Does Earth create gravitational waves on space/time as it turns around itself?

Question

As you know earth turns around itself. At this point can we say earth can create waves on space/time?

Answer 1

By waves on space/time I assume you mean gravitational waves.

The answer is that the rotation of the Earth does not produce any gravitational waves. Have a look at my answer to Is it possible to produce gravitational waves artificially? where I give the equation for the power radiated as gravitational wave emission:

$$ P = \frac{32}{5} \frac{G}{c^5} I_{zz}^2 \epsilon^2 \omega^6 $$

The parameter $I_{zz}$ is the quadrupole moment of the mass distribution about the axis of rotation and $\epsilon=(I_{xx}-I_{yy})/I_{zz}$. For an axially symmetric mass $\epsilon = 0$ and the earth is axially symmetric to a very good approximation. That means when we do the calculation for the Earth we have $\epsilon \approx 0$ and therefore the power radiated as gravitational waves is also approximately zero.

Answer 2

Yes, the rotating Earth does generate gravitational waves (GWs), because it is an example of an asymmetric, rotating triaxial ellipsoid. However, the weakness of the gravitational force means that the emitted power is tiny.

The GW power emitted by a rotating body can be written (as per John Rennie's post) as $$ P = \frac{32}{5} \frac{G}{c^5} I_{zz}^2 \epsilon^2 \omega^6 $$

$I_{zz}$ is the moment of inertia of the Earth measured along its polar axis and is approximately $8\times 10^{37}$ kg m$^2$. $\epsilon$ is the difference in the moments of inertia along two orthogonal axes in the equatorial plane, divided by $I_{zz}$. i.e. $$ \epsilon = \frac{I_{xx} - I_{yy}}{I_{zz}} $$

This has a small (but certainly non-zero) value for the Earth of $\epsilon \simeq 1.77 \times 10^{33}/8.034\times 10^{37} = 2.2\times 10^{-5}$ (see Shen et al. 2012). I believe this asymmetry is considerably larger than that induced by tides, so the latter can safely be neglected.

The resulting GW power is $8 \times 10^{-11}$ Watts.