Potential & Energy Lab
H2 Physics · Gravitation — the well, the gradient, the work, the escape
φ and g on one r-axis — the tangent above is the value below
ⓘ how to read these two graphs
Top: φ = −GM/r, plotted in units of GM/R, so the same curve serves every planet. It sits entirely below the dashed zero line and only reaches it at r = ∞. The pink line is the tangent at r₁; its gradient dφ/dr is the pink bar height on the lower graph, which is g = GM/r². Because φ climbs as r grows, dφ/dr is positive while the field itself points inwards — hence the minus sign in g = −dφ/dr. Drag anywhere on either graph to move the nearest marker.
Escape launcher — real RK4 integration of r̈ = −GM r̂ / r²
Planet
- M
- 5.972 × 10²⁴ kg
- R
- 6.371 × 10⁶ m
- g at surface = GM/R²
- 9.81 N kg⁻¹
- φ at surface = −GM/R
- −6.252 × 10⁷ J kg⁻¹
- v_esc = √(2GM/R)
- 11.18 km s⁻¹
Markers
Two-body mode — the maximum of φ
g = −dφ/dr, checked two ways
- φ(r₁)
- −3.9077 × 10⁷ J kg⁻¹
- gradient of the tangent, dφ/dr (calculus: GM/r₁²)
- 3.83346
- same gradient by finite difference, h = r₁/10⁴
- 3.83346
- the two agree to
- 8 s.f.
- g at r₁ = GM/r₁²
- 3.8335 N kg⁻¹
- g as a radial component = −dφ/dr
- −3.8335 N kg⁻¹
Negative = the field points back toward the planet.
Work done moving m from r₁ to r₂
- Δφ = φ₂ − φ₁
- 2.0688 × 10⁷ J kg⁻¹
- U₁ = mφ₁ = −GMm/r₁
- −3.9077 × 10⁹ J
- U₂ = mφ₂ = −GMm/r₂
- −1.8389 × 10⁹ J
- W = mΔφ
- 2.0688 × 10⁹ J
- GMm(1/r₁ − 1/r₂)
- 2.0688 × 10⁹ J
- Δr = r₂ − r₁
- 1.147 × 10⁷ m
- naive W ≈ mgΔr, g frozen at the surface value
- 1.1254 × 10¹⁰ J
- mgh error against the exact W
- +444.00 %
Sign: W is what you must supply — positive moving outward, negative moving inward, where gravity pays instead.
r₂ > r₁ → W positive: climbing out of the well costs energy.
ⓘ when mgΔr is allowed
mgΔr assumes the full surface g all the way up, so it always overstates the work; starting from the surface the error is exactly Δr/R. Drag r₂ outward and watch it blow up: on Earth 100 m gives +0.0016 %, 1000 km gives +15.7 %, and r₂ = 2R gives +100 %, since g has fallen to a quarter of its surface value by then.
Escape velocity
FALLS BACK
total energy negative → bound, and the path dips below the surface
- v_esc = √(2GM/R)
- 11.182 km s⁻¹
- v_circ = √(GM/R)
- 7.907 km s⁻¹
- ½v² − GM/R at launch
- −2.251 × 10⁷ J kg⁻¹
- apogee
- 2.78 R (h = 1.133 × 10⁷ m)
- live r
- —
- live v
- —
- live ½v² − GM/r
- —
The live energy never moves off its launch value — that constancy is the whole escape argument.
ⓘ where √(2GM/R) comes from
To just reach infinity, a mass m must arrive with zero speed and zero potential energy. Total energy is conserved, so at the surface ½mv² + (−GMm/R) = 0, giving ½mv² = GMm/R and v = √(2GM/R). The m cancels — a marble and a rocket need the same speed. For m = 100 kg on Earth the kinetic energy required is ½mv_esc² = 6.252 × 10⁹ J, exactly the |U| = GMm/R you would have to fill in.
What this confirms
- ›LO (g) — φ is work done per unit mass from infinity — the zero plane in the well is infinity. Drag a ball from the rim to the surface and the depth bar is the work per kilogram gravity has already done on it; the work card turns that into joules for a real mass.
- ›Why φ is always negative — gravity only attracts, so bringing a mass in from infinity releases energy. The well never rises above the zero plane and the φ curve never crosses its dashed line: φ < 0 everywhere, → 0 only as r → ∞.
- ›LO (h) — φ = −GM/r— every planet button feeds real M and R into the same expression, and the φ readout is −GM/r evaluated at the marker. Earth's surface value comes out −6.25 × 10⁷ J kg⁻¹.
- ›LOs (d)(e) — g = −dφ/dr — the tangent gradient from calculus and the same gradient from a finite difference agree to seven or more significant figures at every r you drag to, and both equal GM/r², the bar height on the lower graph.
- ›U = mφ = −GMm/r — the mass slider scales U and W in exact proportion while φ and g sit still, which is the whole distinction between the field quantity and the energy of a particular body.
- ›W = mΔφ = GMm(1/r₁ − 1/r₂) — the two routes print the same joules for any r₁, r₂, and W turns positive the moment r₂ > r₁: moving away from the planet always costs you energy.
- ›The φ maximum IS the neutral point — in two-body mode the hump in φ and the zero of the net field sit on the same green rail, found by two unrelated algorithms: a golden-section maximum of φ that never touches a derivative, and a bisection on dφ/dx = −g. For Earth + Moon both give 0.900188 d, matching d/(1 + √(M₂/M₁)), and φ there is −1.28 × 10⁶ J kg⁻¹ — negative, so a maximum is not an escape point.
- ›mgh is a small-Δr shortcut— the work card prints mgΔr beside the exact GMm(1/r₁ − 1/r₂). From Earth's surface, Δr = 100 m errs by +0.0016 %, Δr = 1000 km by +15.7 %, and r₂ = 2R by exactly +100 %. Always an overestimate, because mgh keeps the surface g at every height.
- ›Escape velocity from energy conservation — the integrator switches from bound to unbound exactly at 1.000 × √(2GM/R), and the live ½v² − GM/r stays pinned at its launch value the whole flight. Earth gives 11.18 km s⁻¹, the Moon 2.37, Mars 5.03, Jupiter 60.2.
- ›Direction does not change the threshold — the angle slider reshapes the path from radial to orbital but never moves the escape point, because energy depends on speed alone. Below v_esc the outcome splits into falls-back or orbits according to whether the perigee clears R.
Modelling notes: G = 6.67 × 10⁻¹¹ N m² kg⁻² and mean radii; the planet is a uniform sphere, so outside it behaves as a point mass at its centre. Inside the sphere the rendered well uses the exact uniform-sphere result φ = −GM(3R² − r²)/2R³ purely so it has a floor instead of an infinity — the graphs and every readout stay in the syllabus region r ≥ R.