USAAAO 2026, I кръг — Задача 10

• D

  The official solution uses T = 86400 s, while the statement gives an orbital period of 86164 s; the printed result h = 3.58 × 10⁷ m is kept as printed.

Solution: In this scenario, gravitational force translates to centripetal force, so we equate these two. Using $M$ for Earth’s mass, $m$ as the satellite’s mass, and $T = 86400$ s for the orbital period, $\left(\frac{GMm}{r^2}\right) = m\omega^2 r$, where the angular speed is $\omega = 2\pi/T$. After substitution and rearrangement, we arrive at $r^3 = \left(\frac{GMT^2}{4\pi^2}\right)$ Plugging in the numbers, we obtain $r \sim 4.216\times10^{7}$ m, the orbital radius from Earth’s center. $h = r - R = 3.58\times10^{7}$ m

Answer:D