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Terminal velocity

Gravity and orbits · level 5, lesson 5 of 6

A question to start

A skydiver on Earth falls at about 55 m/s once they stop speeding up. Mars has 38% of the gravity but much thinner air. Would a skydiver there fall faster or slower, and by how much?

The idea

Two forces act on a falling body: its weight \(mg\) down, and drag up. Drag grows with the square of the speed:

\[F_\text{drag} = \tfrac12 \rho v^2 C_d A,\]

where \(\rho\) is the density of the air, \(A\) the area facing the flow and \(C_d\) a number for the shape (about 1 for a skydiver spread flat, 0.5 for a ball, about 1.3 for a round parachute). The rockets topic explains where this formula comes from in level 7; here we use it.

At first drag is small and the body speeds up. As \(v\) grows, drag catches up. When drag equals weight, the net force is zero and the speed stops changing:

\[\tfrac12 \rho v_t^2 C_d A = mg \qquad\Rightarrow\qquad v_t = \sqrt{\frac{2mg}{\rho C_d A}}.\]

Earth and Mars. Everything else equal, \(v_t \propto \sqrt{g/\rho}\).

Mars Earth
\(g\) 3.71 m/s² 9.81 m/s²
Air density at the ground, \(\rho\) about 0.020 kg/m³ 1.225 kg/m³
\(\sqrt{g/\rho}\) 13.6 2.83

So \(v_t\) on Mars is about \(13.6 / 2.83 = 4.8\) times as big. The skydiver's 55 m/s becomes about 265 m/s, faster than the speed of sound in Mars air (about 240 m/s). A parachute that would land you gently at home would still hit the ground on Mars at the speed of a fast car.

Worked examples

1. The skydiver. Mass 80 kg, \(C_d A = 0.43\) m². Earth: \(v_t = \sqrt{2 \times 80 \times 9.81 / (1.225 \times 0.43)} = \sqrt{2980} = 55\) m/s. Mars: \(\sqrt{2 \times 80 \times 3.71 / (0.020 \times 0.43)} = \sqrt{69\,000} = 263\) m/s.

2. A dust grain. For tiny dust the drag formula changes (the air flows smoothly round it), but the idea holds: light, small things have tiny terminal speeds, which is why fine dust can stay up in Mars storms for weeks.

3. Doubling the area. If a skydiver could double their area, \(v_t\) drops by \(\sqrt2\): from 55 to 39 m/s.

Try it

  1. Write down the balance of forces at terminal velocity.
  2. A 2 kg ball with \(C_d A = 0.02\) m² falls through Earth air. Find its terminal velocity.
  3. The same ball on Mars?
  4. How much bigger must a parachute's area be on Mars than on Earth to land the same load at the same speed?
  5. Why does a heavier skydiver fall faster?
Answers
  1. \(\tfrac12 \rho v_t^2 C_d A = mg\).
  2. \(\sqrt{2 \times 2 \times 9.81 / (1.225 \times 0.02)} = \sqrt{1602} = 40\) m/s.
  3. \(\sqrt{2 \times 2 \times 3.71 / (0.020 \times 0.02)} = \sqrt{37\,100} = 193\) m/s.
  4. \(A \propto g/\rho\) for the same \(v_t\): \((3.71/0.020)/(9.81/1.225) = 185.5/8.0 \approx 23\) times the area.
  5. \(v_t \propto \sqrt{m}\): more weight needs more drag to balance, which needs more speed.

Common mistake

Thinking terminal velocity is only about the air. Gravity matters too: \(v_t\) depends on \(g/\rho\). On Mars both are lower, but \(\rho\) is far lower, so the fall is faster.

Where it leads