This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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To find the speed in orbit and the time for one complete orbit, we need to use the following constants for Earth: • Mass of Earth, M_E = 5.972 × 10^24 kg • Radius of Earth, R_E = 6.371 × 10^6 m • Universal Gravitational Constant, G = 6.674 × 10^-11 N m^2/kg^2 The given altitude of the orbit is h = 230 km = 230 × 10^3 m = 2.30 × 10^5 m. 1. Find its speed in orbit. Step 1: Calculate the orbital radius (r). The orbital radius is the sum of the Earth's radius and the altitude. r = R_E + h r = 6.371 × 10^6 m + 2.30 × 10^5 m r = 6.371 × 10^6 m + 0.230 × 10^6 m r = (6.371 + 0.230) × 10^6 m r = 6.601 × 10^6 m Step 2: Calculate the orbital speed (v). For an object in a circular orbit, the gravitational force provides the centripetal force. (GM_E m)/(r^2) = (mv^2)/(r) Solving for v: v = sqrt((GM_E)/(r)) Substitute the values: v = sqrt((6.674 × 10^-11) N m^2/kg^2) × (5.972 × 10^24 kg)6.601 × 10^6 m v = sqrt(3.9833 × 10^14) m^3/s^26.601 × 10^6 m v = sqrt(6.0344 × 10^7 m)^2/s^2 v ≈ 7768.1 m/s Rounding to three significant figures: v ≈ 7770 m/s The speed in orbit is 7770 m/s. 2. Find the time for one complete orbit. Step 3: Calculate the period (T). The time for one complete orbit (period) is the circumference of the orbit divided by the orbital speed. T = (2 r)/(v) Substitute the values: T = 2 (6.601 × 10^6 m)7768.1 m/s T = 4.1474 × 10^7 m7768.1 m/s T ≈ 5339.2 s Rounding to three significant figures: T ≈ 5340 s To convert to minutes: T = 5339.2 s × 1 min60 s T ≈ 88.987 min Rounding to three significant figures: T ≈ 89.0 min The time for one complete orbit is 5340 s or 89.0 min.