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Physics I course sequence · Interactive lesson

Circular Orbital Motion

Every satellite calculation starts from one line: gravity is the only force on the craft, so gravity must be its centripetal force. From that line come the speed and period at any height, the one radius where a satellite stays over one point on the equator, a planet's mass read from its moon, and why astronauts float in a strong field.

UPI-13.04 · Lesson 4 of 7 in this unit. Position is not a completion score.
Starting model: Circular Orbital MotionTo-scale orbit round Earth. Altitude 20200 km gives r = 26571 km; gravity along the dashed radius is the only force, so v = √(GM/r) = 3.87 km s⁻¹ and T = 11.97 h. After 3.0 h the satellite has made 0.25 laps, the geostationary dot 0.13. g there is 6% of surface g.surfaceorbit6%g at orbit vs surfacet = 3.0 h → 0.25 lapsarrow: velocity vdashed: pull toward Earthfaint ring: geostationary
Start with a picture. Change the model in the lab below.
1

Build the model

An orbit is free fall with enough sideways speed to keep missing the ground.

The key idea

A circular orbit is uniform circular motion in which gravity from the central body is the only force and supplies the entire centripetal force, so the speed at radius r must be v = √(GM/r), whatever the satellite's mass.

Example. 400 km above Earth, r = 6371 km + 400 km = 6.771 × 10⁶ m, so v = √[(3.986 × 10¹⁴ m³ s⁻²)/(6.771 × 10⁶ m)] = 7.67 km s⁻¹ and T = 2πr/v = 5.54 × 10³ s, about 92 minutes.

The complete physical model and assumptions

A satellite in a circular orbit is not balanced between two forces. One force acts on it, gravity, and gravity points at the planet's centre, which is exactly where uniform circular motion needs the acceleration to point. Newton's second law along the radius therefore reads GMm/r² = mv²/r. The satellite's own mass cancels, so everything at the same radius, whatever its mass, must share one speed, v = √(GM/r), and one period, T = 2πr/v = 2π√(r³/GM). Two results run against intuition: a higher orbit is slower, and its period grows faster than its size, with T² ∝ r³. The same cancellation explains apparent weightlessness: crew, cabin and bathroom scale all accelerate at the local g together, so nothing presses on anything and the scale reads zero, in a field only about a tenth weaker than the one at the ground. The model rests on assumptions worth naming: a spherically symmetric central body, a satellite light enough that the planet's own motion can be ignored, no drag, and a velocity that is exactly tangential with magnitude exactly √(GM/r). Change that speed a little and the path becomes an ellipse; make the satellite comparable in mass to the planet and both circle their common centre of mass, so the period measures M + m rather than M.

1Draw the one force and point it inward

Draw the free-body diagram of a satellite and there is exactly one arrow: gravity, of size GMm/r², pointing at the planet's centre. There is no separate centripetal force to add beside it and no outward force to balance it; centripetal is the name of a job, and gravity is what does it. Uniform circular motion needs a net inward force of mv²/r, so Newton's second law along the radius is GMm/r² = mv²/r. Nothing acts along the tangent, so the speed stays constant. Notice that m multiplies both sides: the orbit belongs to the radius and the central mass, not to the satellite. A one-gram screw and a station of several hundred tonnes, released side by side at the same r with the same velocity, stay side by side.

2Speed from radius, and radius from the centre

Solve the radial law for v: v = √(GM/r). For Earth GM = 3.986 × 10¹⁴ m³ s⁻², and r is measured from Earth's centre, so a satellite 400 km up has r = 6.371 × 10⁶ m + 4.00 × 10⁵ m = 6.771 × 10⁶ m and v = 7.67 km s⁻¹. Put the altitude in place of r and you get 31.6 km s⁻¹, four times faster than any circular orbit around Earth can be — the signal that r was measured from the wrong place. The formula says higher orbits are slower: at the GPS radius of 2.656 × 10⁷ m, about four Earth radii, v = 3.87 km s⁻¹, half the low-orbit speed. The fastest circular orbit skims the surface, r = R, where v = √(g₀R) = 7.91 km s⁻¹ with g₀ = GM/R² = 9.82 m s⁻².

3The period grows faster than the orbit

The period is circumference over speed, T = 2πr/v, and substituting v gives T = 2π√(r³/GM), or T² = (4π²/GM) r³. At 400 km that is 5.54 × 10³ s, 92 minutes. Because T² ∝ r³, doubling the radius multiplies the period by 2√2 = 2.83, and the GPS radius, 3.92 times the low orbit's, gives 11.97 h, two orbits per sidereal day. Run the formula backwards for a geostationary satellite, which must turn with the Earth: T = 86 164 s, one rotation relative to the stars, gives r = ∛(GMT²/4π²) = 4.216 × 10⁷ m, about 35 800 km above the equator, at 3.07 km s⁻¹. Using 24 h instead of the sidereal day puts the satellite 77 km too high, and it slowly drifts westward over the ground.

4Run it backwards to weigh a planet

Rearranging T² = 4π²r³/(GM) gives M = 4π²r³/(GT²), and the satellite's mass is nowhere in it: timing any small body on a circular orbit weighs whatever it circles. The Moon has r = 3.844 × 10⁸ m and T = 27.32 days = 2.360 × 10⁶ s, so M = 4π²(3.844 × 10⁸ m)³/[(6.674 × 10⁻¹¹ N m² kg⁻²)(2.360 × 10⁶ s)²] = 6.03 × 10²⁴ kg. That is about 1% more than Earth's accepted 5.97 × 10²⁴ kg, and the gap is itself physics: the Moon has 1.2% of Earth's mass, so Earth also swings round the pair's centre of mass and the formula really returns M + m. For an artificial satellite that correction is below one part in 10¹⁹, which is why the model treats the planet as fixed.

5Weightless is not gravity-free

Stand on a bathroom scale in a station 400 km up. Two forces act on you: gravity, mg(r) with g(r) = GM/r² = 8.69 m s⁻², and the scale's push N. Taking the direction toward Earth as positive, Newton's second law reads mg(r) − N = ma. You share the station's orbit, so your acceleration is its orbital acceleration, v²/r = (7.67 × 10³ m s⁻¹)²/(6.771 × 10⁶ m) = 8.69 m s⁻², which is g(r) itself. So N = 0 and the scale reads zero, although gravity pulls a 70 kg astronaut with 609 N, 89% of the 687 N they weigh on the ground. What you feel as weight is the contact force, not gravity. A released pen floats beside you because it shares your r and v, and therefore your path, not because gravity has been switched off.

6Where the circular model stops

Each step leaned on an assumption, and each has a limit. Spherical symmetry lets the planet act as a point mass; Earth's equatorial bulge breaks it slightly and makes low orbits precess. A fixed planet needs M ≫ m: the Earth–Moon pair already needs M + m, and two stars of similar mass circle a point between them. No drag: at 400 km the thin upper atmosphere removes energy steadily, so stations need periodic reboosts. And a circle needs exactly the right velocity, tangential and of size √(GM/r). Launch from the same point a little faster or slower and the path becomes an ellipse through that point; ellipses, and the energy it costs to move between orbits, come next.

2

Key equations

Read the units, assumptions and limits before substituting.

Radial second law for an orbit

GMm/r² = mv²/r

Gravity is the centripetal force, not a partner to it. The satellite's mass m sits on both sides and cancels.

G = 6.674 × 10⁻¹¹ N m² kg⁻² · M, m in kg · r from the centre, in m · v in m s⁻¹

Orbital speed

v = √(GM/r)

Quadruple the radius and the speed halves: a higher orbit is a slower orbit.

Earth: GM = 3.986 × 10¹⁴ m³ s⁻² · r = R + h, never the altitude h alone

Period

T = 2πr/v = 2π√(r³/GM)

T² ∝ r³, so doubling the radius stretches the period by 2√2 = 2.83, not by 2.

T in s, r in m · r = R gives Earth's shortest orbit, T = 5.06 × 10³ s = 84 min

Weighing the central body

M = 4π²r³/(GT²)

The satellite's mass never enters, so timing a moon weighs the planet it circles.

r and T of any light satellite, M in kg

Geostationary condition

ω = 2π/T = √(GM/r³)

Match ω to Earth's spin and r = 4.216 × 10⁷ m: 35 800 km above the equator, moving at 3.07 km s⁻¹.

ω in rad s⁻¹ · Earth turns at 7.292 × 10⁻⁵ rad s⁻¹, one sidereal day of 86 164 s

Apparent weight in orbit

mg(r) − N = ma, with a = g(r) → N = 0

What vanishes is the contact force, not gravity: floor and feet accelerate together at g(r).

N is the push of floor or scale, in newtons · g(r) = GM/r² = 8.69 m s⁻² at 400 km
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

20200 km
200 to 36000 km; steps of 100.
3.0 h
0 to 24 h; steps of 0.1.

Try this. Start at 20 200 km, the GPS altitude, where T is 12 h; drop to 400 km and the velocity arrow nearly doubles while the g bar leaps to 89% — then set 35 800 km and step t through 24 h to watch the satellite ride with the geostationary dot.

Live results

ORBIT RADIUS r = R + h
26571 km
ORBITAL SPEED v
3.87 km s⁻¹
PERIOD T
11.97 h
g AT THE ORBIT
0.56 m s⁻²

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics I modelTo-scale orbit round Earth. Altitude 20200 km gives r = 26571 km; gravity along the dashed radius is the only force, so v = √(GM/r) = 3.87 km s⁻¹ and T = 11.97 h. After 3.0 h the satellite has made 0.25 laps, the geostationary dot 0.13. g there is 6% of surface g.surfaceorbit6%g at orbit vs surfacet = 3.0 h → 0.25 lapsarrow: velocity vdashed: pull toward Earthfaint ring: geostationary
To-scale orbit round Earth. Altitude 20200 km gives r = 26571 km; gravity along the dashed radius is the only force, so v = √(GM/r) = 3.87 km s⁻¹ and T = 11.97 h. After 3.0 h the satellite has made 0.25 laps, the geostationary dot 0.13. g there is 6% of surface g.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Altitude above surface h: 20200 km; Elapsed time t: 3.0 h

Baseline 20200; choose 200 to 36000, in steps of 100.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

“There is no gravity up there, so astronauts float.” At the space station's height, 400 km, gravity is only 11% weaker than at the ground, and you would have to climb about 2600 km just to halve it. Gravity in orbit is not a leftover to be ignored; it is the reason there is an orbit at all. Switch it off and the station would carry on in a straight line at 7.67 km s⁻¹ and leave. Floating is a statement about contact forces. Everything in the cabin has the same acceleration g(r), because everything is falling around the Earth together, so the floor never needs to push on anyone and a scale reads N = 0. A lift whose cable has snapped gives the same experience at sea level, in a field of full strength.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A mapping probe circles the dwarf planet Ceres (mass 9.38 × 10²⁰ kg, radius 470 km) in a circular orbit 385 km above the surface. Find the probe's orbital speed and its period.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

An Earth-imaging satellite must complete one circular orbit every 100.0 minutes. Find its altitude and speed, then compare the gravitational force on a 75 kg astronaut aboard with the reading of a bathroom scale the astronaut stands on. Use GM = 3.986 × 10¹⁴ m³ s⁻² and R = 6371 km.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

A moonlet and its asteroid are observed 2.40 km apart, centre to centre, circling each other every 16.0 h. A student puts r = 2.40 km and T = 16.0 h into M = 4π²r³/(GT²) and reports the result as the asteroid's mass. Radar later shows the moonlet has one quarter of the asteroid's mass. Find the asteroid's true mass, where the pair's centre of mass lies, and the moonlet's actual orbital speed.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

Satellite P circles a planet at radius r. Satellite Q, with twice P's mass, circles the same planet at radius 4r. Compared with P, how fast does Q move, and how long does Q take to complete one orbit?
7

Review & continue

A circular orbit is uniform circular motion in which gravity from the central body is the only force and supplies the entire centripetal force, so the speed at radius r must be v = √(GM/r), whatever the satellite's mass.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Optional supporting lessons

These chapter resources can help with earlier ideas; they are not additional required Physics I completion tasks.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.