A satellite orbiting a planet at a distance of [math: 8×106m] from its center has a period of 16 hours. What would be the period of a satellite orbiting at a distance of [math: 2×106m] above the planet’s center?

Question

BlackTom AI · Accepted Answer

Kepler's third law for circular orbits gives T ∝ r^(3/2), so the period scales with the 3/2 power of orbital radius.

1) Relevant knowledge points and concepts
- For a satellite in a circular orbit, centripetal force is provided by gravity: m v^2 / r = G M m / r^2.
- The orbital speed v = sqrt(GM / r).
- Period T = circumference / speed = 2π r / v, which leads to T = 2π sqrt(r^3 / (GM)). Therefore for a given planet (constant GM), T ∝ r^(3/2).

2) Complete reasoning step by step
- Let T1 = 16 hr at radius r1 = 8 × 10^6 m.
- Let T2 be the period at radius r2 = 2 × 10^6 m.
- Using the proportionality: T2 / T1 = (r2 / r1)^(3/2).

3) Calculations with intermediate steps
- Compute the radius ratio: r2 / r1 = (2 × 10^6) / (8 × 10^6) = 1/4.
- Raise to the 3/2 power: (1/4)^(3/2) = (1/4)^(1.5) = (1/4)*(1/4)^(0.5) = (1/4)*(1/2) = 1/8.
- Multiply by T1: T2 = 16 hr × (1/8) = 2 hr.

4) Why the correct answer is correct
Because the period depends on the 3/2 power of orbital radius, reducing the radius by a factor of 4 reduces the period by a factor of 4^(3/2) = 8, so 16 hr / 8 = 2 hr.

5) Why the other choices are incorrect (brief)
- a. 4 hr: would imply a period reduction factor of 4 (not the required 8) and does not match T ∝ r^(3/2).
- c. 8 hr and d. 6 hr: neither corresponds to the correct (1/4)^(3/2) scaling from r1 to r2.
- e. 32 hr: increases the period, which would require a larger radius, not a smaller one.

6) Conclusion
The period at 2 × 10^6 m is 2 hours, which matches the provided correct answer (b. 2 hr).

A satellite orbiting a planet at a distance of [math: 8×106m] from its center has a period of 16 hours. What would be the period of a satellite orbiting at a distance of [math: 2×106m] above the planet’s center?单项选择题

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