Motion of Orbiting Satellites: Study Guide (Physics 1 6.6)

This guide is for the algebra-based Physics 1 course. It uses algebra and graphs only. It brings together circular orbits from Topic 2.9, gravitational potential energy from Topic 3.3 and conservation of angular momentum from Topic 6.4. The planets, moons and probes in the examples are fictional.

A two-object system

Take a system of two objects that interact only through gravity: a planet of mass M and a satellite of mass m. No other forces act. By Newton’s third law, the planet pulls the satellite with a force of size GMm/r², and the satellite pulls the planet with the same size of force.

Equal forces, very different masses. The accelerations are GM/r² for the satellite and Gm/r² for the planet, so their ratio is m/M. For a 1000 kg satellite and a planet of 6.0 × 10²⁴ kg, the planet’s acceleration is about 1.7 × 10⁻²² of the satellite’s. The centre of mass of the system sits, for all practical purposes, at the planet’s centre.

So when the satellite’s mass is negligible compared with the central body’s, the central body’s motion is negligible. You treat it as fixed and study the satellite alone, measuring every distance r from the central body’s centre. (If the two masses were similar, as in a pair of stars, both would move about their common centre of mass. That case is not needed here.)

Two conservation laws constrain every orbit

Angular momentum. Gravity on the satellite always points at the centre of the planet. A force that points through a point has no lever arm about it, so its torque about the planet’s centre is zero. With no torque, the satellite’s angular momentum about that centre is constant (Topic 6.4):

L = mvr sin θ = constant

Here θ is the angle between the satellite’s velocity and the line from the planet to the satellite.

Mechanical energy. Gravity is the only force, and it is internal to the satellite–planet system. So the system’s mechanical energy is constant:

E = K + U_g = ½mv² − GMm/r = constant

U_g is defined to be zero when the satellite is infinitely far away. At any finite distance it is negative, and it gets more negative as the satellite gets closer.

Circular orbits: everything is constant

In a circular orbit, gravity alone provides the centripetal force (Topic 2.9):

GMm/r² = mv²/r, so v² = GM/r

Multiply by ½m to get the kinetic energy, then add the potential energy:

• K = ½mv² = GMm/(2r)
• U_g = −GMm/r
• E = K + U_g = −GMm/(2r)

So in a circular orbit, E = −K and U_g = 2E. Because r and v do not change, K, U_g, E and L = mvr are all constant. (The velocity is always perpendicular to the radius, so sin θ = 1.)

Comparing orbits of different radius

Figure 1 shows how the three energies depend on the orbital radius. Read it carefully: it compares different circular orbits. A satellite stays at one value of r.

Moving to a larger circular orbit:

• v and K decrease (the satellite is slower).
• U_g increases (becomes less negative).
• E increases (becomes less negative), so energy must be supplied.
• L = mvr increases, since v ∝ 1/√r and so L ∝ √r. Moving to a new orbit needs an external push, such as a rocket burn.

Elliptical orbits: energy and angular momentum trade

An elliptical orbit brings the satellite closer to the planet and then farther away. Gravity still exerts no torque about the planet’s centre, and it is still the only force. So E and L stay constant, but K and U_g change.

At the closest and farthest points, the velocity is perpendicular to the radius (Figure 2), so L = mvr there and

v_near r_near = v_far r_far

The satellite moves fastest at its closest point and slowest at its farthest point. Energy tells the same story: as the satellite moves away, gravity has a component opposite to its velocity, does negative work, and turns kinetic energy into potential energy. On the way back in, the transfer reverses. (This faster-when-closer behaviour is what Kepler’s second law describes. The course does not ask you to know that law; argue from angular momentum or energy.)

Zero at infinity: bound and unbound

Because U_g = 0 at infinity, the sign of E tells you whether the satellite can get away:

Total energy E	What happens
E < 0	bound: the satellite cannot reach infinity; it stays in orbit or falls back
E = 0	just escapes: its speed falls towards zero as r grows without limit
E > 0	escapes with speed to spare: it still moves at a finite speed very far away

Every circular and elliptical orbit has E < 0.

Escape speed

The escape speed from distance r is the speed that makes the total mechanical energy exactly zero:

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

A satellite launched at exactly this speed, with only gravity acting, moves away forever. Its speed keeps falling and reaches zero only at an infinite distance.

Three things to notice:

• m cancels. A heavy probe and a light probe need the same escape speed (though the heavy one needs more energy).
• Escape speed is √2 times the circular-orbit speed √(GM/r) at the same r. A satellite in a circular orbit needs its speed raised by a factor of 1.41 to escape.
• For Earth, using G, M = 5.97 × 10²⁴ kg and R = 6.37 × 10⁶ m, v_esc = √(2GM/R) ≈ 11.2 km/s from the surface, ignoring air resistance.

Worked example 1: a satellite in a circular orbit

Question. An 800 kg satellite moves in a circular orbit of radius 1.0 × 10⁷ m around a fictional planet of mass 6.0 × 10²⁴ kg. (a) Find its speed, K, U_g, E and L. (b) It is moved to a circular orbit of radius 2.0 × 10⁷ m. Find the change in each energy.

1. GM = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ = 4.00 × 10¹⁴ N·m²/kg.
2. (a) v = √(GM/r) = √(4.00 × 10⁷) = 6.32 × 10³ m/s.
3. K = ½mv² = ½ × 800 × 4.00 × 10⁷ = 1.60 × 10¹⁰ J.
4. U_g = −GMm/r = −4.00 × 10¹⁴ × 800 ÷ 1.0 × 10⁷ = −3.20 × 10¹⁰ J.
5. E = K + U_g = −1.60 × 10¹⁰ J. Check: E = −K. ✓
6. L = mvr = 800 × 6.32 × 10³ × 1.0 × 10⁷ = 5.06 × 10¹³ kg·m²/s.
7. (b) Doubling r halves each energy: K = 8.0 × 10⁹ J, U_g = −1.60 × 10¹⁰ J, E = −8.0 × 10⁹ J.
8. Changes: ΔK = −8.0 × 10⁹ J, ΔU_g = +1.60 × 10¹⁰ J, ΔE = +8.0 × 10⁹ J.

Interpretation. The satellite ends up slower, yet the system has more energy. The 8.0 × 10⁹ J must come from outside the satellite–planet system, for example from rocket fuel. The potential energy rises by twice as much as the kinetic energy falls.

Worked example 2: speed and energy around an elliptical orbit

Question. A 500 kg probe orbits the same planet (GM = 4.00 × 10¹⁴ N·m²/kg) on the ellipse in Figure 2. At its closest point, r = 8.0 × 10⁶ m and v = 8.66 × 10³ m/s. Its farthest point is at r = 2.4 × 10⁷ m. (a) Find its speed at the farthest point. (b) Show that the mechanical energy is the same at both points.

1. (a) Gravity exerts no torque about the planet’s centre, so L is constant. At both points v ⟂ r: v_far = v_near r_near / r_far = 8.66 × 10³ × (8.0 × 10⁶ ÷ 2.4 × 10⁷) = 2.89 × 10³ m/s.
2. (b) Closest point: K = ½ × 500 × (8.66 × 10³)² = 1.875 × 10¹⁰ J; U_g = −4.00 × 10¹⁴ × 500 ÷ 8.0 × 10⁶ = −2.50 × 10¹⁰ J; E = −6.25 × 10⁹ J.
3. Farthest point: K = ½ × 500 × (2.887 × 10³)² = 0.208 × 10¹⁰ J; U_g = −4.00 × 10¹⁴ × 500 ÷ 2.4 × 10⁷ = −0.833 × 10¹⁰ J; E = −6.25 × 10⁹ J.
4. The two totals agree, to the precision of the given speed.

Interpretation. Between the two points, K falls by about 1.67 × 10¹⁰ J and U_g rises by the same amount. E < 0, so the probe is bound. Its closest-point speed (8.66 km/s) is more than the circular speed there (7.07 km/s) but less than the escape speed there (10.0 km/s), which is why it swings out on an ellipse and comes back.

Worked example 3: escaping from a small moon

Question. A fictional moon has mass 9.0 × 10²² kg and radius 1.5 × 10⁶ m, and no atmosphere. (a) Find the escape speed from its surface. (b) A probe is launched straight up at 2.0 km/s. How far from the moon’s centre does it get? (c) Another probe is launched at 3.5 km/s. What speed does it have very far away?

1. GM = 6.67 × 10⁻¹¹ × 9.0 × 10²² = 6.00 × 10¹² N·m²/kg.
2. (a) v_esc = √(2GM/R) = √(2 × 6.00 × 10¹² ÷ 1.5 × 10⁶) = √(8.00 × 10⁶) = 2.83 × 10³ m/s.
3. (b) Energy per kilogram is conserved, and the probe stops at r_max: ½v² − GM/R = −GM/r_max.
4. ½v² = 2.0 × 10⁶ J/kg and GM/R = 4.00 × 10⁶ J/kg, so −GM/r_max = −2.00 × 10⁶ J/kg.
5. r_max = 6.00 × 10¹² ÷ 2.00 × 10⁶ = 3.0 × 10⁶ m, twice the radius: 1.5 × 10⁶ m above the surface.
6. (c) ½v² − GM/R = ½v_far² − 0, so v_far² = v² − v_esc² = (3.5 × 10³)² − 8.00 × 10⁶ = 4.25 × 10⁶. v_far = 2.1 × 10³ m/s.

Check. In (c), v_far is less than the launch speed but not zero, matching E > 0 in the table. A common error in (b) is to use mgh with the surface value of g; that assumes the field stays the same all the way up, which fails for a climb this large.

Common misconceptions

• “The planet does not move at all.” It does move, but by a negligible amount when m ≪ M. That is why you may treat it as fixed.
• “A satellite in a higher orbit has more kinetic energy.” It has less K but more total E (Worked example 1).
• “The total energy of an orbit should be positive.” With U_g = 0 at infinity, every bound orbit has E < 0.
• “A satellite in an elliptical orbit moves at constant speed.” Only K + U_g is constant. Speed is greatest at the closest point.
• “Use L = mvr anywhere on an ellipse.” Only where v is perpendicular to r, at the closest and farthest points. Elsewhere L = mvr sin θ.
• “A heavier rocket needs a higher escape speed.” The mass cancels. It needs more energy, not more speed.
• “At escape speed the probe stops at some large distance.” Its speed approaches zero only as r grows without limit.
• Measuring r from the surface. r is from the centre of the central body.

Where this leads

This topic closes Unit 6. Unit 7 (Oscillations) uses the same energy reasoning for objects that move back and forth. Read the Topic 7.1 study guide next. First, try the practice questions, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.