91Ó°ÊÓ

• The star is 57 light-years from the earth.
• The mass of the star is 0.85 times that of the sun's mass.
• The orbital radius of the detected planet 0.11n times the radius of the earth's orbit around the sun.

This law states that the square of a planet's revolution's period around a star or sun is directly proportional to the "semi-major" axis's cube.

The planet's orbital period is identified by multiplying the Kepler's constant by the orbital radius. Furthermore, dividing the orbital period by the perimeter of the planet provides the planet's orbital speed.

$v=\sqrt{\frac{GM}{r}}$

Here, v is the orbital speed, r is the orbital radius of the planet, $1.65x{10}^{10}m$M is the mass of the star $Rh{o}^{1}Cancri$which is $1.99×{10}^{30}\mathrm{kg}×0.85\text{which is}1.69×{10}^{30}\mathrm{kg}$. G is the gravitational constant, $6.673×{10}^{-11}\frac{N·m^2}{{\mathrm{kg}}^2}$

Substituting the values in the above equation, we get,

localid="1668309502914" $$\begin{gathered} V=\sqrt{\frac{6.673×{10}^{−11}\frac{\mathrm{N}â‹\dots {\mathrm{m}}^2}{{\mathrm{kg}}^2}×1.69×{10}^{30}\mathrm{kg}}{1.65×{10}^{10}\mathrm{m}}} \\ =8.2×{10}^4\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Thus, the orbital speed of the planet is $8.2×{10}^4\frac{m}{s}$.

$T=\frac{2\mathrm{Ï„}\mathrm{Ï„}r}{v}$

Substituting the values in the above equation, we get,

localid="1668309521822" $$\begin{gathered} T=\frac{2\mathrm{π}×1.65×{10}^{10}\mathrm{m}}{8.2×{10}^4\mathrm{m}/\mathrm{s}} \\ T=1252996.409\mathrm{s}×\frac{1\text{hour}}{3600\mathrm{s}}×\frac{1\text{day}}{24\text{hour}} \\ T=14.5\text{day} \end{gathered}$$

Thus, the orbital period of the planet is 14.5 day.