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The maximum and minimum distances of a comet from the Sun are $1.6 \times 10^{12}$ m and $8 \times 10^{10}$ 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 :

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6.0×103m/s

1.5×103m/s

3.0×103m/s

4.5×103m/s

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detailed solution

Correct option is C

Labout sun =consant⇒V1r1=V2r2⇒6×1048×1010=V21.6×1012 ⇒V2=6×1048×10101.6×1012=3000m/s

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The maximum and minimum distances of a comet from the Sun are 1.6×1012 m and 8×1010 m respectively. If the speed of the comet at the nearest point is 6×104ms−1 , the speed at the farthest point is :

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The maximum and minimum distances of a comet from the Sun are $1.6 \times 10^{12}$ m and $8 \times 10^{10}$ 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 :