Asatellite is launched into a circular orbit | vedclass

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Question

When $r=R$

$V=\sqrt{g / R}$

$T_{1}=\frac{2 \pi R}{\sqrt{g / R}}$

$=\frac{2 \pi}{\sqrt{g}}(\sqrt{R})^{3}$

When $r=1.02 R$

$V=\sqrt{1.02

Vedclass · Accepted Answer

$3$

Vedclass · Answer

When $r=R$

$V=\sqrt{g / R}$

$T_{1}=\frac{2 \pi R}{\sqrt{g / R}}$

$=\frac{2 \pi}{\sqrt{g}}(\sqrt{R})^{3}$

When $r=1.02 R$

$V=\sqrt{1.02 g / R}$

$T_{2}=\frac{2 \pi \sqrt{1.02 R}}{g / R}$

$\%$Change in time period

$=\frac{2 \pi}{g}(\sqrt{R})^{3} 	imes(\sqrt{1.02})^{3}-(1)^{3} 	imes 100$

$=\frac{2 \pi}{g}$

$=3 \%$

Asatellite is launched into a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $1.02\,R.$ The period of second satellite is larger than the first one by approximately ........ $\%$

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