1. **State the problem:** We want to find the distance above Earth where a satellite has an orbital period of 125 minutes. 2. **Formula used:** The orbital period $T$ of a satellite is related to its orbital radius $r$ by Kepler's third law: $$T = 2\pi \sqrt{\frac{r^3}{GM}}$$ where $G$ is the gravitational constant and $M$ is the mass of Earth. 3. **Known values:** - $T = 125$ minutes $= 125 \times 60 = 7500$ seconds - $G = 6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2}$ - $M = 5.972 \times 10^{24} \text{ kg}$ - Earth's radius $R_E = 6.371 \times 10^6$ m 4. **Rearrange formula to solve for $r$:** $$T = 2\pi \sqrt{\frac{r^3}{GM}} \implies \frac{T}{2\pi} = \sqrt{\frac{r^3}{GM}} \implies \left(\frac{T}{2\pi}\right)^2 = \frac{r^3}{GM} \implies r^3 = GM \left(\frac{T}{2\pi}\right)^2$$ 5. **Calculate $r^3$:** $$r^3 = (6.674 \times 10^{-11})(5.972 \times 10^{24}) \left(\frac{7500}{2\pi}\right)^2$$ 6. **Calculate intermediate values:** $$GM = 3.986 \times 10^{14}$$ $$\frac{7500}{2\pi} = \frac{7500}{6.2832} \approx 1193.66$$ $$\left(1193.66\right)^2 = 1.425 \times 10^{6}$$ 7. **Calculate $r^3$:** $$r^3 = 3.986 \times 10^{14} \times 1.425 \times 10^{6} = 5.683 \times 10^{20}$$ 8. **Find $r$ by taking cube root:** $$r = \sqrt[3]{5.683 \times 10^{20}} \approx 8.24 \times 10^{6} \text{ m}$$ 9. **Calculate altitude $h$ above Earth's surface:** $$h = r - R_E = 8.24 \times 10^{6} - 6.371 \times 10^{6} = 1.869 \times 10^{6} \text{ m} = 1869 \text{ km}$$ **Final answer:** The satellite must orbit approximately 1869 km above Earth's surface to have a period of 125 minutes.