Circular Orbits & Satellites Lab
H2 Physics · Gravitation — GMm/r² = mv²/r, and everything that falls out of it
Geostationary — all three conditions, or nothing
parked- ✓Orbit lies in the equatorial planei = 0°
- ✓Period = Earth's spin, 23 h 56 minT = 23 h 56 min
- ✓Same direction as the spin (W → E)prograde
- ground-track drift
- 0.000 °/day
- latitude swing
- ±0°
- sub-satellite point
- 0.0°, 0.0°
- elapsed
- 0.00 days
Where 4.22×10⁷ m comes from
T = 86 164 s → r = ∛(GM T² / 4π²) = 4.22×10⁷ m
height = r − R_E = 3.58×10⁷ m ≈ 3.6×10⁴ km above the surface
v = √(GM/r) = 3075 m s⁻¹
Notes usually quote 24 h. Strictly it is the sidereal day, 23 h 56 min 4 s — one spin relative to the stars. Both give r = 4.22×10⁷ m to 3 s.f. Break any one condition and the cyan trace stops being a dot: tilt it and you get a figure-8, change T and it walks in longitude, reverse it and it races backwards.
T² vs r³ — Kepler III builder
Why the line must pass through the origin
GMm/r² = mr(2π/T)² rearranges to T² = (4π²/GM) r³. That is y = mx with no intercept, so the plot is a straight line through the origin whose gradient is a property of the Earth alone — the satellite’s mass cancels. Take logs instead and lg T = 1.5 lg r + lg(2π/√GM): a straight line of gradient 3/2.
Energy of the satellite
Zero is the middle line. Bars shrink towards zero as r grows — every energy tends to 0 at infinity.
- KE
- 4.73×10⁹ J
- −½ PE
- 4.73×10⁹ J
- E total
- -4.73×10⁹ J
- −KE
- -4.73×10⁹ J
Rows pair up at every r: KE = −½PE and E = −KE.
Higher orbit = slower, yet more energy
| quantity | this orbit | +1 R_E higher |
|---|---|---|
| r / m | 4.22×10⁷ | 4.85×10⁷ |
| v / m s⁻¹ | 3075 | 2866 ↓ |
| T | 23 h 56 min | 29 h 34 min |
| E total / J | -4.73×10⁹ | -4.11×10⁹ ↑ |
Slower (v = √(GM/r)) but less negative total energy — you still had to put energy in to get there.
Presets
Controls
This orbit — live
- v = √(GM/r)
- 3075 m s⁻¹
- T = 2π√(r³/GM)
- 23 h 56 min
- T (seconds)
- 8.616×10⁴ s
- orbits per day
- 1.00
- g = GM/r²
- 0.2242 m s⁻²
- a = v²/r
- 0.2242 m s⁻²
Last two rows are always equal — that equality IS the orbit condition.
What this confirms
- ›LO (i) — circular orbits: the “g = GM/r²” and “a = v²/r” readouts stay identical at every radius, which is the derivation GMm/r² = mv²/r running live.
- ›v = √(GM/r): the speed readout falls as r grows, and the LEO racer laps the outer satellite because its period is shorter.
- ›Kepler III, T² ∝ r³: every point dropped by the r slider lands on one straight line through the origin, and the fitted gradient matches 4π²/GM = 9.905×10⁻¹⁴ s² m⁻³ to within rounding.
- ›log–log check: the same data on lg T against lg r gives gradient 3/2 — the standard way an exam asks you to verify a power law.
- ›LO (j) — geostationary orbits: equatorial plane, T = 23 h 56 min, and W→E. Break any one and the ground track stops being a fixed dot; all three give r = 4.22×10⁷ m, about 3.6×10⁷ m up.
- ›Orbital energies: KE = +GMm/2r, PE = −GMm/r, E = −GMm/2r — the paired rows show KE = −½PE and E = −KE hold at every radius, and E is negative because the satellite is bound.
- ›The classic trap: a higher orbit is SLOWER yet has MORE total energy. Both numbers sit side by side in the comparison table.
- ›Orbital decay: with drag on, r falls and v RISES — friction speeds the satellite up. The change columns show why: ΔPE = −2ΔKE, so the total energy the drag removed, ΔE = −ΔKE, is exactly the kinetic energy gained.
- ›Apparent weightlessness: g is far from zero in orbit; the astronaut floats because astronaut and satellite share one acceleration, so the contact force N is zero.