Find the de Broglie wavelength of Earth. Mass of Earth is $6\times {10}^{24} \mathrm{kg}$. Mean orbital radius of Earth and the Sun is $150\times {10}^6 \mathrm{km}$

The Earth revolves around the Sun once a year and its orbit has a mean radius $150\times {10}^6 \mathrm{km}$. Hence its average orbital speed is

$v=\frac{2\mathrm{\pi }\times 150\times {10}^6\times {10}^3}{365\times 24\times 3600}=3\times {10}^4 {\mathrm{ms}}^{-1}$.

As the Earth has mass $6\times {10}^{24} \mathrm{kg}$, its average momentum is

$p=mv=1.8\times {10}^{29} {\mathrm{kgms}}^{-1}$,

and so its de Broglie wavelength is

$\lambda =\frac{h}{p}=\frac{6.63\times {10}^{-34}}{1.8\times {10}^{29}}=3.7\times {10}^{-63} \mathrm{m}$