Heatwaves have been in the news in recent years. I found it amazing to consider that the energy from the Sun is enough to kill you, when only a tiny fraction of its radiation reaches Earth.

Then I wondered, what fraction of the Sun's energy reaches Earth, exactly? I think the way to approach this is to answer: What fraction of the Sun's sky does the Earth take up?

Can anyone tell me what the answer is? How do you calculate it? Is this the right way to approach the question?

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    $\begingroup$ "Is this the right way to approach the question?" One alternative way to get at the amount of energy reaching the Earth would be to multiply the solar constant of 1361 watts/m${}^2$ times the 2D area of Earth's disk $\pi / r_E^2$. The solar constant is the Sun's output at 1 AU and of course Earth's orbit is slightly elliptical so it varies a bit, but since the Sun's output is already contained in the solar constant, you don't need to go there again. $\endgroup$
    – uhoh
    Commented Jan 1, 2023 at 23:28

3 Answers 3


The calculation of the apparent size of Earth as seen from the Sun is pretty straight forward. Just consider the triangle: from your observation point the distance to the center of the object, to the surface. That is a triangle with a 90° at the center of the object you view. You can apply the tangent law (actually a sine, as the Sun is a sphere and not a disk) for the angle between the line to the center and the rim of the object: $$ \tan \alpha^\prime = \frac{r}{d}$$ For small angles $\alpha$ the tangens is nearly identical to the angle itself, so you can approximate the viewing angle $\alpha = 2\alpha^\prime$ of an object with the diameter $2r$ as $$ \alpha \approx \frac{2r}{d}$$.

This angle of course is given in radians. Thus if you want to convert it to degrees you have to apply the typical conversion of $360° = 2\pi$. For the viewing angle of Earth ($r = 6371km$) and the distance to the Sun ($d=149,600,000\mathrm{km}$) you get a viewing angle of $$\alpha = 4.259\cdot 10^{-5} = 0.00244° = 8.78"$$

Compare a disk of radius $\alpha^\prime$ to the overall surface of a sphere ($4\pi$), that is a tiny fraction. But knowing the solar constant on Earth ($S = 1370 W/m^2$) with a cross-section of $A=\pi r^2$, this allows to calculate the overall energy $L$ (called luminosity in astrophysics) radiated by the Sun: $$ L = \frac{4\pi}{\pi(\alpha^\prime)^2} \cdot \left( S\cdot A\right)\\ L = 3.85\cdot 10^{26} \mathrm{W}. $$ That number obtained only using ballpark numbers corresponds surprisingly well with the entry found on Wikipedia of $3.82\cdot 10^{26} \mathrm{W}$.

For perspective: The primary energy production and consumption by the whole world was in the year 2021 around 14800 million tons oil equivalent which corresponds to 11.63TWh each. Break that down it means that Earth alone receives at this tiny viewing angle from the Sun already about 9000x as much energy from the Sun than the whole humankind consumes. And the Sun as a whole produces 2 billion times more energy, every second. Thus if we could tap and store the complete solar energy for one second, we'd have enough to last the world with its current energy consumption as-is about 1500x the current age of the universe.

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    $\begingroup$ Good answer. However, the units of luminosity are W, not W/m2. Also, can you add a link to wikipedia: en.wikipedia.org/wiki/Solar_luminosity. And, for comparison, I would be curious what is the apparent size or luminosity of Mars or Jupiter from Earth. $\endgroup$ Commented Jan 2, 2023 at 19:20
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    $\begingroup$ the fact that humans are now using 1/9000 of the Earth's total energy input should be pretty concerning. Beyond just the fuzzy biological carrying capacity we've already exceeded, if we carry on for another century we'll be hitting hard physical limits to growth that nobody can deny... $\endgroup$ Commented Jan 3, 2023 at 2:09
  • $\begingroup$ You almost state this, but consider explicitly answering the question "What fraction of the Sun's sky does the Earth take up?" -- it's $4.5 \times 10^{-10}$. $\endgroup$
    – nanoman
    Commented Jan 3, 2023 at 2:56
  • $\begingroup$ For comparison, the angular diameter of Mars as viewed from Earth ranges from $3''$—$25''$, and that of Saturn ranges from $15''$–$20''$. $\endgroup$ Commented Jan 3, 2023 at 14:06

According to space.com 109 earths could be lined up across the face of the sun. Therefore, if you were at the sun, the earth would look 109 times smaller than the sun does from the earth. This answer doesn't help calculating the radiation fraction but hopefully addresses the headline question in a meaningful way.
Coincidentally the moon has a similar apparent size (it varies) as the sun so 109 times smaller than the moon is a more convenient comparison. Looking directly at the sun isn't a good idea.

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    $\begingroup$ You could replace the Sun by the Moon. It is visually the same size as the Sun but the size is easier to grasp. $\endgroup$
    – Florian F
    Commented Jan 3, 2023 at 15:21
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    $\begingroup$ Thankyou - I have modified my answer. $\endgroup$
    – Ken Mercer
    Commented Jan 4, 2023 at 22:54
  • $\begingroup$ It's pretty small, we might not notice earth if standing on the sun. $\endgroup$ Commented Jan 6, 2023 at 4:49

The Apparent size of some celestial object viewed from another celestial object is also known as angular diameter . The formula for angular diameter is: $δ=206265∗(actualDiameter/distance)$

Also generally it is in terms of arcseconds. 206265 is the conversion factor for arcsec however if you want the answer to be in degrees then the conversion factor would be 57.3. The actual diameter would be 12742.018 kilometers (equatorial) and the distance would be 1 AU/149597871 kilometers therefore the Earth's angular diameter would be 17.866808 arcsec / 0.00496336 degrees.

However since the Earth travels in an elliptical orbit due to Kepler's law of planetary motion the distance and therefore apparent size may differ depending on the season. The 1 AU distance is the Semi-Major axis of Earth's orbit around the Sun

One can use the Solar constant for the power of the light and multiply it with the cross sectional area/2d area of the earth.

Also the Earth both absorbs and reflects the light so even in that tiny fraction of the light emitted by the Sun so even in that tiny fraction only a few encounter the Earths surface the rest is absorbed by Green(hous) gases releasing Thermal energy/Heat.

So the cross-sectional area of earth is $1.275165×10^8$ square kilometers which when multiplied by the Solar constant gives $174$ petawatts!

To compare according to a 2010 estimate the world uses $18.47$ trillion kilowatt hours per year! but when you divide it by $174$ petawatts it gives $82525$ (years)!

  • $\begingroup$ Thanks, @hobbs. Corrected (seemingly the calculators server i'm using encountered a cosmic ray) $\endgroup$
    – user47732
    Commented Jan 2, 2023 at 9:01

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