.6 × 1040 J
Now we only have to divide the energy contained in the orbit by the energy emitted per second to get the timescale for the orbit to decay. We don’t have to worry about the fact that the total energy is negative. That just means the stars are bound. It’s the absolute value of the energy that matters because it tells us the total amount of energy that must be radiated away.
t = Etot / Egw = 6.6 × 1040 J / 1.9 × 1025 W = 1.1 × 108 yr
2. The average distance of Mercury from the Sun is 5.79 x 107 km. Use this value and the expression from the previous exercise to estimate the amount of energy lost to gravitational waves as Mercury orbits the Sun. Would the amount be larger or smaller for the other planets in the Solar System?
Given: Egw =(2/5)[ GMsun) / (Rorbitc2) ]5 [MMercury/Msun ]2 (c5 /G), Rorbit = 5.79 x 107 m, Msun = 2 x 1030 kg, MMercury = 3.3 x 1023 kg, c =2.99 x 108 m/s and G = 6.67 x 10-11 N m2/kg2
Find: Egw
Solution: Egw=(2/5)[ ( 6.67 x 10-11 N m2/kg2 ) (2 x 1030 kg) / (5.79 x 107 m) (3x108 m/s )2 ]5 * [ ( 3.3 x 1023 kg )/( 2 x 1030 kg ) ]2 ( 3 x108 m/s )5 / ( 6.67 x 10-11 N m2/kg2 ) = 4.3 W
3. Your value above should be much smaller than the result for the "natural" gravitational luminosity in the previous exercise. Why is this the case? Or, more specifically, what is it about Mercury’s orbit that causes its emission of gravitational waves to be so small when the “natural” luminosity expected is so large?
Two effects decrease the luminosity. The first is that Mercury’s orbit is much larger than the Sun’s gravitational radius, so raising their ratio to the fifth power and multiplying greatly reduces the luminosity. Second, Mercury’s mass is much smaller than that of the Sun, so raising their mass ratio to the fifth power and multiplying also decreases the luminosity by a tremendous amount.