A satellite is launched into a circular orbit of radius $$R$$ around the earth. A second satellite is launched into an orbit of radius $$1.03R.$$ The time period of revolution of the second satellite is larger than the first one approximately by:

$$T^2 \propto R^3 \implies T \propto R^{3/2}$$ (Kepler’s Third Law)

Since the increase in radius is small ($$R$$ to $$1.03R$$, a $$3\%$$ change), we can use the approximation method for small percentage changes.

The percentage change in the radius is:

$$\frac{\Delta R}{R} \times 100 = \frac{1.03R - R}{R} \times 100 = 3\%$$

$$\text{Percentage change in } T \approx \frac{3}{2} \times (\text{Percentage change in } R)$$

$$\text{Percentage change in } T \approx \frac{3}{2} \times 3\% = 4.5\%$$