Question

a sattelite is launched into a circular orbit of radius R around the earth while a second satellite is launched into an orbit of radius 1.02R the percentage change in time periods of two sattelites is

Answer 1

$from\ kepler's\ law,\\T^2\ \alpha\ R^3\\ \Rightarrow T\ \alpha\ R^{ \frac{3}{2} }\\ \Rightarrow T=kR^{ \frac{3}{2} };\ k\ is\ some\ constant.\\ \\for\ radius=R,\\ T=kR^{ \frac{3}{2} }\\ \frac{dT}{dR}= \frac{3}{2}kR^{ \frac{1}{2}}\\ \\ when\ radius=1.02R;\ dR=1.02R-R=0.02R\\ \frac{dT}{dR}= \frac{3}{2}kR^{ \frac{1}{2}}\\ \Rightarrow dT=\frac{3}{2}kR^{ \frac{1}{2}} \times dR\\ \Rightarrow dT=1.5kR^{ \frac{1}{2}} \times 0.02R\\ \Rightarrow dT=0.03kR^{ \frac{3}{2}}=0.03T$

% change=$\frac{dT}{T} \times 100= \frac{0.03T}{T} \times 100=\boxed3$

Answer 2

Explanation:

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