Question

A satellite is launched into a circular orbit of radius ' $R$ ' around earth while a second satellite is launched into an orbit of radius $1.02\, R$. The percentage difference in the time periods of two satellites is_________

Answer 1

$T ^{2}= KR ^{3}$
$\Rightarrow 2 \frac{ dT }{ T }=3 \frac{ dR }{ R }$
$\Rightarrow\left(\%\right.$ change in Time period $=\frac{3}{2}(\%$ change in $R )$
$\Rightarrow \frac{3}{2}(2 \%)=3 \%$
Alternate method
$T ^{2} \propto R ^{3}$
$T \propto R ^{\frac{3}{2}}$
$\therefore T = kR ^{\frac{3}{2}}$
For radius $R , T = kR ^{\frac{3}{2}}$
$\therefore \frac{ dT }{ dR }=\frac{3}{2} kR ^{\frac{1}{2}}$
When radius $=1.02\, R$
$\therefore dR =1.02 R - R =0.02 R$
$\therefore \frac{ dT }{ dR }=\frac{3}{2} kR ^{\frac{1}{2}}$
$\therefore dT =\frac{3}{2} kR ^{\frac{1}{2}} \times dR$
$\therefore dT =1.5\, kR ^{\frac{1}{2}} \times 0.02\, R$
$\therefore dT =0.03\, kR ^{\frac{3}{2}}$
$\therefore dT =0.03\, T$
Percentage change $=\frac{ dT }{ T } \times 100$
$=\frac{0.03 T }{ T } \times 100$
$=3$