Question

A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.01)R. The period of the second satellite is large than that of the first one by approximately

• A. 0.5%
• B. 1.0%
• C. 1.5%
• D. 3.0%

Answer 1

Answer: (C)

1.5%

Answer 2

Kepler's Third Law states that $T^2 \propto R^3$, which means $T \propto R^{3/2}$. For the two satellites, we have: $\frac{T_2}{T_1} = \left(\frac{R_2}{R_1}\right)^{3/2}$ Given $R_1 = R$ and $R_2 = 1.01R$. $\frac{T_2}{T_1} = \left(\frac{1.01R}{R}\right)^{3/2} = (1.01)^{3/2}$ Using the binomial approximation $(1+x)^n \approx 1 + nx$ for small $x$: $(1.01)^{3/2} \approx 1 + \frac{3}{2} \times 0.01 = 1 + 1.5 \times 0.01 = 1 + 0.015 = 1.015$ The percentage increase is: $\left(\frac{T_2}{T_1} - 1\right) \times 100\% \approx (1.015 - 1) \times 100\% = 0.015 \times 100\% = 1.5\%$