If due to air drag,the orbital radius of a sa | vedclass

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(A) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{R}} = \sqrt{GM} \cdot R^{-1/2}$.Differentiating both sides with respect to $R

Vedclass · Accepted Answer

$\frac{\Delta R}{2} \sqrt{\frac{GM}{R^3}}$

Vedclass · Answer

(A) The orbital velocity of a satellite is given by $v = \sqrt{\frac{GM}{R}} = \sqrt{GM} \cdot R^{-1/2}$.Differentiating both sides with respect to $R$,we get:$\frac{dv}{dR} = \sqrt{GM} \cdot (-\frac{1}{2}) R^{-3/2} = -\frac{1}{2} \frac{\sqrt{GM}}{R \sqrt{R}} = -\frac{1}{2R} \sqrt{\frac{GM}{R}}$.For small changes,$\Delta v \approx \frac{dv}{dR} \Delta R$.Since the radius decreases by $\Delta R$,the change in radius is $-\Delta R$.Therefore,$\Delta v = (-\frac{1}{2R} \sqrt{\frac{GM}{R}}) \cdot (-\Delta R) = \frac{\Delta R}{2R} \sqrt{\frac{GM}{R}} = \frac{\Delta R}{2} \sqrt{\frac{GM}{R^3}}$.Thus,the change in orbital velocity is $\frac{\Delta R}{2} \sqrt{\frac{GM}{R^3}}$.

If due to air drag,the orbital radius of a satellite decreases from $R$ to $R - \Delta R$,where $\Delta R << R$,the expression for the change in orbital velocity $\Delta v$ is (mass of Earth is $M$):

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