GATE_AE_2027_MOCK_05_Q51 GATE 2027 Aerospace Engineering All-India Master Mock 05 (High-Yield Comprehensive Grand Finale) NAT · 2 Marks
Flight Mechanics & Space Dynamics • Space Dynamics & Orbital Mechanics • Kepler's Laws & Orbital Period

The standard gravitational parameter of Earth is \mu = 3.986 imes 10^5 km^3/s^2. For a circular orbit at radius r = 10000 km, what is the orbital period T in seconds? (Take pi ≈ 3.1416, round off to nearest integer).

Answer Key:
9935

Step-by-Step Verified Solution

### Solution: T = 2 * pi * \sqrt{r^3 / \mu} = 2 * 3.1416 * \sqrt{10^{12} / 398600} = 6.2832 * \sqrt{2508780.7} = 6.2832 * 1583.91 ≈ 9952 s. Let's compute precisely: 2*3.14159265 * sqrt(1e12 / 398600) = 6.283185 * 1583.913 = 9951.98 s ≈ 9952 s.
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