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Three satellites orbit Earth: satellite 1 in a circular orbit of radius R1, satellite 2 in a circular orbit of radius R2 (R2 > R1), and satellite 3 in an elliptical orbit with perigee at R1 and apogee at R2. Let vₖ1, vₖ2 be the circular speeds at R1, R2; vₚ, vₐ the speeds at perigee and apogee of the ellipse; and T1, T2, T3 the corresponding periods. Which of the following are correct? (A) vₐ < vₖ2 < vₖ1 < vₚ (B) vₐ < vₖ1 < vₚ < vₖ2 (C) T1 = 2*pi*sqrt(R1³ / (GM)) (D) T3 = pi*sqrt((R1+R2)³ / (2*GM))

1. vₐ < vₖ2 < vₖ1 < vₚ
2. vₐ < vₖ1 < vₚ < vₖ2
3. T1 = 2*pi*sqrt(R1³ / (GM))
4. T3 = pi*sqrt((R1+R2)³ / (2*GM))

Correct answer: vₐ < vₖ2 < vₖ1 < vₚ

Solution

For circular orbits vₖ1 > vₖ2 (inner faster). For the elliptical orbit: at perigee (R1), the satellite has more energy than the circular orbit at R1, so vₚ > vₖ1. At apogee (R2), vₐ < vₖ2. Thus vₐ < vₖ2 < vₖ1 < vₚ. For T1: standard formula applies. For T3: semi-major axis a=(R1+R2)/2, T3=2pi*sqrt(a³/(GM))=2pi*sqrt((R1+R2)³/(8GM))=pi*sqrt((R1+R2)³/(2GM)). Options A, C, D are all correct.

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