A satellite is revolving in a circular orbit | vedclass

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(D) The orbital velocity of the satellite at height $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$. Since $h \ll R$,we can approximate $R+h \approx R$. T

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$\sqrt{gR}(\sqrt{2}-1)$

Vedclass · Answer

(D) The orbital velocity of the satellite at height $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$. Since $h \ll R$,we can approximate $R+h \approx R$. Thus,$v = \sqrt{\frac{GM}{R}}$.Using the relation $g = \frac{GM}{R^2}$,we get $GM = gR^2$. Substituting this,$v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$.The escape velocity $v'$ required to escape the Earth's gravitational field from height $h$ is $v' = \sqrt{\frac{2GM}{R+h}}$.Again,using $h \ll R$,$v' = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.The minimum increase in speed required is $\Delta v = v' - v$.$\Delta v = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2} - 1)$.

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