Step by Step
1
Orbital velocity — where the formula comes from
For a circular orbit, gravity itself supplies the centripetal force keeping the object on its curved path: GMm/r² = mv²/r. The orbiting mass (m) cancels out completely — meaning orbital speed depends only on the central body's mass (M) and the orbital radius (r), never on the mass of the satellite itself. Solving for v gives v = √(GM/r).
2
Orbital period — how long one trip around takes
Once you know v, the period follows from distance ÷ speed: the circumference of the orbit (2πr) divided by the orbital velocity. Substituting v = √(GM/r) in and simplifying gives T = 2π√(r³/GM) — the same underlying relationship as Kepler's Third Law, just derived from force and motion instead of from observation.
3
Escape velocity — how fast is fast enough to leave
Escape velocity is the speed needed for an object's kinetic energy to exactly cancel out the gravitational potential energy binding it to a body, so it never falls back. Setting ½mv² = GMm/r and solving gives v_esc = √(2GM/r) — exactly √2 times the circular orbital velocity at that same radius. This √2 relationship is worth memorizing on its own.
4
Worked example: the ISS
The ISS orbits at about 400 km altitude, so r = Earth's radius + altitude = 6,371 km + 400 km = 6,771 km = 6.771×10⁶ m. Using GM_Earth = 3.986×10¹⁴ m³/s²: v = √(3.986×10¹⁴ / 6.771×10⁶) ≈ 7,670 m/s (about 7.67 km/s). Period: T = 2πr/v ≈ 5,545 seconds ≈ 92.4 minutes — matching the ISS's real orbital period almost exactly.
5
Worked example: escaping Earth entirely
From Earth's surface, r = 6,371 km = 6.371×10⁶ m. v_esc = √(2 × 3.986×10¹⁴ / 6.371×10⁶) ≈ 11,190 m/s ≈ 11.2 km/s — the well-known figure for Earth's escape velocity, and exactly √2 times the ~7.9 km/s circular orbital velocity at Earth's surface.
Applied Walkthrough
1
Start with what's given: the mass of the body being orbited (M) and the radius of the orbit (r), measured from the center of that body — not from its surface.
2
Plug those into v = √(GM/r) to get orbital velocity — notice the orbiting object's own mass never enters the calculation, which is why a pebble and a space station at the same altitude orbit at the same speed.
3
Use that velocity (or go straight from r and GM) to find the orbital period with T = 2π√(r³/GM) — a satellite closer in always orbits faster and completes its loop sooner than one farther out.
4
If the goal instead is to leave the body's gravity entirely, multiply the circular orbital velocity at that radius by √2 to get escape velocity — the same relationship that gives Earth's orbital speed of ~7.9 km/s and escape speed of ~11.2 km/s at the surface.
Exam Application
Exams test whether you can correctly plug values into v = √(GM/r) and T = 2π√(r³/GM) — including using the correct r (measured from the planet's center, not its surface) — and whether you know that escape velocity is always exactly √2 times the circular orbital velocity at the same radius.
⚠ Common Trap
The most common trap is using altitude instead of orbital radius in these formulas. A satellite's altitude is measured from the planet's surface, but r in every formula here must be measured from the planet's center — for the ISS, that means adding Earth's radius (6,371 km) to its 400 km altitude to get the correct r of 6,771 km, not using 400 km directly.
✓ Quick Self-Check
1. What is the formula for circular orbital velocity, and why does the orbiting object's own mass not appear in it?
v = √(GM/r). The orbiting mass (m) cancels out of the centripetal-force equation, so orbital speed depends only on the central body's mass and the orbital radius.
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2. What is the formula for orbital period, and how is it related to orbital velocity?
T = 2π√(r³/GM), derived from period = circumference (2πr) ÷ velocity, substituting v = √(GM/r).
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3. How does escape velocity relate to circular orbital velocity at the same radius?
Escape velocity is always exactly √2 (about 1.41) times the circular orbital velocity at that same radius: v_esc = √(2GM/r).
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4. For the ISS at 400 km altitude, what value of r should be used in these formulas — and why?
r = 6,771 km (Earth's 6,371 km radius + 400 km altitude), because r must be measured from the planet's center, not its surface.
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5. What is Earth's approximate escape velocity from the surface, and how does it compare to circular orbital velocity there?
About 11.2 km/s — roughly √2 times the ~7.9 km/s circular orbital velocity at Earth's surface.
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