Pick your class, read the steps, then try them right in the lab —
change one slider at a time and watch the rays and graphs respond.
Class 9 — Universal law of gravitation and the value of g
Concept: every mass attracts every other mass with a force F = Gm1m2 ÷ r2. At the surface of a planet this makes every falling body accelerate at g = GM ÷ R2.
- Open Gravity: height, depth, latitude with the Earth. g = GM ÷ R2 = 6.674 × 10−11 × 5.972 × 1024 ÷ (6371 km)2 = 9.82 m/s2. A 60 kg person weighs 589 N.
- Switch to the Moon: g = 1.62 m/s2, about one sixth of the Earth's, so the same person weighs only 97 N (the mass is still 60 kg). On Mars g = 3.73 m/s2.
- Move away from the surface to 2 R from the centre (one Earth radius above the ground): g falls to 9.82 ÷ 4 = 2.455 m/s2. Double the distance and the force becomes a quarter: the inverse-square law.
- Mass is the amount of matter and does not change; weight W = mg is the pull of gravity and depends on where you are.
Takeaway: F = Gm1m2 ÷ r2 and g = GM ÷ R2; weight changes from place to place but mass does not.
Class 11 — Kepler's three laws
Concept: planets move on ellipses with the Sun at one focus; the line to the Sun sweeps equal areas in equal times; and T2 ∝ a3.
- Open Orbits & Kepler's laws with the Earth. a = 1 AU and e = 0.017, so the ellipse is almost a circle. The perihelion is 0.983 AU and the aphelion 1.017 AU.
- The coloured wedges are swept in equal times (T ÷ 12). Their areas, printed at the top, are all equal: 0.2618 AU2. This is the second law, which follows from the conservation of angular momentum.
- The planet is fastest at perihelion (30.28 km/s) and slowest at aphelion (29.29 km/s), since v r = L stays constant.
- Choose Mars: a = 1.524 AU gives T = 1.5241.5 = 1.881 years. Jupiter: a = 5.203 AU, T = 11.87 years. Mercury: T = 0.241 years.
- Halley's comet: a = 17.83 AU, e = 0.967, T = 75.3 years. It rushes past the Sun at 54.45 km/s at 0.588 AU and crawls at 0.91 km/s at 35 AU. The wedges are long and thin but their areas are still equal.
Takeaway: T2 = (4π2 ÷ GM) a3; equal areas in equal times means the speed is greatest nearest the Sun.
Class 11 — The energy of an orbit
Concept: the total energy per unit mass E = ½v2 − GM ÷ r decides the orbit: for E < 0 it is closed with a = −GM ÷ 2E, and for E ≥ 0 the body escapes.
- Choose Your own orbit at r = 1 AU. A speed of 1.00 times the circular speed gives a circle (e = 0). At 1.20 times, E = −11.05 AU2/yr2, a = 1.786 AU, e = 0.44 and T = 2.386 years: the body swings out to 2.571 AU.
- At 1.30 times the speed the body reaches 5.45 AU, with a = 3.226 AU and T = 5.79 years. The energy rises towards zero and the ellipse gets longer.
- At 1.42 times the circular speed (just over √2) the energy is positive: the body escapes and never returns. The path is a hyperbola.
- Look at the lower graph: the curve is the effective potential L2 ÷ 2r2 − GM ÷ r. The planet moves between the two points where the red energy line meets the curve (perihelion and aphelion).
Takeaway: E = −GM ÷ 2a for a closed orbit; the escape speed is √2 times the circular speed.
Class 11 — Satellites and the geostationary orbit
Concept: a satellite is a freely falling body whose sideways speed is so large that it keeps missing the Earth. For a circular orbit, gravity provides the centripetal force: v = √(GM ÷ r).
- Open Satellites & escape speed: launch height 400 km and speed 7.67 km/s. This is the circular speed there, v = √(3.986 × 1014 ÷ 6.771 × 106) = 7.672 km/s, and the period is T = 2π√(r3 ÷ GM) = 92.4 minutes.
- Set the height to 0: the circular speed at the surface is 7.91 km/s and the period 84.4 minutes, the shortest possible orbit period for the Earth.
- Press geostationary: at 35 786 km the speed is 3.075 km/s and the period 23.93 hours, so the satellite stays above one point on the equator. A higher orbit means a slower satellite and a longer period.
- Notice the satellite and everything in it fall together, so astronauts feel weightless even though g is still 8.7 m/s2 at 400 km.
Takeaway: v = √(GM ÷ r), T = 2π√(r3 ÷ GM); a geostationary satellite has T = 24 h at r = 42 164 km.
Class 11 — Newton's cannon and escape speed
Concept: a ball fired sideways from a high mountain falls to the ground, but a faster one falls round the Earth. The escape speed vesc = √(2GM ÷ r) = √2 vc takes it away for ever.
- Keep the launch height at 400 km and lower the speed to 6 km/s: the body falls back to the ground, because its orbit would pass inside the Earth.
- At 7.67 km/s it is a circular orbit. At 9 km/s it moves on an ellipse that rises to 8561 km above the surface and returns every 3.13 hours, passing its lowest point (400 km up) after each lap.
- Press escape speed: 10.85 km/s at 400 km (11.19 km/s from the ground). The path is open and the body never comes back.
- Escape speed does not depend on the mass of the body; it depends on where you launch from. At 35 786 km it is only 4.35 km/s.
Takeaway: vesc = √(2GM ÷ R) = √(2gR) = 11.2 km/s for the Earth, √2 times the orbital speed.
Class 11 — Gravity above and inside the Earth
Concept: outside a planet g ∝ 1 ÷ r2; inside (for uniform density) g ∝ r, so g is zero at the centre and greatest at the surface.
- Open Gravity: height, depth, latitude and move to 0.5 R (half-way down): g = 9.82 × 0.5 = 4.91 m/s2. At 0.9 R: 8.838 m/s2. Only the yellow ball closer to the centre pulls you; the shell above you cancels out (the shell theorem).
- Go up to 1.1 R: g = 9.82 ÷ 1.12 = 8.115 m/s2 (the approximation g(1 − 2h ÷ R) = 7.856 m/s2 is only good for small heights). The fall per unit height outside is twice the fall per unit depth inside.
- At the surface, set the latitude to 0° (equator): the spin of the Earth reduces g to 9.7857 m/s2, while at the poles it stays at 9.8195 m/s2 (the real Earth is also flattened, which adds to the difference).
- The lower graph shows the potential V = −GM ÷ r, with −62.5 MJ/kg at the surface and rising towards zero.
Takeaway: g(h) = g0R2 ÷ (R + h)2 ≈ g0(1 − 2h ÷ R); g(d) = g0(1 − d ÷ R); g is least at the equator.
Class 11 — Two bodies and the centre of mass
Concept: when two bodies attract each other, both move round their common centre of mass. For stars, T2 = d3 ÷ (m1 + m2) in years, AU and solar masses.
- Open Two bodies & centre of mass: m1 = 2, m2 = 1 solar masses, d = 2 AU. The centre of mass divides the line in the inverse ratio of the masses: r1 = 1 × 2 ÷ 3 = 0.667 AU and r2 = 1.333 AU. The heavier star moves on the smaller circle.
- Both take T = √(23 ÷ 3) = 1.633 years for one turn. Their speeds are 12.16 km/s and 24.32 km/s, in the ratio m2 : m1.
- Make the masses equal: the two stars follow the same circle of radius d ÷ 2 about the centre. Make star 1 much heavier: it hardly moves, as the Sun does among its planets.
- Change the starting speed: both stars follow ellipses with the centre of mass at a focus.
Takeaway: m1r1 = m2r2; T2 = 4π2d3 ÷ G(m1 + m2); the total mass sets the period.