91Ó°ÊÓ

Radius of planet = $0.9×{10}^{7}m$, time period =role="math" localid="1648536615774" $402days$, free-fall acceleration =$12.2m/s^2.$

Using the formula of gravitational law we get :

$F=\frac{GmM}{r^2}=ma$

Where, F is gravitational force, M is mass of planet and r is the radius between masses.

Solving for acceleration we get :

$$\begin{gathered} a=\frac{GM}{r^2} \\ 12.2m/s^2=\frac{(6.67×{10}^{-11}N{m}^2/k{g}^2)M}{{\left(0.9×{10}^7\mathrm{m}\right)}^2} \\ M=1.48×{10}^{25}\mathrm{kg} \end{gathered}$$

The mass of the planet is$1.48×{10}^{25}kg.$

Orbit distance = $2.2×{10}^{11}m,$time period =$402days=3.473×{10}^{7}s.$

Using Kepler's third law of planetary motion, the time period is given by :

$T^2=\frac{4{\mathrm{Ï€}}^2{R}^3}{GM}$

Where, T is the time period, R is the radius of orbit of star, G is universal gravitational constant, M is mass of star.

$$\begin{gathered} T^2=\frac{4{\mathrm{π}}^2{R}^3}{GM} \\ M=\frac{4{\mathrm{π}}^2{R}^3}{G{T}^2} \\ M=\frac{4{\mathrm{π}}^2{\left(2.2×{10}^{11}m\right)}^3}{(6.67×{10}^{-11}N{m}^2/k{g}^2){\left(34732800s\right)}^2} \\ M=5.22×{10}^{30}\mathrm{kg} \end{gathered}$$

The mass of the star is$5.22×{10}^{30}kg.$