Physics · Mechanics · Grade 10

Free fall and air

A coin and a feather fall differently in air but together in a vacuum — why?

The situation

Until Galileo's time most people thought that a heavy body falls faster than a light one. And indeed, if you let go of a coin and a feather at the same moment, the coin hits the floor at once, while the feather flutters down through the air. The speed of a parachutist jumping from a plane also does not increase without limit — it levels off at about 200 km/h. What really decides how things fall: mass, air or the planet? We test this with Newton's tube, a parachutist, a thrown ball and light gates in the laboratory.

Topic

Free fall and air resistance. Feather and coin in a vacuum tube, a parachutist's terminal velocity

Key ideas

Acceleration of free fall
Air resistance
Terminal velocity

Formulas

v = g t, h = \frac{g t^{2}}{2}

Free fall: no initial speed, no air resistance.

y = v_{0} t - \frac{g t^{2}}{2}, t_{k} = \frac{v_{0}}{g}, h_{m a x} = \frac{v_{0}^{2}}{2 g}

An object thrown vertically upwards: time of ascent and maximum height.

F_{q} = \frac{1}{2} C ρ S v^{2}, v_{c h} = \sqrt{\frac{2 m g}{C ρ S}}

Air resistance at high speed; the terminal velocity comes from the condition $F_{d} = m g$ .

g = \frac{G M}{{(R + h)}^{2}}

The acceleration of free fall depends on the distance to the planet's centre.

g = \frac{2 (h_{2} - h_{1})}{t_{2}^{2} - t_{1}^{2}}, g = \frac{4 π^{2} L}{T^{2}}

Laboratory: the two-light-gate and simple pendulum methods.

Theory

Why do all objects fall alike?

The Earth pulls every object with a force proportional to its mass: $F = m g$ . But the greater the mass, the harder it is to accelerate (inertia): $a = F / m = g$ . The mass cancels — without air, a coin and a feather fall with the same acceleration. Pump the air out of Newton's tube: from a height of 0.90 m both land in $\sqrt{2 h / g} \approx 0.43$ s. Motion under gravity alone, with no initial speed, is free fall, a special case of uniformly accelerated motion.

Air resistance depends on speed

At low speeds (a dust particle, a small ball in honey) the resistance is proportional to speed — Stokes' law (the “Viscosity and flow” experiment). At high speed it is proportional to the square of the speed: $F_{d} = \frac{1}{2} C ρ S v^{2}$ . As the speed increases, $F_{d}$ grows; when $F_{d} = m g$ the acceleration is zero and the object falls uniformly at its terminal velocity. For a feather this is ≈ 0.6 m/s, for a parachutist ≈ 55–60 m/s, and with an open parachute ≈ 5–6 m/s. If the air becomes thinner ( $ρ$ decreases), $v_{t}$ increases.

The parachutist

Just after the jump, $a \approx g$ ; as the speed increases the resistance grows and the acceleration falls — the $v (t)$ graph bends over and approaches the terminal velocity line (after ~15 s and ~600 m). Falling belly-down the area is large, head-down it is small, so the terminal velocity is higher. When the parachute opens, $C S$ increases about 80 times: for a few seconds the resistance exceeds the weight and slows the person down (≈ 3–4 g), and then a new terminal velocity is established. At altitude the air is thinner, so at 4 km the terminal velocity is about 16 % higher than at 1 km.

Vertically and horizontally launched objects

For a vertical throw, $y = v_{0} t - g t^{2} / 2$ and $v = v_{0} - g t$ . At the highest point $v = 0$ : $t_{k} = v_{0} / g$ , $h_{m a x} = v_{0}^{2} / 2 g$ . Motion without air is symmetric: the fall also takes $v_{0} / g$ , the object returns to the hand with speed $v_{0}$ , and the strobe dots lie at the same heights. With air, the rise is shorter and the fall longer. For an object launched horizontally the two motions are independent: uniform horizontally and free fall vertically — it reaches the ground at the same moment as an object that is simply dropped. Launching at an angle to the horizontal is in the “Basketball: the free throw” experiment.

g far from the Earth

From the law of universal gravitation, $m g = G M m / {(R + h)}^{2}$ , so $g = G M / {(R + h)}^{2}$ : if the distance to the centre increases 2 times ( $h = R$ ), $g$ falls 4 times. On the International Space Station (≈ 400 km), $g \approx 8.7$ m/s² — the weightlessness there is not due to an absence of gravity but to free fall together with the station. Even on the Earth's surface $g$ is not the same everywhere: ≈ 9.78 m/s² at the equator and ≈ 9.83 m/s² at the poles (because of the Earth's rotation and shape), and ≈ 9.80 m/s² in Tashkent.

Measuring g (laboratory work)

Two gates: a ball is released from rest, and a stopwatch shows $t_{1}$ and $t_{2}$ ; from $h = g t^{2} / 2$ , $g = 2 (h_{2} - h_{1}) / (t_{2}^{2} - t_{1}^{2})$ . If you plot the measurements on an $h$ – $t^{2}$ graph, the slope of the straight line is $g / 2$ . Simple pendulum: for a small angle $T = 2 π \sqrt{L / g}$ , $g = 4 π^{2} L / T^{2}$ ; measure the time for 10 oscillations and divide by 10, and keep the angle below 10° (at 20°, $g$ comes out ≈ 1.5 % too low). For a physical (reversible) pendulum, $T = 2 π \sqrt{I / (m g d)}$ ; when the periods about the two pivots are equal, the distance between them equals the reduced length $l_{k}$ and $g = 4 π^{2} l_{k} / T^{2}$ (in detail — “Swing”, “Spring pendulum”). The model: $C$ is constant, there is no wind, the parachute uses the standard atmosphere, and elsewhere $g$ and $ρ$ are constant.

Did you know?

In 1971 David Scott, commander of the Apollo 15 mission, dropped a geological hammer and a falcon feather on the Moon's surface at the same moment — because there is no air, both reached the surface together.
In 2012 Felix Baumgartner jumped from a capsule carried by a stratospheric balloon at about 39 km and reached a speed of about 1358 km/h: the air there is so thin that he exceeded the speed of sound.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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Centre of gravity and equilibrium — Grade 10
Aeroplane wing — Grade 10
Dynamics of a rigid body — Grade 10
Non-inertial frames of reference — Grade 10
Inclined plane and Atwood machine — Grade 10
Jet propulsion and the rocket — Grade 10
Viscosity and flow — Grade 10
Gyroscope and precession — Year 1
Lagrangian mechanics and chaos — Year 2
Hamiltonian mechanics — Year 2