
The mean distance of mars from the sun is 1.524 times that of the earth from the sun. Find the number of years required for mars to make one revolution about the sun.
A. 1.883
B. 2
C. 3.766
D. 4
Answer
572.7k+ views
Hint: Assume that both mars and earth revolve around the sun circular paths with sun at the centre. Then use the Kepler’s law, which gives the relation between the time period of revolution of the planet around the sun and the radius of the path.
Formula used:
$\Rightarrow {{T}^{2}}=k{{r}^{3}}$
Complete answer:
It is given that the mean distance between mars and sun is 1.524 times the distance between the earth and the sun.
Let the mean distance between the mars and the sun be ${{r}_{1}}$ and the mean distance between the earth and the sun be ${{r}_{2}}$.
Let us assume that the mars and the earth are moving in circular paths of radius ${{r}_{1}}$ and ${{r}_{2}}$ respectively.
$\Rightarrow \dfrac{{{r}_{1}}}{{{r}_{2}}}=1.524$
According to Kepler's law, the square of the time period of an object moving in a circular path is directly proportional to the cube of the radius of the circular path.
i.e. ${{T}^{2}}\propto {{r}^{3}}$.
$\Rightarrow {{T}^{2}}=k{{r}^{3}}$, where k is a proportionality constant.
Let the time periods of revolution the mars and the earth be ${{T}_{1}}$ and ${{T}_{2}}$ respectively.
Therefore,
$\Rightarrow T_{1}^{2}=kr_{1}^{3}$ ….. (i).
And
$\Rightarrow T_{2}^{2}=kr_{2}^{3}$ …… (ii).
Divide (i) and (ii).
$\Rightarrow \dfrac{T_{1}^{2}}{T_{2}^{2}}=\dfrac{kr_{1}^{3}}{kr_{2}^{3}}$
$\Rightarrow \dfrac{T_{1}^{2}}{T_{2}^{2}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}$… (iii).
The time period of earth is 1 year.
Hence, ${{T}_{2}}$ = 1yr.
Substitute the values of ${{T}_{2}}$ and $\dfrac{{{r}_{1}}}{{{r}_{2}}}$ in equation (iii).
$\Rightarrow \dfrac{T_{1}^{2}}{{{(1)}^{2}}}={{\left( 1.524 \right)}^{3}}$
$\Rightarrow {{T}_{1}}={{\left( 1.524 \right)}^{\dfrac{3}{2}}}=1.88years$.
So, the correct answer is “Option A”.
Note:
Note that we have assumed that the mars and the earth revolve around the sun in circular paths.
In reality, they revolve in elliptical paths with the sun positioned at one of the foci of the elliptical paths. Therefore, the distance between the sun and the planets is not constant.
Hence, we consider the mean distance between the sun and the planets.
Formula used:
$\Rightarrow {{T}^{2}}=k{{r}^{3}}$
Complete answer:
It is given that the mean distance between mars and sun is 1.524 times the distance between the earth and the sun.
Let the mean distance between the mars and the sun be ${{r}_{1}}$ and the mean distance between the earth and the sun be ${{r}_{2}}$.
Let us assume that the mars and the earth are moving in circular paths of radius ${{r}_{1}}$ and ${{r}_{2}}$ respectively.
$\Rightarrow \dfrac{{{r}_{1}}}{{{r}_{2}}}=1.524$
According to Kepler's law, the square of the time period of an object moving in a circular path is directly proportional to the cube of the radius of the circular path.
i.e. ${{T}^{2}}\propto {{r}^{3}}$.
$\Rightarrow {{T}^{2}}=k{{r}^{3}}$, where k is a proportionality constant.
Let the time periods of revolution the mars and the earth be ${{T}_{1}}$ and ${{T}_{2}}$ respectively.
Therefore,
$\Rightarrow T_{1}^{2}=kr_{1}^{3}$ ….. (i).
And
$\Rightarrow T_{2}^{2}=kr_{2}^{3}$ …… (ii).
Divide (i) and (ii).
$\Rightarrow \dfrac{T_{1}^{2}}{T_{2}^{2}}=\dfrac{kr_{1}^{3}}{kr_{2}^{3}}$
$\Rightarrow \dfrac{T_{1}^{2}}{T_{2}^{2}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}$… (iii).
The time period of earth is 1 year.
Hence, ${{T}_{2}}$ = 1yr.
Substitute the values of ${{T}_{2}}$ and $\dfrac{{{r}_{1}}}{{{r}_{2}}}$ in equation (iii).
$\Rightarrow \dfrac{T_{1}^{2}}{{{(1)}^{2}}}={{\left( 1.524 \right)}^{3}}$
$\Rightarrow {{T}_{1}}={{\left( 1.524 \right)}^{\dfrac{3}{2}}}=1.88years$.
So, the correct answer is “Option A”.
Note:
Note that we have assumed that the mars and the earth revolve around the sun in circular paths.
In reality, they revolve in elliptical paths with the sun positioned at one of the foci of the elliptical paths. Therefore, the distance between the sun and the planets is not constant.
Hence, we consider the mean distance between the sun and the planets.
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