91Ó°ÊÓ

Period is$\text{T}=2.4\text{ h}\left(\frac{3600\text{ }\mathrm{s}}{1\text{ }\mathrm{h}}\right)=8640\text{ }\mathrm{s}$

Radius,$\text{r}=8.0×{10}^6\text{″¾}$

According to Kepler's law of periods, the square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, where M₁ and M₂ are the masses of the two orbiting objects in solar masses.

The formulae are as follows:

$\text{R}=\sqrt{\frac{\text{GM}}{{\text{a}}_{\text{g}}}}$

Here,$\text{R}$ is the radius of the planet,$\text{M}$ is the mass of the planet.

From Kepler's law of periods (where $\text{T}=8640\text{ s}$) find the planet's mass$\mathrm{M}$, using the formula, such that,

$\begin{array}{c}{\text{(8640s)}}^{\text{2}}=\left(\frac{4{\text{Ï€}}^2}{(6.7×{10}^{−11\text{​}}\text{ }\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2)×\mathrm{M}}\right){\left(8.0×{10}^6\mathrm{m}\right)}^3\\ \mathrm{M}=\frac{4{\text{Ï€}}^2×5.12×{10}^{21}\text{ }{\mathrm{m}}^3}{6.7×{10}^{−11\text{​}}\text{ }\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2×7.47×{10}^7\text{ }\mathrm{s}}\\ \mathrm{M}=4.04×{\text{10}}^{\text{25}}\text{ k²µ}\end{array}$

However,

$\begin{array}{c}{\mathrm{a}}_{\mathrm{g}}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}\\ 8.{\text{0m/s}}^2\text{ }=\frac{6.7×{10}^{−11\text{​}}\text{ }\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2×4.04×{\text{10}}^{\text{25}}\text{ k²µ}}{{\mathrm{R}}^2}\\ \mathrm{R}=\sqrt{\frac{6.7×{10}^{−11\text{​}}\text{ }\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2×4.04×{\text{10}}^{\text{25}}\text{ k²µ}}{8.{\text{0m/s}}^2}}\\ \mathrm{R}=1.84×{10}^7\text{ }\mathrm{m}\end{array}$

Therefore, theradius of the planet is $\mathrm{R}=1.84×{10}^7\text{ }\mathrm{m}$.