Short answer first: the sunlight hitting Earth delivers ~1.74×10^17 W (174 petawatts, PW) at the top of the atmosphere. Per square metre at ground under full sun you get about 1,000 W peak; with real PV panels today you can typically get ~150–300 W peak per m² and ~30–60 W/m² on average over day/night and weather.

Below I show the main numbers and quick calculations so you can see where these come from.

Key numbers & concepts

• Solar constant (top of atmosphere): ≈ 1361 W/m².
  Multiply by Earth’s cross-section (πR²) → power intercepted by Earth ≈ 1.74×10^17 W (173.5 PW).

• Average over whole Earth’s surface: solar_const / 4 ≈ 340 W/m² (averaged over day/night and whole globe at TOA). After atmospheric losses the long-term global average reaching the ground is commonly taken ≈ 200–240 W/m² (varies by location/season).

• Peak at surface (clear noon, AM1.5): ≈ 1000 W/m². That’s the usual “1 sun” reference.

• Per-m² electricity (simple conversions):

  • 100% conversion of peak sun: 1000 W electrical / m² (physically unattainable).

  • Typical commercial panels (20% eff) at peak: ≈200 W / m².

  • Lab-record or concentrator systems can reach higher peak efficiencies (hundreds of W/m²), but require optics, cooling, and are not generic rooftop solutions.

  • Average (24-hour, global average) with 20% panels: if average irradiance ≈240 W/m² then electricity ≈ 240 × 0.20 = 48 W/m² (this is the long-term average you’d get per square metre of panels if they were everywhere).

• Global potential on land (simple estimate):
  Earth land area ≈ 1.49×10^14 m² (≈29% of Earth surface).
  Incoming on land (avg ≈240 W/m²) → ≈ 3.55×10^16 W (35.5 PW) incident on land.
  If covered with 20% efficient PV → ≈7.1×10^15 W (7.1 PW) of electricity — ~400× current human global power use (~18 TW ≈ 1.8×10^13 W).
  (So even covering a small fraction of deserts or rooftops could supply humanity many times over.)

Theoretical efficiency limits

• Single-junction detailed-balance (Shockley–Queisser) limit: ~33% for a single p-n junction under unconcentrated sunlight (this is the well-known theoretical limit for one-junction cells).

• Multi-junction cells & concentration: by stacking bandgaps or using solar concentrators you can exceed that single-junction limit; lab multi-junction cells under concentration have reached much higher efficiencies (commercial multi-junction CPV systems are complex and costly).

• There are also thermodynamic/detailed-balance upper bounds for arbitrary converters, but practical engineering, non-ideal optics, temperature and other losses keep real systems well below absolute thermodynamic maxima.

Intuition & practical takeaways

• Per m² numbers are the easiest to use: peak sun ≈1000 W/m²; with good commercial PV (20–25%) you get ≈200–250 W/m² peak, ≈40–60 W/m² averaged over day/night.

• There’s plenty of solar energy — orders of magnitude more than human demand — the limits in practice are land use, storage, transmission, intermittency, material & economic constraints, and siting rather than an absolute shortage of energy from sunlight.

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If you want, I can:

• run the same quick calculations for a specific location (give me latitude / rooftop area) to estimate yearly energy from PV, or

• show a short table comparing panel efficiencies (typical commercial, best lab, concentrator) and expected W/m² (peak and averaged).

Which of those would be most useful?

You said:

the limits in practice are land use, storage, transmission, intermittency, material & economic constraints

Can you take these limits into account in your calculations

ChatGPT said:

Great — I’ll fold the practical limits you named into concrete, careful calculations and show the arithmetic step-by-step so you can see how the numbers change when we account for land-use, transmission, intermittency/storage, and material/economic constraints.

I'll use these baseline numbers (from the earlier message) unless you want different assumptions:

• Peak sun at surface: 1000 W/m².

• Representative panel efficiency: 20% (commercial panels).

• Long-term average ground irradiance (global mean): 240 W/m².

• Global primary electricity demand (approximate baseline to meet): 18 TW = $18\times 10^{12}$ W.

• Earth land area: 1.49×10¹⁴ m².

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1) Basic per-m² and global potential (reminder, step arithmetic)

Panel peak output per m² = $1000\ \text{W/m}^2 \times 0.20 = 200\ \text{W/m}^2.$
Average electrical output per m² (24-hour average) = $240\ \text{W/m}^2 \times 0.20 = 48\ \text{W/m}^2.$

Capacity factor (CF) = $\dfrac{48}{200} = 0.24$ (24%).

Total solar incident on land (from earlier) ≈ $3.55\times 10^{16}$ W.
If every m² of land were covered with 20% PV (purely hypothetical): electricity = $3.55\times 10^{16}\times 0.20 = 7.10\times 10^{15}$ W = 7.10×10¹⁵ W (7.10 PW).
(We are not proposing covering all land — this is just a ceiling.)

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2) Area and installed capacity required to meet current global demand (no extra limits yet)

We want continuous power = 18 TW = $18\times 10^{12}$ W.

Installed peak capacity required (given CF = 0.24):

$$\text{Installed}_\text{peak} = \dfrac{18\times 10^{12}\ \text{W}}{0.24} = \left(\dfrac{18}{0.24}\right)\times 10^{12}\ \text{W} = 75\times 10^{12}\ \text{W} = 75\ \text{TW}.$$

Area needed at 200 W/m² peak:

$$\text{Area} = \dfrac{75\times 10^{12}\ \text{W}}{200\ \text{W/m}^2} = \dfrac{75}{200}\times 10^{12}\ \text{m}^2 = 0.375\times 10^{12}\ \text{m}^2 = 3.75\times 10^{11}\ \text{m}^2.$$

Convert to km²: $3.75\times 10^{11}\ \text{m}^2 \div 10^6 = 375,000\ \text{km}^2.$

Fraction of land: $\dfrac{3.75\times 10^{11}}{1.49\times 10^{14}} \approx 0.00252 = 0.252\%.$

Interpretation: to supply ~18 TW on average with only 20% PV and no other sources, you’d need ~75 TW installed (peak), which requires ~375,000 km² of panels — about 0.25% of Earth’s land area (≈ area of Germany × ~1.1). That’s large but physically small compared with world land area.

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3) Now add the real practical limits

I’ll fold in four constraints you listed: land use (only part of land is usable), transmission losses, intermittency & storage (overbuild factor), and material/economic limits. I'll present three scenarios (conservative, realistic, optimistic) and show arithmetic.

Assumptions for scenarios:

• Transmission losses: 5% (round number; long HVDC lines can be a few %).

• Storage/firming overbuild factor (how much extra installed capacity you need because of night/season and storage losses):

  • Conservative (100% firm with mainly batteries/seasonal storage): ×4 installed.

  • Realistic (mix of grid balancing, wind, hydro, demand response): ×2 installed.

  • Optimistic (high CF panels, tracking, distributed mix; limited storage+other firming): ×1.25 installed.

• Usable land fraction for large PV projects (deserts, rooftops, brownfields): we'll consider 0.25%, 1%, and 2% of land as available in different scenarios.

  • Note: 0.25% of land ≈ $0.0025\times 1.49\times10^{14} = 3.725\times10^{11}$ m² ≈ 372,500 km².

Scenario A — Conservative (firm, high overbuild)

• Overbuild factor = 4. Transmission multiplier = 1/(1 − 0.05) = $1/0.95 \approx 1.0526$.

• Effective installed requirement = $75\ \text{TW} \times 4 \times 1.0526$.

  • $75\times 4 = 300\ \text{TW}.$

  • $300 \times 1.0526 \approx 315.8\ \text{TW}.$

• Area needed = $315.8\times 10^{12}\ \text{W} / 200\ \text{W/m}^2 = 1.579\times 10^{12}\ \text{m}^2$.

  • In km²: $1.579\times 10^{12}/10^6 = 1,579,000\ \text{km}^2.$

• Fraction of land: $1.579\times 10^{12} / 1.49\times10^{14} \approx 0.0106 = 1.06\%.$

Scenario B — Realistic (mix of resources)

• Overbuild factor = 2. Transmission multiplier = 1/0.95 ≈ 1.0526.

• Effective installed = $75\ \text{TW} \times 2 \times 1.0526$.

  • $75\times 2 = 150\ \text{TW}.$

  • $150 \times 1.0526 \approx 157.9\ \text{TW}.$

• Area = $157.9\times10^{12} / 200 = 7.895\times10^{11}\ \text{m}^2 = 789,500\ \text{km}^2.$

• Fraction of land ≈ $0.53\%$.

Scenario C — Optimistic (highly efficient deployment + grid mixes)

• Overbuild factor = 1.25. Transmission multiplier = 1/0.95 ≈ 1.0526.

• Effective installed = $75\ \text{TW} \times 1.25 \times 1.0526$.

  • $75\times 1.25 = 93.75\ \text{TW}.$

  • $93.75 \times 1.0526 \approx 98.7\ \text{TW}.$

• Area = $98.7\times10^{12} / 200 = 4.935\times10^{11}\ \text{m}^2 = 493,500\ \text{km}^2.$

• Fraction of land ≈ $0.33\%$.

Summary (areas in km²):

• Conservative (×4): ~1,579,000 km² (~1.06% land).

• Realistic (×2): ~789,500 km² (~0.53% land).

• Optimistic (×1.25): ~493,500 km² (~0.33% land).

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4) Transmission & storage losses — what actually gets to users

If storage round-trip efficiency is ~85% (batteries) and you need storage for a large fraction of energy, net usable energy becomes further reduced. Example: with overbuild factor 2 and batteries at 85% roundtrip, you’d need somewhat more generation (we crudely baked typical storage losses into the overbuild factor; if you want a more explicit treatment I can model hours of storage, RTE, and diurnal vs seasonal deficits).

Transmission losses (we used 5%) are modest compared to the main drivers (overbuild + CF + land).

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5) Material & economic constraints (qualitative + rough notes)

• Manufacturing scale and ramp rate: installing hundreds of TW of PV is a multi-decade, multi-trillion dollar industrial effort. Capital and factory capacity limit how fast you can build; historically global PV installations grew quickly but sustained ramp to many tens of TW of nameplate will take decades unless massively accelerated.

• Materials: most panels use silicon (abundant), glass, aluminium, copper (for wiring), and silver (for contacts). Silver can be a bottleneck but is commonly reduced via technology (copper plating, less silver per cell) and recycling. Rare earths are not central in silicon PV (more relevant for some inverters, rare magnets). Mining, refining, and recycling flows matter for scale and environmental impact.

• Land use tradeoffs & social acceptance: even if physically feasible area-wise, siting large utility arrays has political/social costs. Rooftops, built environments, and brownfields reduce land conflicts but increase complexity and cost.

• Cost: levelized cost of electricity (LCOE) from PV + storage is falling but depends on finance, supply chains, policy. Economic constraints will shape where/when arrays get built.

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6) Bottom line — is solar “enough” in practice?

• Energetically: yes. Under a wide range of plausible assumptions (even conservative ones), the land area and raw solar resource are more than sufficient to supply current global electricity needs many times over.

• Practically: the real limits are deployment speed, materials/recycling, grid integration, storage & firming strategy, land use choices, and political/economic willingness to invest. Those determine whether solar actually supplies X% of real delivered electricity within a given time horizon.

• Concrete scale: to replace ~18 TW purely with PV you'd need on the order of 0.3–1.6 million km² of panels depending on how aggressively you overbuild for firm power and how much grid/other resources you use. That’s a lot of panels but small as a fraction of land (0.33%–1.06% in our scenarios).

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If you want I can:

• Recompute using a different panel efficiency (e.g., 25% or 15%).

• Model explicit storage (e.g., 12 hours vs seasonal) with round-trip efficiency numbers to show exactly how much overbuild is needed.

• Compute for a country/region (give me rooftop area, desert area, or latitude) and show yearly kWh estimates.