# The time taken by Mars to revolve around the sun is 1.88 years. Find the ratio of the average...

## Question:

The time taken by Mars to revolve around the sun is 1.88 years. Find the ratio of the average distance between Mars and the sun to that between the earth and the sun.

## Time Period:

The time taken by an object to complete one revolution is called its time period. Earth takes one year to make one complete revolution around the Sun. The time period can be calculated using the mass of the Sun and its distance from the Earth.

## Answer and Explanation:

The magnitude of the force of gravitational attraction between two planets of mass *m* and *M* a distance *r* apart is given by

{eq}F \ = \ G \ \dfrac{m \ M}{r^2} {/eq}

where *G* is the Gravitational constant. If the planet with mass *m* is orbiting the other planet with mass *M*, then the magnitude of the centripetal force keeping the first mass in orbit is given by

{eq}F \ = \ \dfrac{m \ v^2}{r} {/eq}

where *v* is the orbital speed of the planet. The two forces are equal, therefore

{eq}\begin{align*} G \ \dfrac{m \ M}{r^2} \ &= \ \dfrac{m \ v^2}{r}\\ \\ \therefore \ v^2 \ &= \ G \ \dfrac{M}{r} \end{align*} {/eq}

The orbital speed can be expressed in terms of the period *T*

{eq}v \ = \ \dfrac{2 \ \pi \ r}{T} {/eq}

Substituting, we get

{eq}\begin{align*} \left ( \dfrac{2 \ \pi \ r}{T} \right )^2 \ &= \ G \ \dfrac{M}{r}\\ \\ \therefore \ \dfrac{4 \ \pi^2 \ r^2}{T^2} \ &= \ G \ \dfrac{M}{r}\\ \\ r^3 \ &= \ G \ \dfrac{M \ T^2}{4 \ \pi^2}\\ \\ \therefore \ r \ &= \ \sqrt[3]{G \ \dfrac{M \ T^2}{4 \ \pi^2}} \end{align*} {/eq}

The last line is an expression for the distance between the centers of the two planets. In our case, Mars takes

{eq}T_1 \ = \ 1.88 \ year \ \times \ 365 \ days/years \ \times \ 24 \ hours/day \ \times \ 3600 \ s/hour \ = \ 5.928768 \ \times \ 10^7 \ s {/eq}

to orbit the Sun, and the Earth takes 1 year, which is

{eq}T_2 \ = \ 1 \ year \ \times \ 365 \ days/years \ \times \ 24 \ hours/day \ \times \ 3600 \ s/hour \ = \ 3.1536 \ \times \ 10^7 \ s {/eq}

Let the distance between the Sun and Mars be {eq}r_1 {/eq} and let the distance between Earth and the Sun be {eq}r_2 {/eq} with *M* as the mass of the Sun. Therefore, the ratio of the distance between Mars and the Sun to that between the Earth and the Sun is given by

{eq}\begin{align*} \dfrac{r_1}{r_2} \ &= \ \sqrt[3]{G \ \dfrac{M \ T_1^2}{4 \ \pi^2}} \ \div \ \sqrt[3]{G \ \dfrac{M \ T_2^2}{4 \ \pi^2}}\\ \\ &= \ \sqrt[3]{\dfrac{T_1^2}{T_2^2}}\\ \\ &= \ \sqrt[3]{\dfrac{(5.928768 \ \times \ 10^7 \ s)^2}{(3.1536 \ \times \ 10^7 \ s)^2}}\\ \\ \therefore \ \dfrac{r_1}{r_2} \ &= \ \mathbf{1.52} \end{align*} {/eq}

correct to three significant figures.

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 9 / Lesson 12