Kepler's laws, satellite orbits, and escape velocity
Gravity follows an inverse-square law (F ∝ 1/r²). The circular orbital velocity v = √(GM/r) is the exact speed for a stable circular orbit — faster means elliptical or escape, slower means the satellite falls. Kepler's 3 laws: (1) orbits are ellipses with the planet at one focus; (2) equal areas in equal times (conservation of angular momentum); (3) T² ∝ a³. Escape velocity is √2 times the circular orbital speed.
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Sign in →Gravitational fields obey an inverse-square law — g = GM/r² pointing toward the source mass — and this single relationship generates all of orbital mechanics. A satellite in circular orbit needs v_circ = √(GM/r); escape requires v_esc = √(2GM/r); and Kepler's third law follows from the same central force. This simulation focuses on the field view: change central mass, test distance, and planet type to see field vectors, potential wells, escape velocity, and binding energy respond in real time. The Earth Surface, Jupiter Gravity, and Near Black Hole presets give quick comparison cases for familiar gravity, strong planetary gravity, and an extreme compact source.
MisconceptionA satellite in orbit has no gravity acting on it — that's why it floats.
CorrectGravity is the only force acting on an orbiting satellite. Its acceleration is g = GM/r² directed toward Earth, roughly 8.7 m/s² at low Earth orbit. The sensation of weightlessness occurs because the satellite and occupants are in free fall together — not because gravity vanishes.
MisconceptionOrbital mechanics requires constant velocity — satellites travel at the same speed around their orbit.
CorrectOnly circular orbits have constant speed. In an elliptical orbit, the satellite moves fastest at periapsis and slowest at apoapsis. This follows from conservation of angular momentum: L = m r v_t (the tangential component) is constant, so v must increase as r decreases. Kepler's second law (equal areas in equal times) is a geometric restatement of this fact.
MisconceptionMoving farther from a planet only makes gravity a little weaker.
CorrectGravitational field strength follows an inverse-square relationship, so distance matters strongly. If the distance from the central mass doubles, g becomes one fourth as large; if distance triples, g becomes one ninth as large. The Distance slider makes that rapid falloff visible in the field vectors and data readouts.
MisconceptionEscape velocity means the rocket has to keep thrusting until it escapes — it can never coast.
CorrectEscape velocity v_esc = √(2GM/r) is the minimum initial speed that allows coasting to infinity against gravity with zero final speed. Once launched at v_esc (or beyond), no further thrust is needed; the object decelerates but never quite stops and never returns. Total orbital energy E = ½mv² − GMm/r ≥ 0 for escape.
MisconceptionKepler's third law T² ∝ a³ only applies to planets orbiting the Sun.
CorrectT² = (4π²/GM)a³ applies to any two-body gravitational system. The proportionality constant 4π²/GM depends on the central body's mass M. For Earth's satellites, replace M_Sun with M_Earth; for the Moon orbiting Earth, use M_Earth. The simulation lets you vary Central Mass to confirm this scaling.
For a circular orbit, ½mv² = GMm/(2r), giving v_circ = √(GM/r). Escape requires total energy ≥ 0: ½mv_esc² − GMm/r = 0, giving v_esc = √(2GM/r) = √2 · v_circ. The factor of √2 arises because you must supply the full gravitational potential energy GMm/r rather than half of it.
Standards 3.C.1 and 3.C.2 address gravitational force, field, and orbital dynamics, while 3.G.1 covers Kepler's laws and gravitational potential energy. Together they form the core of the AP Physics C Mechanics gravity unit. This simulation probes those ideas through the Central Mass, Distance, and Planet Selector controls.
Total orbital energy E = ½mv² − GMm/r. If E is negative the orbit is bound (circle or ellipse); if E = 0 it is a parabolic escape; if E is positive it is a hyperbola. The circle case requires both v_r = 0 (purely tangential velocity) AND v_t = √(GM/r) at the launch radius; any bound orbit failing either condition is an ellipse.
The gravitational force is central — it always points toward the planet — so the torque about the planet is zero, and angular momentum L = mvr·sinφ is conserved. Area swept per unit time equals L/(2m) = constant, so equal time intervals always sweep equal areas regardless of where in the orbit the satellite is.
Launch a stable orbit, then use the simulation's time readout to measure the elapsed time for one complete revolution. For a circular orbit you can also calculate T = 2πr/v_circ. For an ellipse, use Kepler's third law: T² = (4π²/GM)a³, where a = (r_periapsis + r_apoapsis)/2 in meters.
E_orbit = −GMm/(2a) for any elliptical orbit, where a is the semi-major axis. The negative sign means the satellite is bound: you would need to add |E| joules of energy to free it. For a circular orbit of radius r, this simplifies to E = −GMm/(2r). The deeper (lower) the orbit, the more negative the energy and the faster the satellite moves.