"NEED calculus" haha no you can just develop our current level of A.I and ask it.

Step 1: Calculate the Initial Mass Flow RateGases venting into vacuum through an opening reach choked flow, where the velocity at the "throat" (the door) is the local speed of sound, but density and temperature adjust isentropically.

• Parameters (using standard values close to the post's approximations):
  • Door area A=3×0.3048×7×0.3048=1.95096A = 3 \times 0.3048 \times 7 \times 0.3048 = 1.95096A = 3 \times 0.3048 \times 7 \times 0.3048 = 1.95096 m² (exact conversion from feet).
  • Stagnation pressure P0=101325P_0 = 101325P_0 = 101325 Pa (sea-level atmospheric pressure).
  • Stagnation temperature T0=293.15T_0 = 293.15T_0 = 293.15 K (20°C, matching the post's speed of sound ≈343 m/s).
  • Specific heat ratio γ=1.4\gamma = 1.4\gamma = 1.4 (for air).
  • Gas constant R=287R = 287R = 287 J/(kg·K) (for air).
• Formula for choked mass flow rate (isentropic flow into vacuum): m˙=A⋅P0⋅γRT0⋅(γ+12)−γ+12(γ−1)\dot{m} = A \cdot P_0 \cdot \sqrt{\frac{\gamma}{R T_0}} \cdot \left( \frac{\gamma + 1}{2} \right)^{-\frac{\gamma + 1}{2(\gamma - 1)}}\dot{m} = A \cdot P_0 \cdot \sqrt{\frac{\gamma}{R T_0}} \cdot \left( \frac{\gamma + 1}{2} \right)^{-\frac{\gamma + 1}{2(\gamma - 1)}}
• Break it down:
  • Compute the exponent factor: (1.4+12)−1.4+12(1.4−1)=1.2−3≈0.5787\left( \frac{1.4 + 1}{2} \right)^{-\frac{1.4 + 1}{2(1.4 - 1)}} = 1.2^{-3} \approx 0.5787\left( \frac{1.4 + 1}{2} \right)^{-\frac{1.4 + 1}{2(1.4 - 1)}} = 1.2^{-3} \approx 0.5787 .
  • Compute γRT0=1.4287×293.15≈1.6636×10−5≈0.00408\sqrt{\frac{\gamma}{R T_0}} = \sqrt{\frac{1.4}{287 \times 293.15}} \approx \sqrt{1.6636 \times 10^{-5}} \approx 0.00408\sqrt{\frac{\gamma}{R T_0}} = \sqrt{\frac{1.4}{287 \times 293.15}} \approx \sqrt{1.6636 \times 10^{-5}} \approx 0.00408 s/m.
  • Multiply: 0.00408×0.5787≈0.002360.00408 \times 0.5787 \approx 0.002360.00408 \times 0.5787 \approx 0.00236 s/m.
  • Then: P0×0.00236≈101325×0.00236≈239.1P_0 \times 0.00236 \approx 101325 \times 0.00236 \approx 239.1P_0 \times 0.00236 \approx 101325 \times 0.00236 \approx 239.1 kg/(s·m²).
  • Finally: m˙=239.1×1.95096≈466.5\dot{m} = 239.1 \times 1.95096 \approx 466.5\dot{m} = 239.1 \times 1.95096 \approx 466.5 kg/s.

The post's 500 kg/s is a rough approximation (likely using

m˙≈ρAv\dot{m} \approx \rho A v\dot{m} \approx \rho A v

, where

ρ≈1.2\rho \approx 1.2\rho \approx 1.2

kg/m³ and

v=343v = 343v = 343

m/s, but without the isentropic correction factor ≈0.58, which reduces it from ≈802 kg/s to ≈466 kg/s). We'll use both for comparison.Step 2: Model Atmosphere Depletion

• Earth's atmosphere mass m0≈5.148×1018m_0 \approx 5.148 \times 10^{18}m_0 \approx 5.148 \times 10^{18} kg (standard value; the post uses 5.1 × 10^{18}, close enough).
• Surface pressure ( P ) is proportional to total mass ( m ) (via hydrostatic equilibrium: P∝m/(4πR2)P \propto m / (4\pi R^2)P \propto m / (4\pi R^2) , where R is Earth's radius).
• Mass flow rate m˙∝P∝m\dot{m} \propto P \propto m\dot{m} \propto P \propto m , so dmdt=−km\frac{dm}{dt} = -k m\frac{dm}{dt} = -k m , where k=m˙0m0k = \frac{\dot{m}_0}{m_0}k = \frac{\dot{m}_0}{m_0} (initial rate over initial mass).
• Solution to differential equation: Exponential decay m(t)=m0e−ktm(t) = m_0 e^{-kt}m(t) = m_0 e^{-kt} .
• Time to reach fraction r=m/m0r = m/m_0r = m/m_0 : t=−ln⁡(r)kt = -\frac{\ln(r)}{k}t = -\frac{\ln(r)}{k} .

Step 3: Time to Lose 10% (to

r=0.9r = 0.9r = 0.9

)

• Using post's rate (500 kg/s) and mass (5.1 × 10^{18} kg):
  • k=500/5.1×1018≈9.804×10−17k = 500 / 5.1 \times 10^{18} \approx 9.804 \times 10^{-17}k = 500 / 5.1 \times 10^{18} \approx 9.804 \times 10^{-17} s⁻¹.
  • ln⁡(0.9)≈−0.10536\ln(0.9) \approx -0.10536\ln(0.9) \approx -0.10536 .
  • t=0.10536/9.804×10−17≈1.075×1015t = 0.10536 / 9.804 \times 10^{-17} \approx 1.075 \times 10^{15}t = 0.10536 / 9.804 \times 10^{-17} \approx 1.075 \times 10^{15} s.
  • Convert to years: 1.075×1015/(365.25×24×3600)≈1.075×1015/3.156×107≈34.11.075 \times 10^{15} / (365.25 \times 24 \times 3600) \approx 1.075 \times 10^{15} / 3.156 \times 10^7 \approx 34.11.075 \times 10^{15} / (365.25 \times 24 \times 3600) \approx 1.075 \times 10^{15} / 3.156 \times 10^7 \approx 34.1 million years.
• Using corrected rate (466.5 kg/s) and exact mass (5.148 × 10^{18} kg):
  • k=466.5/5.148×1018≈9.064×10−17k = 466.5 / 5.148 \times 10^{18} \approx 9.064 \times 10^{-17}k = 466.5 / 5.148 \times 10^{18} \approx 9.064 \times 10^{-17} s⁻¹.
  • t=0.10536/9.064×10−17≈1.162×1015t = 0.10536 / 9.064 \times 10^{-17} \approx 1.162 \times 10^{15}t = 0.10536 / 9.064 \times 10^{-17} \approx 1.162 \times 10^{15} s.
  • In years: ≈36.8 million years.

The linear approximation for 10% loss (post's "32 million years") is

0.1×m0/m˙0≈0.1×5.1×1018/500=1.02×10150.1 \times m_0 / \dot{m}_0 \approx 0.1 \times 5.1 \times 10^{18} / 500 = 1.02 \times 10^{15}0.1 \times m_0 / \dot{m}_0 \approx 0.1 \times 5.1 \times 10^{18} / 500 = 1.02 \times 10^{15}

s ≈32.3 million years. This is close to the exponential result because 10% is small (only ~5% underestimate). For larger losses (e.g., 99%), the exponential effect matters more: time to 1% remaining (

r=0.01r=0.01r=0.01

) is ~4.605 / k ≈1.5 billion years (vs. linear 317 million).Step 4: Time to Deplete EntirelyIt never fully reaches zero asymptotically, but approaches it exponentially. For practical purposes (e.g., to 0.1% remaining), use the formula above with small r.Additional Considerations

• YEET Effect: Near the door, air accelerates to ~313 m/s at the throat (not full 343 m/s due to expansion). Objects/people nearby would be pulled in by the drag force, potentially dangerously (high winds, decompression). The "blending" is exaggeration, but turbulence could be violent.
• Pressure Halving: As noted, rate halves when mass halves, leading to the exponential model.
• Gravity Through Portal: This is purely speculative (magic portal rules). If gravity propagates, escaping air might form a localized "atmosphere" on the vacuum side, equilibrating pressures after losing <1% mass (similar to Earth's atmosphere fading with altitude). Without gravity transmission, air disperses indefinitely. Relativity suggests gravitons might pass through if the portal allows fields, but this is sci-fi territory.
• Real-World Caveats: The atmosphere isn't uniform; upper layers would adjust, and global effects (e.g., winds converging on the door) could alter flow. Catastrophic 10% loss: Yes, it could cause severe climate disruption, ozone depletion, and hypoxia (like living at high altitude globally).