Satellite Orbit: What the Data Actually Show
A small satellite of mass 420 kg is in a circular orbit at an altitude of 550 km above Earth's surface. Earth's radius is 6.37 × 10⁶ m and its mass is 5.97 × 10²⁴ kg. A student calculates the orbital period and gets 1.6 hours. A second student uses the same data but forgets to add Earth's radius to the altitude and gets a period of about 1.5 hours. The teacher asks the class to decide which is correct and to explain why the difference is small. Later, the satellite's thrusters fire briefly to raise the orbit to 600 km.
1. Calculate the orbital radius, speed, and period for the 550 km orbit, and evaluate the claim that the difference between the two students' answers is 'negligible'.
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Orbital radius r = 6.37 × 10⁶ + 5.50 × 10⁵ = 6.92 × 10⁶ m.
Speed: v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.92 × 10⁶). GM = 3.98 × 10¹⁴. So v = √(3.98 × 10¹⁴ / 6.92 × 10⁶) = √(5.75 × 10⁷) = 7.58 × 10³ m/s.
Period: T = 2πr / v = 2π × 6.92 × 10⁶ / 7.58 × 10³ = 5.74 × 10³ s ≈ 1.59 h.
So the first student is correct: about 1.6 h. The second student's value uses r = 5.50 × 10⁵ m, which is physically wrong (inside Earth).
The difference between 1.6 h and 1.5 h is about 6%, which is not negligible for orbit prediction - over one day it accumulates to more than an hour of timing error. A correct answer must reject the second method, not merely call the difference small.
2. Explain, using energy ideas, what happens to the satellite's speed and period when the thrusters raise the orbit to 600 km.
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Raising the orbit increases r from 6.92 × 10⁶ m to 6.97 × 10⁶ m.
Orbital speed v = √(GM/r) decreases as r increases, so the satellite slows down in the higher orbit. This is counter-intuitive: the thrusters add energy, but the extra energy goes into gravitational potential energy, and the kinetic energy falls.
Total mechanical energy E = −GMm/(2r) becomes less negative (increases), consistent with energy being added.
Period T = 2π√(r³/GM) increases with r. Using the ratio: T₂/T₁ = (6.97/6.92)^1.5 = (1.0072)^1.5 ≈ 1.011, so the period increases by about 1.1%, roughly 1 minute.
The key judgement is that 'speeding up' the satellite with thrusters does not mean it moves faster in the new orbit; the burn raises the orbit, and the new circular speed is lower.
3. Identify one assumption in the circular-orbit model and describe a consequence if it fails.
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Assumption: Earth is a point mass and its mass is distributed spherically, so the gravitational field outside is exactly GM/r². In reality Earth is oblate and has mass anomalies, so the field is not perfectly central.
Consequence: the orbit is not a perfect circle; the satellite drifts, and the period calculated from GM/r³ is only an average. For precise prediction over many orbits, this matters.
Another assumption is that the orbit is circular. A brief thruster burn usually produces an elliptical orbit, so the speed varies around the orbit and the 'period' is the time for one full ellipse, given by Kepler's third law with the semi-major axis, not the current radius. If the burn is not tangential, the orbit also changes inclination, which the simple model ignores.