Can superhero physics make the impossible look merely improbable? In The Boys, Homelander can’t save a plummeting airliner, prompting a viral debate: could Superman do better? We tested the scenario with hard numbers. A Boeing 737‑800 at 75 tons (75,000 kg) falling at 600 mph (≈268 m/s) carries enormous momentum and kinetic energy that must be bled off gently and spread across the airframe to avoid tearing it apart. The math tells us what it would take—and why Hollywood usually cheats.

Key Takeaways

– Shows a 75,000 kg 737 at 600 mph holds ≈2.7 GJ kinetic energy and ≈2.01×10^7 N·s momentum, requiring careful dissipation to avoid catastrophe. – Reveals a 30 s vertical slowdown demands total upward force ≈1.41 MN and ≈4.0 km stopping distance, keeping loads near ~1.9 g on the airframe. – Demonstrates shortening deceleration to 10 s raises force to ≈2.75 MN and loads to ~3.7 g; 5 s needs ≈4.76 MN and ~6.5 g—likely fatal to structures. – Indicates point pushing is deadly: applying meganeutons over a hand-sized 0.02 m² creates 70–240 MPa local stress, far beyond thin skin tolerance. – Suggests without a wide contact area or fictional “force field,” localized forces would puncture the fuselage; experts say distributed deceleration is essential.

Crunching the numbers for a 75‑ton 737 at 600 mph

Start with the givens. Mass m = 75,000 kg. Speed v = 600 mph ≈ 268 m/s. Momentum p = m·v ≈ 75,000×268 ≈ 2.01×10^7 kg·m/s. Kinetic energy KE = ½ m v² ≈ 0.5×75,000×(268)² ≈ 2.7×10^9 J, or about 2.7 gigajoules.

That energy is equivalent to roughly 0.65 tons of TNT (2.7 GJ ÷ 4.184 GJ/ton). It all must go somewhere: into Superman’s work on the air, heat, acoustic shock, and structural deflection—ideally not into shearing the fuselage.

To stop the plane smoothly in time t, the required total upward force is Ftotal = mg + m·(v/t). Gravity mg ≈ 75,000×9.81 ≈ 735,750 N.

– If Superman takes 30 s: a = v/t ≈ 268/30 ≈ 8.94 m/s². Ftotal ≈ 735,750 + 75,000×8.94 ≈ 1,406,250 N ≈ 1.41 MN. Stopping distance s ≈ v·t/2 ≈ 268×30/2 ≈ 4,020 m. – In 20 s: a ≈ 13.41 m/s²; Ftotal ≈ 1.74 MN; s ≈ 2,680 m. – In 10 s: a ≈ 26.82 m/s²; Ftotal ≈ 2.75 MN; s ≈ 1,340 m. – In 5 s: a ≈ 53.64 m/s²; Ftotal ≈ 4.76 MN; s ≈ 670 m. – In 3 s: a ≈ 89.4 m/s²; Ftotal ≈ 7.44 MN; s ≈ 402 m.

These forces translate into airframe “load factor” n = Ftotal/(mg) = 1 + a/g:

– 30 s: n ≈ 1.91 g – 20 s: n ≈ 2.37 g – 10 s: n ≈ 3.74 g – 5 s: n ≈ 6.47 g – 3 s: n ≈ 10.11 g

Passengers can tolerate ~2 g for short periods; beyond ~4–6 g without support, injuries mount quickly. More critically, a transport-category airliner’s structure is designed for distributed aerodynamic loads, not concentrated external pushes. So the deceleration must be both gradual and spread over a large area to keep the airplane intact.

Superhero physics 101: momentum, impulse, and g‑loads

The central constraint is impulse: to change the plane’s momentum by Δp ≈ 2.01×10^7 N·s, Superman must deliver that impulse via a large force over time, or a smaller force over longer time. Short, violent pushes spike forces and g‑loads; long, gentle pushes keep forces manageable. Real-world physics doesn’t care about intent—only the product F·t. As one physicist puts it, the required opposite impulse is enormous, and concentrated loads would shred the fuselage long before the momentum is safely bled off [3].

If he arrives at a large relative speed and “catches” the airplane in a fraction of a second, forces escalate into tens of meganeutons, guaranteed to cause structural failure. The right first move is to match the plane’s velocity vector, then begin a controlled, time-stretched deceleration.

Superhero physics meets aerostructures: contact area kills

Airliner skins are thin, optimized for aerodynamic pressure and distributed lift/drag—not for a point load under the nose or belly. The difference between safe and catastrophic is contact area. Suppose Superman uses a hand-sized 0.02 m². At 30 s deceleration, Ftotal ≈ 1.41×10^6 N, so average contact stress is ≈70 MPa; at 10 s it’s ≈138 MPa; at 5 s ≈238 MPa. That’s far beyond what a thin skin and local frames can carry without denting, buckling, or tearing.

By contrast, spread the same 1.41 MN over 10 m² and average pressure falls to ≈0.14 MPa (≈1.4 atmospheres). That’s still a heavy local surge for doors, joints, and floor beams, but it moves from instantly catastrophic to maybe survivable for a short interval if the load is spread even more widely and tied into primary structure like wing roots. Popular analyses warn that “hand stops” punch through skins; only a widely distributed force—or a fictional field/aura—avoids a can‑opener outcome [1].

A worked benchmark from the canon of superhero physics

A published back‑of‑the‑envelope model of a similar movie scene (a 200,000 kg jet at 600 mph slowed over 29.7 s) finds an upward force of roughly 3.77×10^6 N—nearly 3.8 MN—consistent with our equations scaled to a heavier aircraft and nearly identical speed and time assumptions [2]. Our 75‑ton scenario needs proportionally less: about 1.41 MN for 30 s, 2.75 MN for 10 s, 4.76 MN for 5 s. The physics is linear in mass and roughly inverse in stop time, so doubling the time halves the added force term m·v/t.

The more you stretch time and spread force, the more “doable” the numbers look, with two giant caveats: the required altitude (kilometers) and the absolute need to avoid localized loads.

What a “safe” rescue demands in practice

Put the pieces together, and a realistic recipe emerges:

– Match velocity first. If Superman arrives with near‑zero relative speed, he avoids the destructive spike of an impulsive catch. – Spread the load. He must distribute force across many square meters—say, under both wings or via a broad belly cradle—so local stresses stay low. – Stretch the deceleration. Aim for 30 s or more, keeping the load factor under ~2 g and buying a 4 km stopping distance. – Manage pitch and yaw. Even small offsets create torque; a 1.41 MN push 1 m off‑center adds a 1.41 MN·m moment, risking structural torsion. Use symmetric application points. – Bleed energy mostly into the air. Angle slightly to use aerodynamic drag as an ally; any energy “eaten” by wake and heating is energy not driven into frames and skins.

Real experts stress the same three rules: match speed, distribute force, decelerate gradually, because thin skins and light frames are allergic to point loads [4]. If any of those criteria are missed—especially the second—failure starts locally (puncture, buckling, fastener pull‑through) and cascades.

How much area is “enough” to avoid punching through?

There’s no single threshold, because airliner structure isn’t a uniform shell; it’s a semi‑monocoque network of skin, frames, stringers, floor beams, and spars. But a sanity check helps.

– 30 s case (F ≈ 1.41 MN): To keep average pressure below 0.2 MPa (≈2 atmospheres), needed area A ≈ F/P ≈ 1.41×10^6 / 2×10^5 ≈ 7.05 m². – 10 s case (F ≈ 2.75 MN): A ≈ 13.7 m² for 0.2 MPa. – 5 s case (F ≈ 4.76 MN): A ≈ 23.8 m² for 0.2 MPa.

Those areas rival the size of large belly sections and still ignore local stiffener spacing, cutouts, and attachment limits. Safer still is to route force into the wing/spar system—the airplane’s strongest structure—rather than the fuselage skin. In other words, the “right way” is not palms on the nose; it’s a big, stiff cradle under both wings or a field that couples to the entire airframe.

Don’t forget altitude, speed regime, and airflow

A 600 mph descent near sea level implies dynamic pressure q ≈ ½ρv² ≈ 0.5×1.225×268² ≈ 44 kPa (≈0.44 atm). Even if Superman cancels vertical speed, the air is still slamming the airframe with high transonic loads if the airplane maintains that forward speed. Coordinating the recovery means trimming pitch to keep the wing within survivable angles of attack, avoiding sudden pull‑ups that could spike g‑loads toward structural limits, and burning off speed steadily. It also means you need altitude: about 4 km for a 30 s vertical slowdown from 600 mph, more if you also want to scrub forward speed gently.

What if Superman just “grabs and holds”?

If he instantaneously applies a meganeuton‑plus force while the plane is still hurtling by, two problems erupt:

– Relative motion shears the contact point. A hand on a fast‑moving skin is a knife. – Impulse over milliseconds means forces in the tens of meganeutons. With a hand‑sized area, that’s hundreds to thousands of megapascals—far above yield even ignoring buckling.

This is the fatal flaw in cinematic “catch the nose” shots. Without a broad, conformal contact that couples into primary structure, the first milliseconds of the rescue are the last milliseconds of the airplane.

The big picture: could he do it—safely?

Yes, but only under strict conditions. The numbers show a narrow lane where the physics isn’t instantly self‑destructive:

– He matches the plane’s velocity vector first. – He applies a broad, symmetric, high‑area interface (think many square meters), ideally braced under the wings. – He stretches deceleration to ~30 s or more, keeping loads near 2 g and requiring ~4 km of altitude. – He uses airflow and drag to help, not fight, the energy dissipation.

Even then, it’s razor’s‑edge engineering. Miss on area or time and you get puncture, shear, or g‑overload. That’s why analysts of Superman tropes often stipulate an “aura” or force‑field effect that couples his push to the whole airframe rather than to a palm‑sized patch [1].

Why movies cheat—and what extra “powers” would make it plausible

On screen, the audience sees a hand on aluminum and a jet magically slowing. In real life, the jet rips first. An oft‑cited benchmark calculation for a heavier 200,000 kg jet at 600 mph over 29.7 s yields ≈3.77 MN, showing the orders of magnitude involved when you’re forced to do it “by hand” [2]. Physicists routinely caution that feats like catching planes safely need more than strength: they need mechanisms to distribute force and soak up momentum without destroying the target. That can mean a force field, telekinetic coupling, or altered mass properties—explicitly beyond standard physics [5].

Bottom line: With only muscle and invulnerability, Superman can certainly generate the force, but he can’t safely deliver it to a fragile airframe unless he slows the event way down and spreads the push way out. That, more than raw strength, is the heart of credible superhero physics.

Sources: [1] Gizmodo – The physics of Superman: https://gizmodo.com/the-physics-of-superman-5810945 [2] The Physics Factbook / Hypertextbook – Force of a Superhero: Superman Returns: https://hypertextbook.com/facts/2007/superman.shtml [3] AskThePhysicist.com – Can Superman stop a plane?: www.askthephysicist.com/ask_phys_q%26a_old12.html” target=”_blank” rel=”nofollow noopener noreferrer”>https://www.askthephysicist.com/ask_phys_q%26a_old12.html [4] Bryant University News – Up, up, and awry: Movie superheroes’ never-ending battle with real-world physics: https://news.bryant.edu/and-awry-movie-superheroes-never-ending-battle-real-world-physics [5] IGN – Neil deGrasse Tyson on superhero physics (Superman example): www.ign.com/articles/neil-degrasse-tyson-explains-which-superhero-film-has-the-most-unrealistic-physics” target=”_blank” rel=”nofollow noopener noreferrer”>https://www.ign.com/articles/neil-degrasse-tyson-explains-which-superhero-film-has-the-most-unrealistic-physics

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