Motion of Orbiting Satellites: Study Guide (Physics C: Mechanics 6.6)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 6.6 called Motion of Orbiting Satellites. This guide is the calculus-based one. It derives the potential energy and the circular-orbit energy relations, and uses both conservation laws together to solve elliptical orbits. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

The system: a massive body and a light satellite

Take a system of two objects that interact only through gravity: a central body of mass M (a planet or star) and a satellite of mass m. No outside forces act, so the system’s total momentum is constant and its centre of mass does not accelerate. Both bodies actually orbit that centre of mass.

The momenta of the two bodies are equal and opposite in the centre-of-mass frame, so Mv_M = mv_m and

v_M ÷ v_m = m ÷ M

When m ≪ M, the central body’s speed is tiny and the centre of mass sits almost exactly at the central body’s centre. So we treat the central body as fixed and study only the satellite. In Worked example 1, the planet’s speed works out at about 2 × 10⁻¹⁸ m/s.

Gravitational potential energy with the zero at infinity

The gravitational force on the satellite points towards the central body. Taking r outward, its radial component is F_r = −GMm/r². Potential energy is defined by the work done by this force. Choose U = 0 when the separation is infinite. Then

U(r) = −∫ from ∞ to r of F_r dr′ = −∫ from ∞ to r of (−GMm/r′²) dr′ = −[GMm/r′] from ∞ to r

U_g = −GMm/r

Three things to notice:

• U is negative at every finite r. It rises towards zero as r grows. Moving the satellite outward always increases U.
• The sign comes from the choice of zero. Only changes in U have physical meaning: ΔU = GMm(1/r₁ − 1/r₂) when the satellite moves out from r₁ to r₂.
• U belongs to the satellite–central-body system, not to the satellite alone. Kinetic energy, by contrast, is the satellite’s (the central body’s is negligible).

A system whose total mechanical energy E = K + U is negative is bound: the satellite cannot reach r = ∞, where it would need K = E < 0, which is impossible.

What is conserved, and why

Two conservation laws control every orbit:

1. Energy. Gravity is the only force, it is internal to the system and it is conservative. So E = ½mv² − GMm/r is constant.
2. Angular momentum about the central body. The force on the satellite points along the line to the centre, so its torque about the centre, τ = r × F, is zero. So the satellite’s angular momentum L = mvr sin φ is constant (φ is the angle between r and v).

What this means for each shape of orbit:

Quantity	Circular orbit	Elliptical orbit
Total mechanical energy E	constant	constant
Angular momentum L of the satellite	constant	constant
Gravitational potential energy U	constant (r is fixed)	changes: lowest at the closest point
Kinetic energy K (and speed)	constant	changes: greatest at the closest point

In an elliptical orbit, energy flows back and forth between K and U while their sum stays fixed.

Energy in a circular orbit

For a circular orbit of radius r, gravity alone supplies the centripetal force (Topic 2.10):

GMm/r² = mv²/r, so mv² = GMm/r

Then the kinetic energy is

K = ½mv² = GMm/(2r) = −½U

and the total energy is

E = K + U = GMm/(2r) − GMm/r = −GMm/(2r) = ½U = −K

Each result has a meaning you can check:

• E is negative, so the orbit is bound.
• A larger orbit has a less negative E, so moving to a higher orbit needs energy input.
• But K = GMm/(2r) gets smaller as r grows. The satellite in a higher orbit moves more slowly. The energy you add, plus the kinetic energy it loses, all goes into U: ΔU = −2ΔK and ΔE = −ΔK.

Elliptical orbits: using both conservation laws

In an elliptical orbit the distance r changes. The closest point is called periapsis (perigee for an Earth orbit) and the farthest apoapsis. At these two points, and only these two, the velocity is at right angles to the radius line, so L = mvr exactly. Conservation of angular momentum then gives

r_p v_p = r_a v_a

So the satellite moves fastest where it is closest. Combine this with energy conservation,

½mv_p² − GMm/r_p = ½mv_a² − GMm/r_a

and you have two equations for two unknowns. A useful result (background: you can derive it, but it is not a required equation) is that eliminating v_a gives

v_p² = 2GM r_a ÷ [r_p(r_p + r_a)] and E = −GMm ÷ (r_p + r_a)

For a circle, r_p = r_a = r, and E = −GMm/(2r) as before. ✓

Escape velocity

A satellite escapes if it can reach r → ∞. There U = 0, so it needs K ≥ 0 there, which means E ≥ 0. The escape velocity is the launch speed that makes E exactly zero:

½mv_esc² − GMm/r = 0, so v_esc = √(2GM/r)

• It does not depend on the satellite’s mass, or on the direction of launch (if nothing is in the way).
• A body launched at exactly v_esc keeps moving away and slows down. Its speed at distance r′ is v_esc√(r/r′), which tends to zero as r′ → ∞. It never turns back.
• A faster launch leaves the body with speed v_∞ = √(v₀² − v_esc²) far away. A slower one gives E < 0: the body stops at some r_max and falls back, or stays in a bound orbit.
• At the same r, v_esc = √2 × v_circular. A satellite in a circular orbit escapes if its speed increases by a factor of √2 (about 41 %).

Worked example 1: raising a circular orbit

Question. Corvan is a fictional planet of mass 3.0 × 10²⁴ kg. A 1200 kg satellite moves in a circular orbit of radius 8.0 × 10⁶ m. (a) Find its speed, K, U, E and L. (b) It is moved to a circular orbit of radius 1.2 × 10⁷ m. Find the new values and the energy that had to be supplied. (c) How much must its speed increase, in the original orbit, for it to escape? (d) How fast does Corvan move about the system’s centre of mass?

(a) GM = 6.67 × 10⁻¹¹ × 3.0 × 10²⁴ = 2.00 × 10¹⁴ N·m²/kg.

1. v = √(GM/r) = √(2.00 × 10¹⁴ ÷ 8.0 × 10⁶) = 5.0 × 10³ m/s.
2. U = −GMm/r = −(2.00 × 10¹⁴ × 1200) ÷ 8.0 × 10⁶ = −3.0 × 10¹⁰ J.
3. K = −½U = 1.5 × 10¹⁰ J (check: ½ × 1200 × 5000² = 1.5 × 10¹⁰ J ✓).
4. E = ½U = −1.5 × 10¹⁰ J.
5. L = mvr = 1200 × 5.0 × 10³ × 8.0 × 10⁶ = 4.8 × 10¹³ kg·m²/s.

(b) At r = 1.2 × 10⁷ m: v = 4.1 × 10³ m/s, U = −2.0 × 10¹⁰ J, K = 1.0 × 10¹⁰ J, E = −1.0 × 10¹⁰ J, L = 5.9 × 10¹³ kg·m²/s.

• Energy supplied: ΔE = −1.0 × 10¹⁰ − (−1.5 × 10¹⁰) = +5.0 × 10⁹ J.
• K fell by 5.0 × 10⁹ J while U rose by 1.0 × 10¹⁰ J. The satellite is slower, by a factor √(8.0/12) = 0.82, in the higher orbit.
• L changed because the move needed thrust, which is an outside force on the satellite. In each circular orbit L is constant.

(c) v_esc = √2 × 5.0 × 10³ = 7.07 × 10³ m/s, so the speed must rise by about 2.1 × 10³ m/s. That raises K by 1.5 × 10¹⁰ J and brings E to zero.

(d) v_M = (m/M)v = (1200 ÷ 3.0 × 10²⁴) × 5.0 × 10³ = 2 × 10⁻¹⁸ m/s. Treating the planet as fixed is an excellent model.

Worked example 2: an elliptical orbit from its two ends

Question. A 600 kg probe orbits Corvan (GM = 2.00 × 10¹⁴ N·m²/kg) on an ellipse. Its closest distance from the planet’s centre is r_p = 7.0 × 10⁶ m and its farthest is r_a = 2.1 × 10⁷ m (Figure 2). (a) Find its speeds at both ends. (b) Find its total mechanical energy. (c) Describe how K and U change around the orbit.

(a) Angular momentum: r_p v_p = r_a v_a, so v_a = v_p r_p/r_a = v_p/3.

Energy: ½v_p² − GM/r_p = ½(v_p/3)² − GM/r_a (m cancels). Rearranging:

½(1 − 1/9)v_p² = (4/9)v_p² = GM(1/r_p − 1/r_a) = 2.00 × 10¹⁴ × (1/7.0 × 10⁶ − 1/2.1 × 10⁷) = 1.91 × 10⁷ m²/s²

So v_p = 6.5 × 10³ m/s (6547 m/s) and v_a = 2.2 × 10³ m/s (2182 m/s).

(b) E = ½ × 600 × 6547² − (2.00 × 10¹⁴ × 600) ÷ 7.0 × 10⁶ = 1.286 × 10¹⁰ − 1.714 × 10¹⁰ = −4.3 × 10⁹ J. Check at apoapsis: 1.43 × 10⁹ − 5.71 × 10⁹ = −4.3 × 10⁹ J ✓, and −GMm/(r_p + r_a) = −4.3 × 10⁹ J ✓.

(c) From periapsis to apoapsis, U rises from −1.71 × 10¹⁰ J to −5.7 × 10⁹ J, and K falls by the same 1.14 × 10¹⁰ J, from 1.29 × 10¹⁰ J to 1.4 × 10⁹ J. K at periapsis is 9 times K at apoapsis, because the speed is 3 times larger. Then the process reverses on the way back in.

Checks. At r_p, the circular-orbit speed would be √(GM/r_p) = 5.3 × 10³ m/s and the escape speed 7.6 × 10³ m/s. The probe’s 6.5 × 10³ m/s lies between them: faster than circular, so it swings out, but not fast enough to escape, consistent with E < 0.

Common misconceptions

• “U = −GMm/r is negative, so it is less than nothing.” The sign comes from putting the zero at infinity. Only changes in U matter.
• “Higher orbit, faster satellite.” In circular orbits v = √(GM/r): higher means slower, even though E is larger.
• “Kinetic energy is constant in every orbit.” Only in circular orbits. In an ellipse gravity has a component along the velocity and does work.
• Using L = mvr everywhere on an ellipse. Only at the two ends, where v is at right angles to r. Elsewhere use mvr sin φ.
• “Escape velocity depends on the satellite’s mass.” Mass cancels.
• “At escape velocity, the body keeps a steady speed.” It slows towards zero as r → ∞.
• Using altitude for r. r is measured from the centre of the central body.
• Forgetting that L changes when thrusters fire. Conservation laws hold only while gravity is the only force.

Where this leads

This topic closes Unit 6. You have now used energy and angular momentum together for rolling bodies and orbits. Unit 7 turns to oscillations: Topic 7.1, Defining Simple Harmonic Motion, again starts from a force law and uses energy to describe the motion. Next, try the practice questions, then use the revision notes and the checklist. You can also go back to Topic 6.5, Rolling, or return to the course roadmap.