
The maximum and minimum distances of a comet from the sun are \[V=3.0\times {{10}^{3}}m{{s}^{-1}}\] m and $8.0\times {{10}^{10}}$m respectively. If the speed of the comet at the nearest point is \[6\times {{10}^{4}}m{{s}^{-1}}\], the speed at the farthest point is:
a. \[1.5\times {{10}^{3}}m{{s}^{-1}}\]
b. \[4.5\times {{10}^{3}}m{{s}^{-1}}\]
c. \[3.0\times {{10}^{3}}m{{s}^{-1}}\]
d. \[6.0\times {{10}^{3}}m{{s}^{-1}}\]
Answer
163.2k+ views
Hint: As this problem comes under mechanics, it is possible to solve it if you know conservation laws. Here, conservation of angular momentum is used to solve the speed of a comet at the farthest distance. Conservation of angular momentum says that total angular momentum at any point remains constant.
Formula used:
\[V=\frac{rv}{R}\]
Where,
V=speed of comet at farthest distance.
R=maximum distance.
r=minimum distance.
Complete answer:
The orbit around the sun is elliptical. Let ‘r’ denote the minimum distance and ‘v’ denote the speed of the comet at the minimum distance. Let ‘R’ be the maximum distance and ‘V’ be the speed of the comet at farthest distance from sun and ‘m’ be the mass of the comet
From the conservation of angular momentum
Angular momentum at nearest point = Angular momentum at farthest point
\[mvr=mVR\]
Therefore, speed of comet at farthest distance is
\[V=\frac{rv}{R}\]
That is,
\[V=\frac{8\times {{10}^{10}}\times 6\times {{10}^{4}}}{1.6\times {{10}^{12}}}\]
On calculating we get the speed of comet as
\[V=3.0\times {{10}^{3}}m{{s}^{-1}}\]
Therefore, answer is (c).
Note: Here Kepler’s first law and second law is used. Kepler’s first law states that the path of revolution of the planet around the sun is elliptical. That is every revolving object has a point closer to the sun and farther to the sun. Kepler’s second law states that the radius vector sweeps equal area in equal intervals of time. That is for conserving angular momentum the speed of the planet nearer to the sun is high and at a point farther from the sun the speed of the planet is low. So this can be used to check your answer whether the answer we got is right or wrong.
Formula used:
\[V=\frac{rv}{R}\]
Where,
V=speed of comet at farthest distance.
R=maximum distance.
r=minimum distance.
Complete answer:
The orbit around the sun is elliptical. Let ‘r’ denote the minimum distance and ‘v’ denote the speed of the comet at the minimum distance. Let ‘R’ be the maximum distance and ‘V’ be the speed of the comet at farthest distance from sun and ‘m’ be the mass of the comet
From the conservation of angular momentum
Angular momentum at nearest point = Angular momentum at farthest point
\[mvr=mVR\]
Therefore, speed of comet at farthest distance is
\[V=\frac{rv}{R}\]
That is,
\[V=\frac{8\times {{10}^{10}}\times 6\times {{10}^{4}}}{1.6\times {{10}^{12}}}\]
On calculating we get the speed of comet as
\[V=3.0\times {{10}^{3}}m{{s}^{-1}}\]
Therefore, answer is (c).
Note: Here Kepler’s first law and second law is used. Kepler’s first law states that the path of revolution of the planet around the sun is elliptical. That is every revolving object has a point closer to the sun and farther to the sun. Kepler’s second law states that the radius vector sweeps equal area in equal intervals of time. That is for conserving angular momentum the speed of the planet nearer to the sun is high and at a point farther from the sun the speed of the planet is low. So this can be used to check your answer whether the answer we got is right or wrong.
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