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A satellite is launched into a circular orbit of radius $R$ around earth while a second satellite is launched into a orbit of radius $1.02 R$. The percentage difference in the time periods of the two satellites is :
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From Kepler's law,
$T^2 \propto R^3 \Rightarrow \frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}$
$\frac{T_1}{T_2}=\left(\frac{R_1}{R_2}\right)^{3 / 2}=\left(\frac{R}{1.02 R}\right)^{3 / 2}$
$\Rightarrow \quad \frac{T_1}{T_2}=\left(\frac{50}{51}\right)^{3 / 2}$
$\Rightarrow \quad \frac{T_1}{T_2}=\frac{50}{51}\left(\frac{50}{51}\right)^{\frac{1}{2}}$
$\Rightarrow \quad \frac{T_1}{T_2}=\frac{50}{51} \times 0.99$
$\frac{T_1}{T_2}=\frac{0.97}{1} \Rightarrow T_1=0.97 x, T_2=x$
Difference $=T_2-T_1$
$=x-0.97 x$
$=0.03 x$
$\%$ difference $=\frac{0.03 x}{x} \times 100$
$=3 \%$
$T^2 \propto R^3 \Rightarrow \frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}$
$\frac{T_1}{T_2}=\left(\frac{R_1}{R_2}\right)^{3 / 2}=\left(\frac{R}{1.02 R}\right)^{3 / 2}$
$\Rightarrow \quad \frac{T_1}{T_2}=\left(\frac{50}{51}\right)^{3 / 2}$
$\Rightarrow \quad \frac{T_1}{T_2}=\frac{50}{51}\left(\frac{50}{51}\right)^{\frac{1}{2}}$
$\Rightarrow \quad \frac{T_1}{T_2}=\frac{50}{51} \times 0.99$
$\frac{T_1}{T_2}=\frac{0.97}{1} \Rightarrow T_1=0.97 x, T_2=x$
Difference $=T_2-T_1$
$=x-0.97 x$
$=0.03 x$
$\%$ difference $=\frac{0.03 x}{x} \times 100$
$=3 \%$
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