Why Do Satellites Shift from Circular to Elliptical Orbits?

This lesson explains how a small change in a satellite's speed affects its orbit, specifically how slowing down turns a circular orbit into an elliptical one. You'll learn how this impacts kinetic energy, potential energy, and the orbital period, answering key questions like: what happens when a satellite slows down?, why does a satellite follow an elliptical path?, and how do you calculate orbital period using Kepler’s law?

📘 What You’ll Learn

• How to calculate orbital period and speed in low Earth orbit
• What happens when a satellite reduces its velocity mid-orbit
• How to compute kinetic energy and gravitational potential energy of satellites
• How total energy defines orbital shape and period duration
• How elliptical orbits form from speed changes
• How to calculate satellite arrival time differences using orbital mechanics

🔑 Key Concepts Covered

• orbital speed calculation
• satellite in circular orbit
• elliptical orbit physics
• Kepler’s third law
• gravitational potential energy satellite
• kinetic energy in orbit
• orbital maneuver Δv
• time period of satellite
• satellite motion physics
• how elliptical orbits form

🧠 Why This Lesson Matters Understanding satellite motion and orbital mechanics is crucial for success in AP Physics, IB, and JEE. This lesson builds your ability to answer questions such as how to calculate orbital period, how energy changes affect orbits, and why does satellite follow elliptical path. These concepts are key to mastering topics in astrophysics, aerospace engineering, and competitive exam preparation.

🔗 Prerequisite or Follow-Up Lessons

• Gravitational Potential Energy and Orbits
• Kepler’s Laws of Planetary Motion

📖 Full Lesson: Orbital Maneuver and Energy in Satellite Motion

Step 1: Define the Orbital Radius For a satellite 400 km above Earth's surface:

r = (6370 + 400) km = 6770 km = 6.77 × 10⁶ m

Step 2: Find the Orbital Period (T₀) Using Kepler's third law:

T₀ = √[ (4π² r³) / (G Mᴇ) ] = √[ (4π² (6.77 × 10⁶)³) / ((6.67 × 10⁻¹¹)(5.98 × 10²⁴)) ] = 5.54 × 10³ s

Step 3: Compute Orbital Speed (v₀)

v₀ = (2πr) / T₀ = [2π (6.77 × 10⁶)] / (5.54 × 10³) = 7.68 × 10³ m/s

Step 4: New Kinetic Energy After Speed Reduction Blipp reduces speed by 1%:

v = 0.99 v₀ K = ½ m v² = ½ (2000)(0.99 × 7.68 × 10³)² = 5.78 × 10¹⁰ J

Step 5: Potential Energy Remains Unchanged

U = – (G Mᴇ m) / r = – [(6.67 × 10⁻¹¹)(5.98 × 10²⁴)(2000)] / (6.77 × 10⁶) = –11.8 × 10¹⁰ J

Step 6: Total Mechanical Energy (E)

E = K + U = 5.78 × 10¹⁰ – 11.8 × 10¹⁰ = –6.02 × 10¹⁰ J

Step 7: Find the Semi-Major Axis (a)

E = –G Mᴇ m / (2a) ➔ a = –(G Mᴇ m)/(2E) a = –[(6.67 × 10⁻¹¹)(5.98 × 10²⁴)(2000)] / (2 × –6.02 × 10¹⁰) = 6.63 × 10⁶ m

Step 8: Orbital Period in Elliptical Orbit (T)

T = √[ (4π² a³) / (G Mᴇ) ] = √[ (4π² (6.63 × 10⁶)³) / (6.67 × 10⁻¹¹ × 5.98 × 10²⁴) ] = 5.37 × 10³ s

Step 9: Time Advantage at Return to Point P Zapp is 90 s ahead initially. Blipp's new period is 170 s shorter. So Blipp returns 80 s earlier to point P than Zapp:

 Δt = 170 s – 90 s = 80 s

Key Takeaways

• Orbital radius fixes both speed and period
• Even a 1% Δv can shift an orbit significantly
• Total mechanical energy determines orbit shape and timing
• how to apply Kepler’s third law
• what happens when a satellite slows down
• how to calculate orbital period
• how elliptical orbits form
• satellite velocity and orbit shape
• energy change in satellite motion