JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 5)
A particle of mass m moves in a circular orbit under the central potential field, $$U(r) = - {C \over r}$$, where C is a positive constant. The correct radius $$-$$ velocity graph of the particle's motion is :
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Explanation
Given potential field U(r) = $${{ - C} \over r}$$
$$ \because $$ $$F = {{dU} \over {dr}} = {C \over {{r^2}}}$$
& $${F_c} = {{m{v^2}} \over r}$$
$$ \therefore $$ $${{m{v^2}} \over r} = {C \over {{r^2}}}$$
$$ \Rightarrow r = {C \over {m{v^2}}}$$
$$ \Rightarrow r$$ $$ \propto $$ $$ {1 \over {{v^2}}}$$
$$ \therefore $$ the graph between r & v will be hyperbolic.
$$ \because $$ $$F = {{dU} \over {dr}} = {C \over {{r^2}}}$$
& $${F_c} = {{m{v^2}} \over r}$$
$$ \therefore $$ $${{m{v^2}} \over r} = {C \over {{r^2}}}$$
$$ \Rightarrow r = {C \over {m{v^2}}}$$
$$ \Rightarrow r$$ $$ \propto $$ $$ {1 \over {{v^2}}}$$
$$ \therefore $$ the graph between r & v will be hyperbolic.
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