TVvoodoo
Ambassador of The Comfortably Slung
I was feeling a little sporty today and in my head made a guess of approx 25 minutes. Then I asked Grok AI... make your own guess before reading the answer!
How long would it take for a 1 lb. stone to sink to the very bottom of the Marianas trench in the Western Pacific Ocean?
To estimate how long it would take for a 1-pound stone to sink to the bottom of the Mariana Trench, we need to consider the physics of the stone's descent through water, including its terminal velocity, the depth of the trench, and relevant environmental factors. The Mariana Trench’s deepest point, the Challenger Deep, is approximately 10,984 meters (36,037 feet) below sea level. Let’s break this down step by step.Assumptions and Simplifications
- Stone properties: The stone is assumed to be roughly spherical with a mass of 1 pound (0.453592 kg). We’ll estimate its density as that of typical rock, such as granite (~2700 kg/m³), to calculate its volume and size.
- Water properties: Seawater density is approximately 1025 kg/m³ at the surface, increasing slightly with depth due to pressure, but we’ll use a constant 1025 kg/m³ for simplicity.
- Descent dynamics: The stone will reach a terminal velocity quickly due to drag forces balancing its buoyant weight. We assume it falls straight down, unaffected by currents.
- Drag model: We’ll use Stokes’ law for a spherical object at low Reynolds numbers, but since the stone may be larger, we’ll also consider a quadratic drag model (more realistic for larger objects).
- No initial velocity: The stone is dropped from rest at the surface.
- Neglecting compressibility: We assume the stone’s density and water’s properties remain relatively constant, though pressure increases significantly with depth.
- Mass:
m=0.453592 kgm = 0.453592 \, \text{kg}m = 0.453592 \, \text{kg}
. - Density of stone:
ρstone=2700 kg/m3\rho_{\text{stone}} = 2700 \, \text{kg/m}^3\rho_{\text{stone}} = 2700 \, \text{kg/m}^3
. - Volume of stone:
V=mρstone=0.4535922700≈1.68×10−4 m3.V = \frac{m}{\rho_{\text{stone}}} = \frac{0.453592}{2700} \approx 1.68 \times 10^{-4} \, \text{m}^3.V = \frac{m}{\rho_{\text{stone}}} = \frac{0.453592}{2700} \approx 1.68 \times 10^{-4} \, \text{m}^3. - Radius of spherical stone: Assuming a sphere,
V=43πr3V = \frac{4}{3}\pi r^3V = \frac{4}{3}\pi r^3
.
r=(3V4π)1/3=(3×1.68×10−44×3.14159)1/3≈0.0342 m (3.42 cm).r = \left( \frac{3V}{4\pi} \right)^{1/3} = \left( \frac{3 \times 1.68 \times 10^{-4}}{4 \times 3.14159} \right)^{1/3} \approx 0.0342 \, \text{m} \, (3.42 \, \text{cm}).r = \left( \frac{3V}{4\pi} \right)^{1/3} = \left( \frac{3 \times 1.68 \times 10^{-4}}{4 \times 3.14159} \right)^{1/3} \approx 0.0342 \, \text{m} \, (3.42 \, \text{cm}).
The stone has a radius of ~3.42 cm, so a diameter of ~6.84 cm (about the size of a small rock).
- Weight of stone:
Fg=mg=0.453592×9.81≈4.45 N.F_g = m g = 0.453592 \times 9.81 \approx 4.45 \, \text{N}.F_g = m g = 0.453592 \times 9.81 \approx 4.45 \, \text{N}. - Buoyant force:
Fb=ρwaterVg=1025×1.68×10−4×9.81≈1.69 N.F_b = \rho_{\text{water}} V g = 1025 \times 1.68 \times 10^{-4} \times 9.81 \approx 1.69 \, \text{N}.F_b = \rho_{\text{water}} V g = 1025 \times 1.68 \times 10^{-4} \times 9.81 \approx 1.69 \, \text{N}. - Net force at terminal velocity:
Fnet=Fg−Fb=4.45−1.69=2.76 N.F_{\text{net}} = F_g - F_b = 4.45 - 1.69 = 2.76 \, \text{N}.F_{\text{net}} = F_g - F_b = 4.45 - 1.69 = 2.76 \, \text{N}.
This net force is balanced by drag at terminal velocity.
Fd=6πηrv,F_d = 6 \pi \eta r v,F_d = 6 \pi \eta r v,
where \eta = 1.0 \times 10^{-3} \, \text{Pa·s} (viscosity of water at 20°C),
r=0.0342 mr = 0.0342 \, \text{m}r = 0.0342 \, \text{m}
, and ( v ) is terminal velocity. At terminal velocity:
Fd=Fg−Fb=2.76 N.F_d = F_g - F_b = 2.76 \, \text{N}.F_d = F_g - F_b = 2.76 \, \text{N}.
6πηrv=2.76.6 \pi \eta r v = 2.76.6 \pi \eta r v = 2.76.
v=2.766π×1.0×10−3×0.0342≈2.766.43×10−4≈4290 m/s.v = \frac{2.76}{6 \pi \times 1.0 \times 10^{-3} \times 0.0342} \approx \frac{2.76}{6.43 \times 10^{-4}} \approx 4290 \, \text{m/s}.v = \frac{2.76}{6 \pi \times 1.0 \times 10^{-3} \times 0.0342} \approx \frac{2.76}{6.43 \times 10^{-4}} \approx 4290 \, \text{m/s}.
This velocity (4.3 km/s) is unrealistically high for an object sinking in water, indicating Stokes’ law is inappropriate because the Reynolds number is too high for laminar flow.Quadratic Drag (Turbulent Flow)For larger objects or higher speeds, quadratic drag is more appropriate:
Fd=12ρwaterCdAv2,F_d = \frac{1}{2} \rho_{\text{water}} C_d A v^2,F_d = \frac{1}{2} \rho_{\text{water}} C_d A v^2,
where:
- Cd≈0.47C_d \approx 0.47C_d \approx 0.47
(drag coefficient for a sphere), - A=πr2=π×(0.0342)2≈0.00367 m2A = \pi r^2 = \pi \times (0.0342)^2 \approx 0.00367 \, \text{m}^2A = \pi r^2 = \pi \times (0.0342)^2 \approx 0.00367 \, \text{m}^2
, - ρwater=1025 kg/m3\rho_{\text{water}} = 1025 \, \text{kg/m}^3\rho_{\text{water}} = 1025 \, \text{kg/m}^3
.
12ρwaterCdAv2=Fg−Fb=2.76.\frac{1}{2} \rho_{\text{water}} C_d A v^2 = F_g - F_b = 2.76.\frac{1}{2} \rho_{\text{water}} C_d A v^2 = F_g - F_b = 2.76.
12×1025×0.47×0.00367×v2=2.76.\frac{1}{2} \times 1025 \times 0.47 \times 0.00367 \times v^2 = 2.76.\frac{1}{2} \times 1025 \times 0.47 \times 0.00367 \times v^2 = 2.76.
0.883v2=2.76.0.883 v^2 = 2.76.0.883 v^2 = 2.76.
v2=2.760.883≈3.125,v≈3.125≈1.77 m/s.v^2 = \frac{2.76}{0.883} \approx 3.125, \quad v \approx \sqrt{3.125} \approx 1.77 \, \text{m/s}.v^2 = \frac{2.76}{0.883} \approx 3.125, \quad v \approx \sqrt{3.125} \approx 1.77 \, \text{m/s}.
Reynolds Number CheckTo confirm the flow regime:
Re=ρwatervdη=1025×1.77×(2×0.0342)1.0×10−3≈1.24×105.\text{Re} = \frac{\rho_{\text{water}} v d}{\eta} = \frac{1025 \times 1.77 \times (2 \times 0.0342)}{1.0 \times 10^{-3}} \approx 1.24 \times 10^5.\text{Re} = \frac{\rho_{\text{water}} v d}{\eta} = \frac{1025 \times 1.77 \times (2 \times 0.0342)}{1.0 \times 10^{-3}} \approx 1.24 \times 10^5.
Since
Re≫1000\text{Re} \gg 1000\text{Re} \gg 1000
, the flow is turbulent, and the quadratic drag model is appropriate. The terminal velocity is ~1.77 m/s.Step 4: Time to Reach the BottomAssuming the stone reaches terminal velocity almost instantly (common for dense objects in water), the time to sink to the bottom of the Mariana Trench (10,984 m) is:
t=depthv=109841.77≈6206 seconds.t = \frac{\text{depth}}{v} = \frac{10984}{1.77} \approx 6206 \, \text{seconds}.t = \frac{\text{depth}}{v} = \frac{10984}{1.77} \approx 6206 \, \text{seconds}.
Convert to hours:
6206÷3600≈1.72 hours.6206 \div 3600 \approx 1.72 \, \text{hours}.6206 \div 3600 \approx 1.72 \, \text{hours}.
Step 5: Considerations and Adjustments
- Depth variation: The Mariana Trench’s depth is often cited as ~11,000 m, but 10,984 m is a precise measurement for the Challenger Deep. Variations in depth estimates would slightly adjust the time.
- Water density increase: At great depths, water density increases to 1050 kg/m³ due to compression, slightly increasing buoyancy and reducing terminal velocity. This effect is small (2–3% change in density), so our estimate remains reasonable.
- Currents and non-spherical shape: Ocean currents or an irregular stone shape could affect descent, but we assume a straight path and spherical shape for simplicity.
- Initial acceleration: The stone takes a short time to reach terminal velocity, but this is negligible over 11 km.
If you’d like me to refine the calculation with specific stone dimensions, account for variable water density, or consider ocean currents, let me know!
