Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. AP Physics 1
  2. Motion of Orbiting Satellites

AP PHYSICS 1: ALGEBRA-BASED • ENERGY AND MOMENTUM OF ROTATING SYSTEMS

Motion of Orbiting Satellites

Discover how gravity, energy, and momentum govern the paths of every object in orbit.

SECTION 1

Historical Context & Motivation

The question of how celestial bodies move has captivated thinkers for millennia, from ancient Greek astronomers who imagined crystalline spheres to Renaissance scientists who dared to place the Sun at the center of the solar system. The physics of orbiting satellites rests on a chain of discoveries that began with careful naked-eye observations and culminated in Newton's universal law of gravitation. Understanding this history reveals how a single force—gravity—can explain both the fall of an apple and the orbit of the Moon.

1609
Kepler's Laws of Planetary Motion
Johannes Kepler published his first two laws—planets move in ellipses with the Sun at one focus, and a line from the Sun to a planet sweeps equal areas in equal times—providing the first accurate mathematical description of orbital paths.
1687
Newton's Principia
Isaac Newton demonstrated that an inverse-square gravitational force produces Kepler's elliptical orbits, unifying terrestrial and celestial mechanics in a single framework and enabling quantitative predictions of satellite motion.
1798
Cavendish Measures G
Henry Cavendish used a torsion balance to measure the gravitational constant G, making it possible to calculate the mass of the Earth and predict precise orbital parameters for artificial satellites.
1957
Sputnik 1 — First Artificial Satellite
The Soviet Union launched Sputnik 1 into low Earth orbit, confirming centuries of theoretical predictions and inaugurating the space age. Its orbital period matched Newtonian calculations to high precision.

The central question this lesson addresses is deceptively simple: what determines the speed, period, and energy of a satellite in orbit? The answer ties together Newton's law of gravitation, circular-motion dynamics, and conservation of energy—three pillars that appear repeatedly on the AP Physics 1 exam.

SECTION 2

Core Principles & Definitions

An orbiting satellite is in perpetual free fall toward the central body; it simply moves forward fast enough that the curvature of its path matches the curvature of the planet beneath it. Three foundational ideas govern this motion and connect directly to the energy and momentum framework of rotating systems.

1

Gravitational Force as Centripetal Force

For a satellite in circular orbit, the gravitational attraction toward the central body is the net centripetal force. No tangential force acts, so the satellite's speed remains constant.
2

Orbital Speed Depends Only on Radius

For a given central body, the orbital speed of a satellite is uniquely determined by the orbital radius—heavier satellites orbit at the same speed as lighter ones at the same altitude.
3

Mechanical Energy Is Negative and Conserved

A bound satellite has negative total mechanical energy. The gravitational potential energy is negative and twice the magnitude of the kinetic energy, keeping the satellite trapped in orbit.
4

Angular Momentum Conservation

With no external torque about the central body, a satellite's angular momentum L = mvr is conserved, which explains Kepler's equal-area law and velocity changes in elliptical orbits.
✦ KEY TAKEAWAY
KEY TAKEAWAY
SECTION 3

Visual Explanation — Circular Orbit Force & Velocity

M (Earth)Satellite (m)FgvrSatellite at later timeFgvVelocity is always tangent to the orbit; gravitational force always points radially inward.The speed |v| remains constant for a circular orbit.
A satellite of mass m orbits a central body of mass M at radius r. The red arrows show the gravitational force (always radially inward), while the cyan arrows show the velocity (always tangent to the orbit). At two different positions along the orbit, the force and velocity directions change, but their magnitudes stay constant.

The diagram above captures the essential geometry of a circular orbit. Notice that the velocity vector is always perpendicular to the gravitational force vector. Because gravity does no work on the satellite—the force is perpendicular to the displacement at every instant—the kinetic energy and speed remain constant throughout the orbit. This perpendicularity is a hallmark of uniform circular motion and is central to deriving the orbital speed equation in the next section.

SECTION 4

Mathematical Framework

Setting Newton's law of gravitation equal to the centripetal force requirement yields all the key orbital relationships for circular orbits. Each equation below can be derived from the single condition Fg = Fc.

GRAVITATIONAL–CENTRIPETAL CONDITION
GMm / r² = mv² / r
G = gravitational constant (6.674 × 10⁻¹¹ N·m²/kg²), M = central body mass, m = satellite mass, r = orbital radius (center to center), v = orbital speed. The satellite mass m cancels, confirming that orbital speed is independent of satellite mass.
ORBITAL SPEED
v = √(GM / r)
Orbital speed decreases as the orbital radius increases. A satellite farther from Earth orbits more slowly—this is a direct consequence of the inverse-square nature of gravity.
ORBITAL PERIOD
T = 2πr / v = 2π√(r³ / GM)
T is the time for one complete orbit. Squaring both sides gives T² = (4π²/GM)r³, which is Kepler's Third Law for circular orbits: the square of the period is proportional to the cube of the radius.
TOTAL MECHANICAL ENERGY
E = K + U = GMm/(2r) + (−GMm/r) = −GMm/(2r)
K = ½mv² = GMm/(2r) is the kinetic energy. U = −GMm/r is the gravitational potential energy (zero at infinity). The total energy is negative, indicating a bound orbit. Note that |U| = 2K for any circular orbit.
AP Exam Tip
SECTION 5

Energy & Radius Relationships

One of the most conceptually challenging aspects of satellite motion is the relationship between energy and orbital radius. Increasing a satellite's altitude requires adding energy (a thruster burn), yet the satellite ends up moving slower in the higher orbit. This apparent paradox resolves when you consider kinetic and potential energy separately.

Energy vs. Orbital Radius for a Circular OrbitOrbital Radius r →Energy0KUE = K + UK = GMm/(2r)U = −GMm/rE = −GMm/(2r)As r increases: K decreases, U increases (less negative), E increases (less negative).
Energy versus orbital radius for a satellite in circular orbit. The kinetic energy K (cyan, positive) decreases with radius. The potential energy U (red, negative) increases toward zero. The total energy E (violet, dashed) is always negative and equals −K, approaching zero as r → ∞.

The graph reveals three crucial relationships. First, kinetic energy is always positive and equals exactly half the magnitude of the potential energy: K = |U|/2. Second, the total mechanical energy is always negative for a bound orbit and equals −K. Third, moving to a higher orbit (larger r) means the total energy becomes less negative—the satellite gains energy overall—even though it slows down. This is because the increase in potential energy exceeds the decrease in kinetic energy by a factor of two.

Summary of how key orbital quantities scale with radius for circular orbits.
QuantityFormulaAs r increases
Orbital speed v√(GM/r)Decreases (∝ 1/√r)
Period T2π√(r³/GM)Increases (∝ r³ᐟ²)
Kinetic energy KGMm/(2r)Decreases (∝ 1/r)
Potential energy U−GMm/rIncreases (less negative)
Total energy E−GMm/(2r)Increases (less negative)
Angular momentum Lm√(GMr)Increases (∝ √r)
SECTION 6

Worked Example — Geostationary Orbit

A geostationary satellite orbits Earth with a period of exactly 24 hours so that it remains above the same point on the equator. Determine the orbital radius, orbital speed, and total mechanical energy of a 500 kg geostationary satellite. Use ME = 5.97 × 10²⁴ kg and G = 6.674 × 10⁻¹¹ N·m²/kg².

Step 1 — Identify Given Values

T = 24 h = 86 400 s, m = 500 kg, ME = 5.97 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N·m²/kg².

Step 2 — Solve for Orbital Radius

From T = 2π√(r³/(GM)), square both sides: T² = 4π²r³/(GM). Rearranging: r³ = GMT²/(4π²). Substitute: r³ = (6.674 × 10⁻¹¹)(5.97 × 10²⁴)(86 400)² / (4π²) = (3.986 × 10¹⁴)(7.465 × 10⁹) / 39.48 = 7.534 × 10²² m³.
r = (7.534 × 10²²)¹ᐟ³ ≈ 4.22 × 10⁷ m ≈ 42 200 km from Earth's center.

Step 3 — Calculate Orbital Speed

v = 2πr / T = 2π(4.22 × 10⁷) / 86 400.
v ≈ 3 070 m/s (about 3.07 km/s).

Step 4 — Calculate Total Mechanical Energy

E = −GMm/(2r) = −(6.674 × 10⁻¹¹)(5.97 × 10²⁴)(500) / (2 × 4.22 × 10⁷).
E ≈ −2.36 × 10⁹ J (negative, confirming a bound orbit).

Step 5 — Verify with Alternate Method

Check: K = ½mv² = ½(500)(3070)² ≈ 2.36 × 10⁹ J. U = −GMm/r ≈ −4.72 × 10⁹ J. E = K + U ≈ −2.36 × 10⁹ J ✓. Also note |U| = 2K as expected for a circular orbit.
SECTION 7

Common Misconceptions & Pitfalls

Frequent misconceptions about satellite motion and their corrections.
MisconceptionReality
Satellites in orbit are beyond Earth's gravity.Gravity provides the centripetal force for the orbit. At the ISS altitude (~400 km), g is still about 8.7 m/s², roughly 89% of its surface value.
Heavier satellites need to orbit faster.Satellite mass cancels in the derivation. Orbital speed depends only on M and r, not on m.
Moving to a higher orbit means the satellite speeds up.Higher orbits have lower speeds (v ∝ 1/√r). The total energy increases, but the kinetic energy decreases.
There is an outward centrifugal force on the satellite.In an inertial frame, only gravity (inward) acts. Centrifugal force is a fictitious force that appears only in a rotating reference frame.
Orbital radius equals altitude above the surface.Orbital radius r = R (radius of planet) + h (altitude). Confusing r with h is one of the most common algebraic errors.
✦ KEY TAKEAWAY
KEY TAKEAWAY
SECTION 8

Connection to Advanced Topics

The circular orbit model presented here is a powerful starting point, but real satellite dynamics extend into richer territory. The AP Physics 1 framework focuses on circular orbits and energy conservation; AP Physics C and college-level mechanics introduce elliptical orbits, escape velocity, and transfer orbits. Recognizing where the AP 1 model ends helps you know when simplifying assumptions are valid.

How the AP Physics 1 treatment compares with more advanced orbital mechanics.
TopicAP Physics 1 TreatmentAdvanced Treatment
Orbit shapeCircular only; constant speed and radiusElliptical (Kepler's 1st law); speed varies with position
EnergyE = −GMm/(2r), K and U fixedE = −GMm/(2a) where a is the semi-major axis; K and U vary along the orbit
Escape velocityConceptual: E ≥ 0 means unboundv_esc = √(2GM/r); derivable via energy conservation
Orbital transfersNot coveredHohmann transfer orbits; Δv budgets; bi-elliptic transfers
Angular momentumL = mvr constant for circular orbitL conserved in elliptical orbits; explains v_max at periapsis, v_min at apoapsis

Even within the scope of AP Physics 1, the satellite model connects beautifully to broader themes. The fact that gravitational potential energy is defined as zero at infinity mirrors convention choices you will encounter in electrostatics (AP Physics C) and quantum mechanics. The conservation of angular momentum you see in orbits reappears in spinning ice skaters, collapsing stars, and rotating galaxies—making this lesson a gateway to understanding rotating systems across many scales.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL
A satellite is in a stable circular orbit around Earth. If the satellite's mass were doubled but its orbital radius remained the same, what would happen to its orbital speed?
PROBLEM 2 — BASIC CALCULATION
The International Space Station orbits at an altitude of 408 km above Earth's surface. Given RE = 6.37 × 10⁶ m, ME = 5.97 × 10²⁴ kg, and G = 6.674 × 10⁻¹¹ N·m²/kg², what is the orbital speed of the ISS?
PROBLEM 3 — INTERMEDIATE
Satellite A orbits a planet at radius r with speed v. Satellite B orbits the same planet at radius 4r. What is the ratio of Satellite B's period to Satellite A's period, TB / TA?
PROBLEM 4 — APPLIED
A space agency wants to move a 1 000 kg satellite from a circular orbit of radius r₁ = 7.0 × 10⁶ m to a higher circular orbit of radius r₂ = 1.4 × 10⁷ m around Earth. How much energy must the propulsion system add to the satellite? Use GME = 3.986 × 10¹⁴ N·m²/kg.
PROBLEM 5 — CRITICAL THINKING
A student claims: 'If we place a satellite in a circular orbit at exactly the same radius as another satellite but give it twice the mass, it will need twice the centripetal force, so it must orbit at a higher speed to generate that force.' Design an experiment or theoretical argument to test this claim. In your response: (a) state whether the claim is correct or incorrect, (b) derive or cite the relevant equations to justify your answer, (c) explain the physical reason the student's reasoning fails (if it does), and (d) describe how you could verify your conclusion experimentally using two satellites of different masses.
SUMMARY

Lesson Summary

Varsity Tutors • AP Physics 1: Algebra-Based • Motion of Orbiting Satellites