Motion of Orbiting Satellites — Worked Solutions

These solutions belong to Motion of Orbiting Satellites. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

For U = −GMm/r with zero at infinity, what happens to U as a satellite moves closer to the central object?

1. Use U = −GMm/r, with zero at infinity. Why: The orbital potential energy is for the two-object system and uses center distance.

2. Smaller positive r makes U more negative. Why: Inward motion decreases potential energy and can increase kinetic energy at fixed total E.

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Question 2

Which quantities stay constant in an unpowered elliptical orbit with gravity as the only interaction?

1. Gravity is conservative and radial about the central object. Why: A conservative interaction keeps mechanical energy constant, and a radial force has zero torque there.

2. E and L stay constant while K, U, speed, and radius can change. Why: An elliptical orbit is not uniform circular motion.

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Question 3

How does threshold escape speed compare with circular-orbit speed at the same radius?

1. Circular motion gives v_c² = GM/r; threshold escape energy gives v_e² = 2GM/r. Why: The circular condition comes from force, while escape comes from E = 0.

2. Taking the ratio and positive square root gives v_e/v_c = √2. Why: The two speeds are compared at the same starting radius and under the same gravity-only model.

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Question 4

For a model planet, GM = 3.20 × 10¹⁴ m³/s². A satellite is at center-to-center distance 8.0 × 10⁶ m. Find its circular speed and the threshold escape speed there.

1. Use center distance r = 8.0 × 10⁶ m and the supplied GM. Why: Orbital expressions use center-to-center distance, not surface height.

2. v_c = √(GM/r) = √(4.0 × 10⁷) ≈ 6.32 × 10³ m/s. Why: Circular motion requires gravity to supply the centripetal force.

3. v_e = √(2GM/r) ≈ 8.94 × 10³ m/s. Why: Threshold escape sets total mechanical energy to zero, with zero speed at infinity in the ideal model.

4. The ratio is √2 and satellite mass cancels from both. Why: The same gravitational and inertial mass appears in the force or energy balance.

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Question 5

A 100 kg satellite has a gravity-only elliptical orbit around a model central object with GM = 1.20 × 10¹⁴ m³/s². At its nearest point, r = 2.0 × 10⁷ m and speed is 3.0 × 10³ m/s; at its farthest point, r = 6.0 × 10⁷ m. Predict the far-point speed and build labeled kinetic/potential energy bars at both points using zero gravitational potential energy at infinity. Decide whether the orbit is bound. Explain why the same satellite can change speed while keeping angular momentum constant.

1. At the near and far turning points, velocity is perpendicular to radius. Use r_near v_near = r_far v_far to get v_far = (2.0 × 10⁷)(3.0 × 10³)/(6.0 × 10⁷) = 1.0 × 10³ m/s. Why: Gravity is radial, so its torque about the central object is zero; the simple mrv form applies at these turning points.

2. Near the central object, K = ½(100)(3.0 × 10³)² = 4.5 × 10⁸ J and U = −(1.20 × 10¹⁴)(100)/(2.0 × 10⁷) = −6.0 × 10⁸ J. Why: Kinetic energy is positive, while the infinity-zero gravitational potential energy is negative and lower at smaller radius.

3. At the far point, K = 5.0 × 10⁷ J and U = −2.0 × 10⁸ J. Both bar charts sum to E = −1.5 × 10⁸ J, so the orbit is bound in the stated gravity-only model. Why: The decrease in K offsets the increase in U. Negative total energy is below the threshold needed to reach infinity with zero speed.

4. Draw force toward the central object at both points and show its line of action through that object. Why: The zero lever arm explains constant angular momentum even while conservative gravity exchanges kinetic and potential energy and changes speed.

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