An asteroid, whose mass is $\mathbf{2}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is twice Earth’s distance from the Sun.

(a) Calculate the period of revolution of the asteroid in years.

(b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?

The mass of the earth is ${\text{M}}_{\text{e}}$

The mass of the asteroid $\left({\text{M}}_{\text{a}}\right)=(2\times {10}^{-4}){\text{M}}_{\text{e}}$

The distance between the earth and the sun ${\text{r}}_{\text{es}}$

The distance between the asteroid and the sun $\left({\text{r}}_{\text{a}}\right)=2{\text{r}}_{\text{es}}$

We can use Kepler’s law of period to find the time period of a satellite.Usingtheformula for the kinetic energy of orbiting object, we can find the ratio $\frac{{\mathbf{\text{K}}}_{\mathbf{\text{a}}}}{{\mathbf{\text{K}}}_{\mathbf{\text{e}}}}$

Formula:

${\text{T}}^{\text{2}}=\frac{{\text{4π}}^{\text{2}}{\text{r}}^{\text{3}}}{\text{Gm}}$

$\text{K}=\frac{\text{GMm}}{\text{2}\left(\text{r}\right)}$

The period of revolution of the asteroid

$\begin{array}{l}{{\text{T}}_{\text{a}}}^{\text{2}}\text{=}\frac{{\text{4π}}^{\text{2}}{\text{r}}_{\text{a}}^{\text{3}}}{\text{GM}}\\ {\text{T}}_{\text{a}}\text{=}\sqrt{\frac{{\text{4π}}^{\text{2}}{\text{r}}_{\text{a}}^{\text{3}}}{\text{GM}}}\end{array}$

The radius of the orbit is twice the radius of the Earth’s orbit.

$\begin{array}{c}{\mathrm{r}}_{\mathrm{a}}=2\times {\mathrm{r}}_{\mathrm{e}}\\ =2(150\times {10}^9\text{\hspace{0.17em}}\mathrm{m})\\ =300\times {10}^9\mathrm{m}\end{array}$

$\begin{array}{c}{\text{T}}_{\text{a}}=\sqrt{\frac{4{\mathrm{\pi }}^2{(300\times {10}^9\mathrm{m})}^3}{(6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2)(1.99\times {10}^{30}\mathrm{kg})}}\\ =8.96\times {10}^7\mathrm{s}\\ =2.8\mathrm{years}\end{array}$

The period of revolution is $2.8\mathrm{years}$

The kinetic energy of the asteroid

${\text{K}}_{\text{a}}=\frac{\text{GMma}}{\text{2}\left({\text{r}}_{\text{a}}\right)}$

The kinetic energy of the earth

${\text{K}}_{\text{e}}=\frac{{\text{GMm}}_{\text{e}}}{\text{2}\left({\text{r}}_{\text{e}}\right)}$

$\frac{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{e}}}=\frac{{\text{GMm}}_{\text{a}}}{\text{2}\left({\text{r}}_{\text{a}}\right)}\times \frac{\text{2}\left({\text{r}}_{\text{e}}\right)}{{\text{GMm}}_{\text{e}}}$

Using given values of ${\text{m}}_{\text{a}}$and role="math" localid="1661165028548" ${\text{r}}_a$

$\frac{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{e}}}=\frac{\text{GM}\left({\text{2×10}}^{\text{-4}}\right){\text{m}}_{\text{e}}}{\text{2}\left({\text{2r}}_{\text{e}}\right)}\times \frac{\text{2}\left({\text{r}}_{\text{e}}\right)}{{\text{GMm}}_{\text{e}}}$

$\begin{array}{c}\frac{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{e}}}=\frac{2\times {10}^{-4}}{2}\\ =\frac{1}{{10}^4}\\ =\frac{1}{10000}\end{array}$

The ratio of the kinetic energy of the asteroid to the kinetic energy of Earth is$\frac{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{e}}}=\frac{1}{10000}$.