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7.Gravitation
medium
The maximum and minimum distances of a comet from the Sun are $1.6 \times 10^{12}\, m$ and $8.0 \times 10^{10}\, m$ respectively. If the speed of the comet at the nearest point is $6 \times 10^{4}\, ms ^{-1},$ the speed at the farthest point is ......... $\times 10^{3}\, m / s$
A
$1.5$
B
$6.0$
C
$3.0$
D
$4.5$
(JEE MAIN-2021)
Solution
By angular momentum conservation
$mv _{1} r _{1}= mv _{2} r _{2}$
$v _{1}=\frac{48 \times 10^{14}}{1.6 \times 10^{12}}=3000 m / sec$
$=3 \times 10^{3} m / sec$
Standard 11
Physics