Preview

Gravity and Orbits

Q: What happens if you give a planet exactly circular orbit speed but point it slightly off?

Still an ellipse — slight angle change preserves energy but changes shape.

Q: Earth orbits at 30 km/s. What is Earth's escape velocity from the Sun?

V_escape = √2 × v_circular = √2 × 30 ≈ 42.4 km/s.

Q: Verify Kepler's Third Law: compare orbital periods for planets at 1 AU and 4 AU.

T² ∝ r³ → (T₂/T₁)² = (4/1)³ → T₂ = 8 years.

Model planetary motion and orbital mechanics

Newton explained orbits with a thought experiment that still works better than any equation: put a cannon on a very tall mountain and fire horizontally. Too slow, and the ball arcs to the ground. Faster, and it lands farther away. Now fire it fast enough that the ground CURVES AWAY beneath it as quickly as it falls — 'the surface curves away from them at exactly the same rate as they fall', as the textbook puts it — and the cannonball falls forever without landing. That is an orbit: not a balance, not an escape from gravity, but free fall with enough sideways speed to keep missing. Every instinct people bring to this lab says otherwise. They expect an under-speed satellite to spiral helplessly into the star — it doesn't; it dips into an ellipse, speeds up as it falls, and swings right back out, forever. They expect gravity to be absent or cancelled up there — the rose force arrow in this sim never once blinks off. They expect a heavier planet to need a different path — drag the planet's mass across three orders of magnitude mid-orbit and watch the trajectory ignore it completely. And running through all of it is the cleanest law in celestial mechanics: T² ∝ a³, Kepler's third, which this panel displays as a live ratio that refuses to leave 1.0000 no matter which bound orbit you fly — until you change the STAR, because the constant was always the star's. One slider crossing √2 ends the story: energy touches zero, the ellipse opens into a hyperbola, and the planet leaves and does not come back.

What you'll be able to do

Read an orbit as perpetual falling with the two arrows: the rose gravity arrow points at the star at full strength through every instant of every orbit — nothing cancels it, it is busy turning the cyan velocity arrow. Launch at 1.00×v_circ and the fall closes into a circle (e = 0.0000, badge Circular); at 0.75× it dips into an ellipse with e = 0.4375, perihelion 1.565 AU — closer, faster, and back out, never a spiral
Operate the energy frontier: total E < 0 is bound (E = −GMm/2a, −6.62×10³² J on the default circle), and the Launch Speed slider walks E monotonically upward until it crosses zero at exactly √2 — the Escape preset's 1.42 tips the badge to Hyperbolic (e = 1.016) and the period to ∞
Audit Kepler's Third Law on its own panel row: T²/a³ reads 1.0000 at every bound orbit around a 1 M☉ star — the default 4 AU/8.000 yr, the Inner preset's 1.5 AU/1.837 yr, the 0.75× ellipse's a = 2.783 AU/4.642 yr — and flips to 0.2500 = 1/M around the 4 M☉ Massive Star; meanwhile the Planet Mass slider (0.1 → 318 M⊕) moves nothing but the energy books — the trajectory never asks the planet how heavy it is

Formulas

v_{orbital} = \sqrt{\frac{GM}{r}}

Circular orbit speed

T^2 \propto r^3

Kepler's Third Law

v_{escape} = \sqrt{\frac{2GM}{r}}

Escape velocity (√2 × circular)

E = -\frac{GMm}{2a}

Total energy of a bound orbit — negative means bound

v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)

Vis-viva: speed anywhere on the orbit

Make a prediction

Default circular orbit: launched at 1.00×v_circ from 4 AU, the planet circles forever (e = 0.0000, T = 8.000 yr). Now drop the Launch Speed to 0.75×v_circ — too slow for a circle. What does the planet do?

No grading here — commit to a guess, then scroll down and test it yourself.

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Explore

Run the gate live: hit the Elliptical preset (0.75×v_circ) and watch the whole refutation — the badge flips to Elliptical, the path dips to r = 1.565 AU while the speed row SURGES (falling = accelerating), then climbs back out to 4 AU. Let it lap twice: same ellipse, same numbers, no decay, no spiral.
Cross the frontier: nudge Launch Speed up through 1.40, then hit Escape (1.42 ≈ √2). The Total energy row climbs through zero, the badge flips to Hyperbolic (e = 1.016), the period reads ∞, and the planet leaves for good. Escape was never about outrunning gravity — it's about your energy account reaching zero.
Audit Kepler like an astronomer: read T²/a³ = 1.0000 on the default circle, then on Inner Planet (1.5 AU, T = 1.837 yr), then on the 0.75× ellipse — the ratio refuses to move. Now hit Massive Star: 0.2500, exactly 1/M for 4 M☉. The 'constant' in Kepler's law is the star's signature — which is literally how stars get weighed.
Interrogate the passenger: mid-orbit, drag Planet Mass from 1 to 318 M⊕ (Jupiter). The path, period, and eccentricity do not flinch — only the energy rows rescale. The trajectory never asks the planet how heavy it is; m cancels from a = F/m, the same cancellation the pendulum pulled last batch.

Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It falls into an ELLIPSE — dips to 1.57 AU, speeds up as it falls, and swings back out to 4 AU, repeating forever (e = 0.44, T = 4.642 yr): a closed orbit, not a doom spiral

Nothing in a gravity-only sky can spiral: a spiral means losing energy every lap, and there is no drag, no friction, nothing to lose it to — the conservation assertion in this sim's own verify suite certifies the energy books close orbit after orbit. What actually happens is a trade: launched at 0.75×v_circ the planet is short on speed, so it falls inward — and falling makes it FASTER (altitude for speed, the vis-viva account v² = GM(2/r − 1/a)), until at perihelion 1.565 AU it is moving too fast to fall further and climbs back out, returning exactly to its launch point. Ellipse, e = 0.4375, forever. Option B is the balance-scale picture of orbits — as if gravity and speed were in a tug-of-war that speed just lost. But an orbit was never a balance: it is free fall that keeps missing, and a slower launch just means missing by less. Option C breaks the geometry: a circle at 4 AU REQUIRES exactly √(GM/4) of tangential speed — with less, the pull immediately wins the first instant's argument and the path bends inside the circle. Fly the Elliptical preset and watch the speed row surge through perihelion — falling means accelerating, which is precisely why it never crashes.

Quiz (0/3)

What happens if you give a planet exactly circular orbit speed but point it slightly off?

Earth orbits at 30 km/s. What is Earth's escape velocity from the Sun?

Verify Kepler's Third Law: compare orbital periods for planets at 1 AU and 4 AU.

You can now

Read an orbit as perpetual falling with the two arrows: the rose gravity arrow points at the star at full strength through every instant of every orbit — nothing cancels it, it is busy turning the cyan velocity arrow. Launch at 1.00×v_circ and the fall closes into a circle (e = 0.0000, badge Circular); at 0.75× it dips into an ellipse with e = 0.4375, perihelion 1.565 AU — closer, faster, and back out, never a spiral
Operate the energy frontier: total E < 0 is bound (E = −GMm/2a, −6.62×10³² J on the default circle), and the Launch Speed slider walks E monotonically upward until it crosses zero at exactly √2 — the Escape preset's 1.42 tips the badge to Hyperbolic (e = 1.016) and the period to ∞
Audit Kepler's Third Law on its own panel row: T²/a³ reads 1.0000 at every bound orbit around a 1 M☉ star — the default 4 AU/8.000 yr, the Inner preset's 1.5 AU/1.837 yr, the 0.75× ellipse's a = 2.783 AU/4.642 yr — and flips to 0.2500 = 1/M around the 4 M☉ Massive Star; meanwhile the Planet Mass slider (0.1 → 318 M⊕) moves nothing but the energy books — the trajectory never asks the planet how heavy it is