🌍Orbital Motion
Orbits Are Just Free-Fall
Here's a mind-bending idea: an orbiting satellite is always falling toward Earth — it just keeps missing!
If you throw a ball horizontally fast enough, it falls toward Earth but Earth's surface curves away at the same rate. The ball is in orbit.
For a circular orbit, gravity provides the centripetal force:
GMm/r² = mv²/r
Solving for orbital speed: v = √(GM/r)
Notice: orbital speed does NOT depend on the mass of the satellite! A 1 kg and 1000 kg satellite at the same height orbit at the same speed.
Kepler's Third Law
For any orbit around the same central body:
T² ∝ r³ (Kepler's Third Law)
More precisely: T² = (4π²/GM) × r³
This means: - Closer orbits have shorter periods (move faster) - Farther orbits have longer periods (move slower) - The ISS (low orbit) takes 90 min per orbit; the Moon (far away) takes 27 days
This applies to planets around the Sun, moons around planets, and any satellite.
Think About It
Astronauts on the ISS appear "weightless." But the ISS orbits just 400 km up, where g ≈ 8.7 m/s² — close to Earth's surface value. Are they really weightless?
✏️ Worked Example
Problem: Find the orbital speed of the ISS at altitude 400 km above Earth. (R_Earth = 6.4×10⁶ m, M_Earth = 6×10²⁴ kg)
📐 Key Equations
Orbital motion
v_orbit = √((GM)/(r))T² = (4π²)/(GM) r³(GMm)/(r²) = (mv²)/(r) (circular orbit)⚠️ Common Mistakes
Misconception: Thinking a more massive satellite needs a higher orbit or faster speed to stay in orbit.
✓ Correct thinking: Orbital speed v = √(GM/r) depends only on the mass of the central body and the orbital radius, not the satellite's mass.
Why: The satellite's mass cancels out when you set F_gravity = F_centripetal. A 1 kg and 1000 kg object at the same orbit have identical speeds.
Misconception: Using altitude instead of orbital radius in Kepler's Third Law or the orbital speed formula.
✓ Correct thinking: r in T² = (4π²/GM)r³ and v = √(GM/r) is the orbital radius measured from the center of the planet, not the altitude above the surface.
Why: Altitude h and orbital radius r differ by one Earth radius: r = R_Earth + h. Using h alone gives an incorrect (too small) r.
Misconception: Believing that to slow down a satellite you fire rockets forward, and it will drop to a lower orbit moving slower.
✓ Correct thinking: This is actually correct! Firing retrograde (backward) rockets reduces orbital energy, the satellite spirals to a lower orbit and — counterintuitively — speeds up at the lower orbit.
Why: Lower orbits have smaller r, so v = √(GM/r) is larger. The satellite loses total energy but gains kinetic energy. This is one of the most surprising results of orbital mechanics.
📝 Practice Problems
Try these problems. Check your answer when ready.
A satellite orbits Earth at a radius of 8×10⁶ m. Find its orbital speed. (G = 6.67×10⁻¹¹ N·m²/kg², M_Earth = 6×10²⁴ kg)
v = √((GM)/(r))If satellite A orbits at radius r and satellite B orbits at radius 4r (same planet), how do their orbital periods compare?
T² ∝ r³Earth orbits the Sun with period 1 year at radius 1.5×10¹¹ m. Mars has an orbital radius of 2.28×10¹¹ m. Find Mars's orbital period in Earth years.
T² = (4π²)/(GM) r³A geosynchronous satellite must have a period of 24 hours. Find its orbital radius above Earth. (G = 6.67×10⁻¹¹ N·m²/kg², M_Earth = 6×10²⁴ kg)
r = \left((GMT²)/(4π²)\right)^1/3Show that the orbital speed of a satellite decreases as its orbital radius increases. Use v = √(GM/r) to justify the claim.
A moon orbits a planet of mass 4×10²³ kg with a period of 3×10⁵ s. Find the orbital radius.
r = \left((GMT²)/(4π²)\right)^1/3An astronaut orbits Earth in a circular orbit just above the surface (r ≈ R_Earth = 6.4×10⁶ m). Find the orbital speed and period. (g_surface = 9.8 m/s²)
v = √(gR), T = (2π R)/(v)Finished reading through this lesson?