Answer
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Hint: Assume that there is no collision. For no collision 2a > d, where a is the length of the semi major axis of the comet’s elliptical orbit and d is the distance between the sun and the earth. The relation between the time period T and a is ${{T}^{4}}=\dfrac{4{{\pi }^{2}}}{GM}{{a}^{3}}$. Use this formula to find a. Check whether the assumed relation is correct and hence, conclude your answer.
Formula used:
${{T}^{4}}=\dfrac{4{{\pi }^{2}}}{GM}{{a}^{3}}$
Complete step by step answer:
When a celestial body or any object revolves around the sun, it moves in an elliptical path with the Sun at one of the foci of the ellipse. The time period of revolution of the revolving body is given as
${{T}^{4}}=\dfrac{4{{\pi }^{2}}}{GM}{{a}^{3}}$ …… (i).
Where T is the time period of the body, G is the universal gravitational constant, M is the mass of the Sun and a is the length of the semi- major axis of the elliptical path of the body.
You can think of the semi-major axis as the radius of the ellipse if it was a circular path.
Let the mean distance between the sun and the earth be ‘d’.
Now, we know that both earth and the comet are revolving around the sun. If both of them were revolving in a circular path, then they would have collided if the radius of the path of the comet was equal to the radius of the path of earth.
However, the comet and the earth are revolving in elliptical orbits.
Let us assume that the comet does not collide with the earth. For this to happen, the comet must orbit within the orbit of the earth. In other words, the orbit of the comet must be smaller than the orbit of the earth.
That means, the major axis of the elliptical path of the comet should be less than the mean distance (d) between the sun and the earth.
The length of the semi major axis of an ellipse is equal to 2a.
Let us calculate the value of the semi major axis (a). Use equation (i).
Therefore, we get that
${{a}^{3}}={{T}^{4}}\dfrac{GM}{4{{\pi }^{2}}}$
$\Rightarrow a={{\left( {{T}^{4}}\dfrac{GM}{4{{\pi }^{2}}} \right)}^{\dfrac{1}{3}}}$
It is given that the value of $T=90days=90\times 24hours=90\times 24\times 3600\sec $ and M= $2\times {{10}^{30}}kg$. And we know that $G=6.67\times {{10}^{-11}}k{{g}^{-1}}{{m}^{3}}{{s}^{-2}}$.
Therefore,
$\Rightarrow a={{\left( {{T}^{4}}\dfrac{GM}{4{{\pi }^{2}}} \right)}^{\dfrac{1}{3}}}={{\left( {{\left( 90\times 24\times 3600 \right)}^{4}}\dfrac{\left( 6.67\times {{10}^{-11}} \right)\left( 2\times {{10}^{30}} \right)}{4{{\pi }^{2}}} \right)}^{\dfrac{1}{3}}}=5.89\times {{10}^{10}}m$
$\Rightarrow 2a=2\times 5.89\times {{10}^{10}}=1.178\times {{10}^{10}}m$
And $d=1.5\times {{10}^{11}}m$
This implies that 2a < d.
This means that our assumption was true and the comet revolves within the orbit of the earth. Therefore, collision between the comet and the earth is not possible.
Hence, the correct option is B.
Note: If the length of major axis of the elliptical orbit of the comet is equal to or greater than the mean distance between sun and earth (i.e. $2a\ge d$), then it does not mean that collision will take place. It so happens that the orbit of the comet is parallel to that of earth and both never intersect or they may intersect and collision may take place.
Therefore, in this case we cannot tell whether the collision will be there or not.
Formula used:
${{T}^{4}}=\dfrac{4{{\pi }^{2}}}{GM}{{a}^{3}}$
Complete step by step answer:
When a celestial body or any object revolves around the sun, it moves in an elliptical path with the Sun at one of the foci of the ellipse. The time period of revolution of the revolving body is given as
${{T}^{4}}=\dfrac{4{{\pi }^{2}}}{GM}{{a}^{3}}$ …… (i).
Where T is the time period of the body, G is the universal gravitational constant, M is the mass of the Sun and a is the length of the semi- major axis of the elliptical path of the body.
You can think of the semi-major axis as the radius of the ellipse if it was a circular path.
Let the mean distance between the sun and the earth be ‘d’.
Now, we know that both earth and the comet are revolving around the sun. If both of them were revolving in a circular path, then they would have collided if the radius of the path of the comet was equal to the radius of the path of earth.
However, the comet and the earth are revolving in elliptical orbits.
Let us assume that the comet does not collide with the earth. For this to happen, the comet must orbit within the orbit of the earth. In other words, the orbit of the comet must be smaller than the orbit of the earth.
That means, the major axis of the elliptical path of the comet should be less than the mean distance (d) between the sun and the earth.
The length of the semi major axis of an ellipse is equal to 2a.
Let us calculate the value of the semi major axis (a). Use equation (i).
Therefore, we get that
${{a}^{3}}={{T}^{4}}\dfrac{GM}{4{{\pi }^{2}}}$
$\Rightarrow a={{\left( {{T}^{4}}\dfrac{GM}{4{{\pi }^{2}}} \right)}^{\dfrac{1}{3}}}$
It is given that the value of $T=90days=90\times 24hours=90\times 24\times 3600\sec $ and M= $2\times {{10}^{30}}kg$. And we know that $G=6.67\times {{10}^{-11}}k{{g}^{-1}}{{m}^{3}}{{s}^{-2}}$.
Therefore,
$\Rightarrow a={{\left( {{T}^{4}}\dfrac{GM}{4{{\pi }^{2}}} \right)}^{\dfrac{1}{3}}}={{\left( {{\left( 90\times 24\times 3600 \right)}^{4}}\dfrac{\left( 6.67\times {{10}^{-11}} \right)\left( 2\times {{10}^{30}} \right)}{4{{\pi }^{2}}} \right)}^{\dfrac{1}{3}}}=5.89\times {{10}^{10}}m$
$\Rightarrow 2a=2\times 5.89\times {{10}^{10}}=1.178\times {{10}^{10}}m$
And $d=1.5\times {{10}^{11}}m$
This implies that 2a < d.
This means that our assumption was true and the comet revolves within the orbit of the earth. Therefore, collision between the comet and the earth is not possible.
Hence, the correct option is B.
Note: If the length of major axis of the elliptical orbit of the comet is equal to or greater than the mean distance between sun and earth (i.e. $2a\ge d$), then it does not mean that collision will take place. It so happens that the orbit of the comet is parallel to that of earth and both never intersect or they may intersect and collision may take place.
Therefore, in this case we cannot tell whether the collision will be there or not.
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