91影视

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

Radius,$\text{r}=8.0脳{10}^6\text{鈥尘}$

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 $\text{T}=8640\text{鈥塻}$) find the planet's mass$\mathrm{M}$, using the formula, such that,

$\begin{array}{c}{\text{(8640s)}}^{\text{2}}=\left(\frac{4{\text{蟺}}^2}{(6.7脳{10}^{鈭�11\text{鈥�}}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2)脳\mathrm{M}}\right){\left(8.0脳{10}^6\mathrm{m}\right)}^3\\ \mathrm{M}=\frac{4{\text{蟺}}^2脳5.12脳{10}^{21}\text{鈥�}{\mathrm{m}}^3}{6.7脳{10}^{鈭�11\text{鈥�}}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2脳7.47脳{10}^7\text{鈥�}\mathrm{s}}\\ \mathrm{M}=4.04脳{\text{10}}^{\text{25}}\text{鈥塳驳}\end{array}$

However,

$\begin{array}{c}{\mathrm{a}}_{\mathrm{g}}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}\\ 8.{\text{0m/s}}^2\text{鈥�}=\frac{6.7脳{10}^{鈭�11\text{鈥�}}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2脳4.04脳{\text{10}}^{\text{25}}\text{鈥塳驳}}{{\mathrm{R}}^2}\\ \mathrm{R}=\sqrt{\frac{6.7脳{10}^{鈭�11\text{鈥�}}\text{鈥�}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{kg}}^2脳4.04脳{\text{10}}^{\text{25}}\text{鈥塳驳}}{8.{\text{0m/s}}^2}}\\ \mathrm{R}=1.84脳{10}^7\text{鈥�}\mathrm{m}\end{array}$

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