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(A) The areal speed $v_a$ is defined as the rate of change of area swept by the radius vector,given by $v_a = \frac{dA}{dt} = \frac{1}{2} r v$.Squarin

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$\frac{4v_a^2}{gR^2} - R$

Vedclass · Answer

(A) The areal speed $v_a$ is defined as the rate of change of area swept by the radius vector,given by $v_a = \frac{dA}{dt} = \frac{1}{2} r v$.Squaring both sides,we get $v_a^2 = \frac{1}{4} r^2 v^2$.For a satellite in a circular orbit,the orbital velocity is $v = \sqrt{\frac{GM}{r}}$. Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Thus,$v^2 = \frac{gR^2}{r}$.Substituting $v^2$ into the areal speed equation: $v_a^2 = \frac{1}{4} r^2 \left( \frac{gR^2}{r} \right) = \frac{1}{4} r g R^2$.Solving for the orbital radius $r$: $r = \frac{4v_a^2}{gR^2}$.The height $h$ from the surface of the Earth is $h = r - R$.Therefore,$h = \frac{4v_a^2}{gR^2} - R$.

$A$ satellite is orbiting around the Earth with an areal speed $v_a$. At what height from the surface of the Earth is it rotating,if the radius of the Earth is $R$?

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