A 750 kg spacecraft circles a 6.0 × 10²⁴ kg planet at radius 8.0 × 10⁶ m. Use G = 6.67 × 10⁻¹¹ N m²/kg². (a) Find field strength, orbital speed and period. (b) Find angular speed and verify centripetal acceleration. (c) Find energy needed to move to a circular orbit of radius 1.20 × 10⁷ m using E = −GMm/(2r). (d) Explain why one forward burn first makes an ellipse.

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g = 6.25 m/s²; v = 7.07 km/s; T = 7.11 × 10³ s; ω = 8.84 × 10⁻⁴ rad/s; energy increase 6.25 × 10⁹ J. Field strength GM/r² = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴/(8.0 × 10⁶)² = 6.25 m/s². Circular balance gives v = √(GM/r) = 7.07 × 10³ m/s. Period 2πr/v = 7.11 × 10³ s. Angular speed is 2π/T = 8.84 × 10⁻⁴ rad/s. Check ω²r = 6.25 m/s², equal to gravitational acceleration in the circular orbit. Energy increase is GMm/2(1/r₁ − 1/r₂) = 6.25 × 10⁹ J. The less negative final energy represents added mechanical energy. A short forward burn changes velocity at one location, not position everywhere, so it creates an ellipse with the burn point at lower apsis. A second burn at the raised side circularises; the final larger circular speed is lower despite added energy.

Circular orbit, energy and transfer

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