Lessons · gravitation

Circular orbit: why the satellite falls around, not down

Regime 1 — the algebra is the calculus, evaluated. Step the algebra, step the calculus, and watch the algebra formula fall out of the integral. SymPy proves the two registers agree.

Regime 1 · algebra is calculus, evaluated Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A satellite circles a planet ( $\mu = GM = 4\times10^{14}$ m³/s²) at radius $R = 7\times10^{6}$ m. Gravity pulls it straight toward the centre the whole time — so why doesn't it fall in? Because at orbital speed the inward pull is exactly the centripetal acceleration a circle needs: the satellite is in continuous free fall, falling around the planet. Drag the radius outward and watch the orbit widen while the speed drops as $1/\sqrt{R}$ and the period stretches as $R^{3/2}$ — Kepler's third law, live.

Orbital speed

v = \sqrt{\mu/R}

7559 m/s

Orbital period

T = 2\pi\sqrt{R^3/\mu}

5818 s

Gravity at the orbit

g = \mu/R^2

(the centripetal pull)

8.163 m/s²

Angular rate

\omega = \sqrt{\mu/R^3}

0.00108 1/s

Gravity is a central force — the orbit solves a vector ODE

m\,\ddot{\mathbf r} = -\frac{GMm}{r^2}\,\hat{\mathbf r} \ \Longrightarrow\ \ddot{\mathbf r} = -\frac{\mu}{r^3}\,\mathbf r

Newton's second law with the inverse-square force is a second-order differential equation for the position. The satellite's mass cancels, leaving the path governed by $\mu = GM$ alone. The orbit is whatever curve solves this equation.

check $x^2+y^2=R^2$ ; check $\ddot{\mathbf r} = -\mu\,\mathbf r/R^3$ ; show $v=\sqrt{\mu/R}$ and $T^2=4\pi^2R^3/\mu$ fall out

✓ The path is a circle of radius $R$ : $x^2+y^2 = R^2$ (the distance to the centre never changes). [simplify]
✓ It solves the inverse-square equation of motion: with $r=R$ , $\ddot x = -\mu x/R^3$ (gravity points inward and is the centripetal pull). [structural]
✓ And the same vertically: $\ddot y = -\mu y/R^3$ . [structural]
✓ The orbital speed falls out: $\dot x^2+\dot y^2 = \mu/R$ , so $v = \sqrt{\mu/R}$ — the memorized circular-orbit speed. [structural]
✓ Kepler's third law falls out of the period: $T^2 = 4\pi^2 R^3/\mu$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A satellite floats in orbit because there is no gravity that far out in space.”

There is plenty of gravity — at this orbit it is about $\mu/R^2 \approx 8$ m/s², roughly 80% of its surface value. That inward pull is exactly what bends the satellite's straight-line motion into a circle: $\mu/R^2 = v^2/R$ . The astronauts feel weightless not because gravity is absent but because they and the station are in continuous free fall together — falling around the Earth fast enough that the ground curves away beneath them. Cut the gravity and the satellite would fly off in a straight line (Newton's first law), not float.

Modeling assumptions — author-asserted, disclosed not discharged

Spherically symmetric planet, so the field outside is $GMm/r^2$ (point-mass equivalent); the satellite's own mass is negligible and cancels.
A perfectly circular orbit (constant radius); no atmospheric drag, no other bodies. $\mu = GM$ is used so the path is unit-clean.

The trajectory, fully annotated

A static rendering (Matplotlib) at the default launch — the path y vs x, with the apex and range marked. The interactive version with launch-angle and speed sliders is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):