Braking the earth

For this problem we assume the following:

1)The gravitational radiation is all emitted by the Earth due to the fact that the sun remains stationary.

2)The center of mass of the two bodies is located almost at the center of the sun because $M\gg{m}$ , making $r\simeq{1.5\times10^{11} meters}$ .

The emitted radiation results in decrease of the mechanical energy of the orbiting earth. Its effect on the orbit can be modeled as a decelerating tangential force. From Newton's 2nd law: $P=\overrightarrow{F}\cdot\overrightarrow{v}=F\cdot{v}=ma\cdot{v}$ (1)

since $\overrightarrow{F}//\overrightarrow{v}$ for circular orbits.

$v=\omega{R}$ (2)

$\omega^2=\frac{GM}{R^3}$ (3, from Kepler's law).

Solving for the acceleration and plugging data in (1) yields: $a=\frac{P}{mωR}=\frac{32G^{7/2}m^{1/2}M^{2}(M+m)}{5c^5r^{9/2}}=1.1\times{10^{-27}} m/s^2$ where $G=6.673\times10^{-11} m^3kg^{-1}s^{-2}$ , $m$ the mass of the Earth and $M$ the mass of the sun.

The radiation of gravitational waves reduces the earth's kinetic energy and slows the earth down. This would also make it appear that there is a force opposite the earth's motion in order to slow it down. Looking at the relationship for power, we have:

$P=\frac{W}{t}=\frac{F d}{t}= F v$

And from Newton's 2nd Law, we know that:

$F = ma$

So substituting the second equation into the first we have:

$P = mav$

And solving for acceleration:

$a = \frac{P}{mv}$

We already know the mass of the earth, so we just need to calculate the power and the speed. Using the power equation given in the problem statement, and substituting in the values given,

$P = 197.7 W$

Next, we need to calculate the speed of the earth. Assuming the orbit is circular and the speed is constant:

$v = \frac{d}{t} = \frac{2 \pi R }{T}$

Substituting in values and assuming the period is one year:

$v = \frac{2 \pi (1.5\times10^{11} m) }{3.15\times10^7 s} = 29,866 \frac{m}{s}$

Substituting these values back into the acceleration equation:

$a = \frac{197.7 W }{(29,866 \frac{m}{s})(6 \times 10^{24} kg)} = 1.1\times 10^{-27} \frac{m}{s^2}$

Since power is being emitted, this means that energy must be lost in some way from the system. For a planet orbiting a stationary star (the Sun and the Earth), the total energy is given by the sum of the gravitational potential between the two masses and the kinetic energy from the motion of the Earth. We can't lost gravitational potential energy, so the energy loss must come from the orbital velocity of the Earth. Specifically, for some time $t$ , we have:

$E(t) \ = \ \frac{1}{2} m_{Earth} \ v(t)^2 \ - \ \frac{GM_{Sun} \ m_{Earth}}{R}$

Notice how we have assumed that the radius between the Sun and Earth doesn't change with time, even though the Earth is slowing down (the effects will turn out to be so small that we can make this approximation). Now, we know the following will hold true:

$E(0) \ - \ E(t) \ = \ P t$

So we now have:

$E(t) \ = \ E(0) \ - \ Pt \ \Rightarrow \ \frac{1}{2} m_{Earth} \ v(t)^2 \ - \ \frac{GM_{Sun} \ m_{Earth}}{R} \ = \ \frac{1}{2} m_{Earth} \ v(0)^2 \ - \ \frac{GM_{Sun} \ m_{Earth}}{R} \ - \ P t$

$\Rightarrow \ \frac{1}{2} m_{Earth} \ v(t)^2 \ = \ \frac{1}{2} m_{Earth} \ v(0)^2 \ - \ P t \ \Rightarrow \ v(t)^2 \ = \ \ v(0)^2 \ - \ \frac{2 P t}{m_{Earth}}$

We now take the derivative, in order to get the instantaneous acceleration:

$2v(t) \ \frac{d v(t)}{dt} \ = \ - \frac{2P}{m_{Earth}}$

In order to initially maintain orbit, the initial velocity must be given as:

$\frac{m_{Earth} \ v(0)^2}{R} \ = \ \frac{ G \ M_{sun} \ m_{Earth}}{R^2} \ \Rightarrow \ v(0) \ = \ \sqrt{\frac{G \ M_{sun}}{R}}$

Since the velocity decreases slowly, we approximate $v(t)$ to be equal to $v(0)$ , thus we have:

$a(t) \ = \ -\frac{P}{v(0) \ m_{Earth}} \ = \ \frac{32 G^4 m_{Earth} \ M_{sun}^2 \ (m_{Earth} \ + \ M_{sun})}{\sqrt{\frac{G \ M_{sun}}{R}} \ 5 c^5 R^5} \ = \ 1.1073426 \ \times \ 10^{-27} \ \approx \ 1.1 \ \times \ 10^{-27} \ m/s^2$

Use the fact that the rate of decrease of the total mechanical energy of the system appears as the power emmited by gravitational waves.....And since this is quite a slow process circular approximation of the orbits holds good. Such systems are very common..like Supermassive Black Holes(SBH)..In Binary SBH due to gravitational radiations energy is lost the rate of loss is given by $dE/dt=-(1024/5Gc^5) (\omega^3 I)^2$ . $I$ denotes the Quadrapule Moment $I=2Ma^2$ .of the binary system each rotating in a radius $a$ .One could use the same approach and say calculate the time taken for the system to collapse.