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(B) The acceleration due to gravity $g$ at a height $h$ from the surface of the earth is given by:$g = g_0 \left( \frac{R}{R+h} \right)^2 = \frac{16}{

Vedclass · Accepted Answer

$\frac{343}{16}$

Vedclass · Answer

(B) The acceleration due to gravity $g$ at a height $h$ from the surface of the earth is given by:$g = g_0 \left( \frac{R}{R+h} \right)^2 = \frac{16}{49} g_0$Taking the square root on both sides:$\frac{R}{R+h} = \frac{4}{7}$$7R = 4R + 4h \implies 4h = 3R \implies h = \frac{3R}{4}$The orbital radius $r$ is:$r = R + h = R + \frac{3R}{4} = \frac{7R}{4}$The time period $T$ of a satellite is given by:$T = 2\pi \sqrt{\frac{r^3}{GM}}$Squaring both sides:$T^2 = \frac{4\pi^2 r^3}{GM} = \frac{4\pi^2}{GM} \left( \frac{7R}{4} \right)^3 = \frac{4\pi^2}{GM} \left( \frac{343 R^3}{64} \right) = \frac{343}{16} \left[ \frac{\pi^2 R^3}{GM} \right]$Comparing this with $K \left[ \frac{\pi^2 R^3}{GM} \right]$,we get $K = \frac{343}{16}$.

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