Asked by .Yuki.
2 skydivers of different masses jump from a plane. Explain how their falls compare including the effects of gravity mass and air resistance
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Short answer
- In vacuum both skydivers fall identically (same acceleration g) regardless of mass.
- In air, both feel gravity mg downward and a drag force upward that grows roughly like 1/2 ρ Cd A v^2. The net acceleration is a = g − D/m. Because D is divided by m, the heavier diver (same shape/area) has a larger net acceleration at a given speed and therefore a higher terminal speed. The lighter diver reaches a lower terminal velocity sooner.
Explanation and formulas
- Weight: Fg = m g.
- Aerodynamic drag (approximate for turbulent flow): D = 1/2 ρ Cd A v^2, where ρ is air density, Cd the drag coefficient, A the cross‑sectional area, and v the speed.
- Newton’s 2nd law: m a = m g − D, so a = g − (D/m) = g − (ρ Cd A v^2)/(2 m).
- Early in the fall v is small so D ≈ 0 and a ≈ g for both.
- As v increases, D grows ~v^2 and reduces the acceleration. Because the drag term is divided by m, the same drag slows a lighter person more.
- Terminal velocity (when m g = D):
v_t = sqrt(2 m g / (ρ Cd A)).
So, for the same Cd and A, v_t increases with the square root of mass. If the heavier person also has a larger area, that offsets some of the effect; if mass and area scale like m ∝ L^3 and A ∝ L^2 (geometrically similar bodies), v_t ∝ m^(1/6) — a weak dependence.
Practical consequences
- Two skydivers with the same posture (same Cd and A): the heavier one falls faster after drag becomes significant and has a higher terminal velocity.
- If one adopts a different posture (spread-eagle vs head-first) the change in A and Cd can dominate mass effects; posture and parachute size are the main controls for safe descent speed.
- Typical behavior: initial fall nearly g for both, then lighter reaches a lower terminal speed sooner; heavier continues to accelerate longer and levels off at a higher speed.
If you want, I can plug in numbers for two specific masses and a realistic Cd/A to show the quantitative difference in terminal speeds.
- In vacuum both skydivers fall identically (same acceleration g) regardless of mass.
- In air, both feel gravity mg downward and a drag force upward that grows roughly like 1/2 ρ Cd A v^2. The net acceleration is a = g − D/m. Because D is divided by m, the heavier diver (same shape/area) has a larger net acceleration at a given speed and therefore a higher terminal speed. The lighter diver reaches a lower terminal velocity sooner.
Explanation and formulas
- Weight: Fg = m g.
- Aerodynamic drag (approximate for turbulent flow): D = 1/2 ρ Cd A v^2, where ρ is air density, Cd the drag coefficient, A the cross‑sectional area, and v the speed.
- Newton’s 2nd law: m a = m g − D, so a = g − (D/m) = g − (ρ Cd A v^2)/(2 m).
- Early in the fall v is small so D ≈ 0 and a ≈ g for both.
- As v increases, D grows ~v^2 and reduces the acceleration. Because the drag term is divided by m, the same drag slows a lighter person more.
- Terminal velocity (when m g = D):
v_t = sqrt(2 m g / (ρ Cd A)).
So, for the same Cd and A, v_t increases with the square root of mass. If the heavier person also has a larger area, that offsets some of the effect; if mass and area scale like m ∝ L^3 and A ∝ L^2 (geometrically similar bodies), v_t ∝ m^(1/6) — a weak dependence.
Practical consequences
- Two skydivers with the same posture (same Cd and A): the heavier one falls faster after drag becomes significant and has a higher terminal velocity.
- If one adopts a different posture (spread-eagle vs head-first) the change in A and Cd can dominate mass effects; posture and parachute size are the main controls for safe descent speed.
- Typical behavior: initial fall nearly g for both, then lighter reaches a lower terminal speed sooner; heavier continues to accelerate longer and levels off at a higher speed.
If you want, I can plug in numbers for two specific masses and a realistic Cd/A to show the quantitative difference in terminal speeds.
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