Orbits — Circular Satellite Motion

We now put everything from the last three lessons to work. A satellite in orbit is a perfect laboratory for applying Newton's law of gravitation, the centripetal force equation, and conservation of energy all at once. In this lesson we derive the formulae for orbital speed, period, and energy, and work through a range of exam-style orbital problems. We also clarify one of the most persistent misunderstandings in physics — why astronauts in orbit appear weightless, even though gravity is pulling them just as hard as it pulls you right now.

This is OCR A-Level Physics A Module 5.4 (Gravitational fields).

---

The Condition for a Circular Orbit

A satellite of mass m in a circular orbit of radius r around a body of mass M must obey two constraints:

1. Gravity provides the centripetal force, so GMm/r² = mv²/r.
2. The gravitational force is perpendicular to the velocity, so it does no work, and the kinetic energy stays constant — the satellite moves at constant speed.

From the first constraint, cancelling m and rearranging:

v = √(GM / r)

This is the orbital speed at radius r. Some immediate consequences:

• Orbital speed depends only on M and r, not on the satellite's mass.
• Higher orbits have slower speeds — counter to many people's intuition.
• The closer you are to the central body, the faster you must move.

Worked Check — ISS Orbital Speed

Using M_E = 5.97 × 10²⁴ kg and r = 6.77 × 10⁶ m (ISS orbit):

v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.77 × 10⁶)
  = √(3.98 × 10¹⁴ / 6.77 × 10⁶)
  = √(5.88 × 10⁷)
  = 7.67 × 10³ m s⁻¹
  ≈ 7.67 km s⁻¹

The ISS zips around the Earth at nearly eight kilometres per second. At that speed you would cross London in less than five seconds.

---

Orbital Period — Kepler's Third Law Revisited

From the last lesson we derived

T² = 4π² r³ / (GM)

so

T = 2π √(r³ / (GM))

An equivalent form uses v = √(GM/r) and circumference 2πr:

T = 2πr / v

Both forms give the same answer.

Exam Tip: Choose whichever form matches your data. If you know r and M, use the T² = 4π²r³/GM form. If you already have v, use T = 2πr/v.

---

Orbital Energy

The kinetic energy of a satellite in a circular orbit is

E_k = ½mv² = ½m × (GM/r) = GMm/(2r)

The gravitational potential energy (from Lesson 9) is

E_p = −GMm/r

The total mechanical energy is therefore

E_total = E_k + E_p = GMm/(2r) − GMm/r = −GMm/(2r)

Several remarkable facts follow:

1. The total energy is negative, just like the potential energy. Negative total energy means the satellite is bound to its parent body — it cannot escape to infinity (where the energy would be at least zero).
2. E_k = −½ E_p — a general result known as the virial theorem for inverse-square force fields.
3. Higher orbits have larger (less negative) total energy. To move a satellite to a higher orbit, you must do positive work on it — even though its kinetic energy decreases in the higher orbit.

This last point is surprising and worth pausing on. A satellite in a high orbit moves slowly, so its kinetic energy is low. But its potential energy is much higher (less negative), so the total energy is greater. Boosting a satellite to a higher orbit costs more PE than you save in KE — hence the need for propulsion.

---

Worked Example 1 — Speed and Period at a Given Altitude

A geostationary satellite orbits at an altitude of 36 000 km above the Earth's surface. Calculate (a) its orbital radius, (b) its orbital speed, (c) its period.

(a)

r = R_E + h = 6.37 × 10⁶ + 3.60 × 10⁷ = 4.24 × 10⁷ m

(b)

v = √(GM/r) = √(3.98 × 10¹⁴ / 4.24 × 10⁷)
  = √(9.39 × 10⁶)
  = 3.07 × 10³ m s⁻¹
  ≈ 3.07 km s⁻¹

(c)

T = 2πr/v = 2π × 4.24 × 10⁷ / 3.07 × 10³
  = 2.66 × 10⁸ / 3.07 × 10³
  = 8.67 × 10⁴ s
  ≈ 24.1 hours

Just a shade under one sidereal day, as expected for a geostationary orbit (more on this in the next lesson).

---

Worked Example 2 — Orbital Decay

A low-orbit satellite is at 300 km altitude. Its orbit is decaying due to air resistance, and it falls to 200 km altitude. By how much does its orbital speed change? Does it speed up or slow down?

At 300 km (r₁ = 6.67 × 10⁶ m):

v₁ = √(3.98 × 10¹⁴ / 6.67 × 10⁶) = √(5.97 × 10⁷) = 7.73 × 10³ m s⁻¹

At 200 km (r₂ = 6.57 × 10⁶ m):

v₂ = √(3.98 × 10¹⁴ / 6.57 × 10⁶) = √(6.06 × 10⁷) = 7.78 × 10³ m s⁻¹