Question

A $\mathbf{20}\mathbf{\text{kg}}$satellite has a circular orbit with a period of$\mathbf{2.4}\mathbf{\text{h}}$and a radius of$\mathbf{8.6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{m}}$ around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is$\mathbf{8.0}{\mathbf{\text{m/s}}}^{\mathbf{2}}$ , what is the radius of the planet?

Short answer

Theradius of the planet is $\mathrm{R}=1.84\times {10}^7\mathrm{m}$.

Step-by-step solution

Given

Period is$\text{T}=2.4\text{\hspace{0.17em}h}\left(\frac{3600\mathrm{s}}{1\mathrm{h}}\right)=8640\mathrm{s}$

Radius,$\text{r}=8.0\times {10}^6\text{\hspace{0.17em}m}$

Determining the concept

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.

(a) Determining the radius of the planet

From Kepler's law of periods (where $\text{T}=8640\text{\hspace{0.17em}s}$) find the planet's mass$\mathrm{M}$, using the formula, such that,

$\begin{array}{c}{\text{(8640s)}}^{\text{2}}=\left(\frac{4{\text{π}}^2}{(6.7\times {10}^{-11\text{}}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2)\times \mathrm{M}}\right){\left(8.0\times {10}^6\mathrm{m}\right)}^3\\ \mathrm{M}=\frac{4{\text{π}}^2\times 5.12\times {10}^{21}{\mathrm{m}}^3}{6.7\times {10}^{-11\text{}}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2\times 7.47\times {10}^7\mathrm{s}}\\ \mathrm{M}=4.04\times {\text{10}}^{\text{25}}\text{\hspace{0.17em}kg}\end{array}$

However,

$\begin{array}{c}{\mathrm{a}}_{\mathrm{g}}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}\\ 8.{\text{0m/s}}^2=\frac{6.7\times {10}^{-11\text{}}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2\times 4.04\times {\text{10}}^{\text{25}}\text{\hspace{0.17em}kg}}{{\mathrm{R}}^2}\\ \mathrm{R}=\sqrt{\frac{6.7\times {10}^{-11\text{}}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2\times 4.04\times {\text{10}}^{\text{25}}\text{\hspace{0.17em}kg}}{8.{\text{0m/s}}^2}}\\ \mathrm{R}=1.84\times {10}^7\mathrm{m}\end{array}$

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