Question

35.A satellite is launched into a circular orbit of radius R around the earth.A second satellite is launched into an orbit of radius 1.01R.The time peiriod of the second satellite is larger than that of the first one by approximately--------

Answer 1

Kepler's third law (law of periods):

$$\begin{gathered} T^2\alpha r^3 \\ {T}^2=k{r}^3 \\ \mathrm{Here},\mathrm{k}\mathrm{is}\mathrm{a}\mathrm{proportionality}\mathrm{constant} \end{gathered}$$

Apply logarithm on both sides:

$$\begin{gathered} \log T^2=\log k+\log r^3 \\ 2\log T=\log k+3\log r \\ \log T=\frac{\log k}{2}+\frac{3}{2}\log r \end{gathered}$$


Differentiate on both sides with respect to r on both sides:

$$\begin{gathered} \frac{1}{T}\frac{dT}{dr}=0+\frac{3}{2}\left(\frac{1}{r}\right) \\ \frac{dT}{T}=\frac{3}{2}\left(\frac{dr}{r}\right) \\ =\frac{3}{2}\left(\frac{1.01R-R}{R}\right) \\ =0.015 \\ =1.5\% \end{gathered}$$