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Gravitation Part 1

Section: Physics

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Question : 25 of 100

Marks: +1, -0

The maximum and minimum distance of a comet from the Sun are

1.6 \times 10^{12} \text{ m}

8.0 \times 10^{10} \text{ m}

, respectively. If the speed of the comet at the nearest point is

6 \times 10^{4} \text{ ms}^{-1}

, then the speed at the farthest point is

[16 Mar 2021 Shift 1]

$1.5 \times 10^{3} \text{ m/s}$
$6.0 \times 10^{3} \text{ m/s}$
$3.0 \times 10^{3} \text{ m/s}$
$4.5 \times 10^{3} \text{ m/s}$

Solution:

Given, maximum distance of comet from the Sun,

r_{1}=1.6 \times 10^{12} \text{ m}

Minimum distance of comet from the Sun,

r_{2}=8.0 \times 10^{10} \text{ m}

Speed of the comet at nearest point,

v_{2}=6 \times 10^{4} \text{ ms}^{-1}

Applying law of conservation of angular momentum,

m v_{1} r_{1}=m v_{2} r_{2} (\because

mass of comet will remain same

)

\Rightarrow \;\; V_{1}= \frac{V_{2} r_{2}}{r_{1}}

= \frac{6 \times 10^{4} \times 8 \times 10^{10}}{1.6 \times 10^{12}} = \frac{48 \times 10^{14}}{1.6 \times 10^{12}}

=3 \times 10^{3} \text{ m/s}

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