JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 16)
The minimum and maximum distances of a planet revolving around the sun are x1 and x2. If the minimum speed of the planet on its trajectory is v0 then its maximum speed will be :
$${{{v_0}x_1^2} \over {x_2^2}}$$
$${{{v_0}x_2^2} \over {x_1^2}}$$
$${{{v_0}x_1^{}} \over {x_2^{}}}$$
$${{{v_0}x_2^{}} \over {x_1^{}}}$$
Explanation
Angular momentum conservation equation $${v_0}{x_2} = {v_1}{x_1}$$
$${v_1} = {{{v_0}{x_2}} \over {{x_1}}}$$
$${v_1} = {{{v_0}{x_2}} \over {{x_1}}}$$
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