<p>According to <a href="https://www.space.com/17001-how-big-is-the-sun-size-of-the-sun.html" rel="nofollow noreferrer">space.com</a> 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.<br />
Coincidentally the moon has a <a href="https://earthsky.org/space/coincidence-that-sun-and-moon-seem-same-size/" rel="nofollow noreferrer">similar apparent size</a> (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.</p>

CC BY-SA 4.0
2023-01-03T15:21:39.010Z
You could replace the Sun by the Moon.  It is visually the same size as the Sun but the size is easier to grasp.
Florian F
CC BY-SA 4.0
2023-01-04T22:54:17.107Z
Thankyou - I have modified my answer.
Ken Mercer
CC BY-SA 4.0
2023-01-06T04:49:54.527Z
It's pretty small, we might not notice earth if standing on the sun.
Zeeshan Ahmad Khalil
CC BY-SA 4.0
2023-01-02T22:04:27.280Z
2023-01-04T22:53:42.960Z
2023-01-04T22:53:42.960Z
Ken Mercer
<p>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 (<a href="https://astronomy.stackexchange.com/a/47731/16685">actually a sine</a>, 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: <span class="math-container">$$ \tan \alpha^\prime = \frac{r}{d}$$</span>
For small angles <span class="math-container">$\alpha$</span> the tangens is nearly identical to the angle itself, so you can approximate the viewing angle <span class="math-container">$\alpha = 2\alpha^\prime$</span> of an object with the diameter <span class="math-container">$2r$</span> as
<span class="math-container">$$ \alpha \approx \frac{2r}{d}$$</span>.</p>
<p>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 <span class="math-container">$360° = 2\pi$</span>. For the viewing angle of Earth (<span class="math-container">$r = 6371km$</span>) and the distance to the Sun (<span class="math-container">$d=149,600,000\mathrm{km}$</span>) you get a viewing angle of <span class="math-container">$$\alpha = 4.259\cdot 10^{-5} = 0.00244° = 8.78"$$</span></p>
<p>Compare a disk of radius <span class="math-container">$\alpha^\prime$</span> to the overall surface of a sphere (<span class="math-container">$4\pi$</span>), that is a tiny fraction. But knowing the solar constant on Earth (<span class="math-container">$S = 1370 W/m^2$</span>) with a cross-section of <span class="math-container">$A=\pi r^2$</span>, this allows to calculate the overall energy <span class="math-container">$L$</span> (called luminosity in astrophysics) radiated by the Sun: <span class="math-container">$$
L = \frac{4\pi}{\pi(\alpha^\prime)^2} \cdot \left( S\cdot A\right)\\
L = 3.85\cdot 10^{26} \mathrm{W}.
$$</span>
That number obtained only using ballpark numbers corresponds surprisingly well with the entry found on <a href="https://en.m.wikipedia.org/wiki/Solar_luminosity" rel="nofollow noreferrer">Wikipedia</a> of <span class="math-container">$3.82\cdot 10^{26} \mathrm{W}$</span>.</p>
<p>For perspective: The primary energy production and consumption by the <em>whole world</em> was <a href="https://en.wikipedia.org/wiki/World_energy_supply_and_consumption" rel="nofollow noreferrer">in the year 2021 around 14800 million tons oil equivalent</a> 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.</p>

CC BY-SA 4.0
2023-01-02T19:20:19.003Z
Good answer. However, the units of luminosity are W, not W/m2.

Also, can you add a link to wikipedia: https://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.
Julian Hyde
CC BY-SA 4.0
2023-01-03T02:56:08.547Z
You almost state this, but consider explicitly answering the question "What fraction of the Sun's sky does the Earth take up?" -- it's $4.5 \times 10^{-10}$.
nanoman
CC BY-SA 4.0
2023-01-03T14:06:39.283Z
For comparison, the angular diameter of Mars as viewed from Earth ranges from $3''$—$25''$, and that of Saturn ranges from $15''$–$20''$.
Michael Seifert
CC BY-SA 4.0
2023-01-01T18:05:49.653Z
2023-01-03T01:35:19.367Z
2023-01-03T01:35:19.367Z
planetmaker
<p>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.</p>
<p>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?</p>
<p>Can anyone tell me what the answer is? How do you calculate it? Is this the right way to approach the question?</p>

CC BY-SA 4.0
2023-01-01T23:28:58.140Z
"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](https://en.wikipedia.org/wiki/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.
uhoh
<the-sun><earth><distances><radiation><size>
What is the apparent size of Earth from the Sun?
albertoeda
# What is the apparent size of Earth from the Sun?

(Asked by: username_0 on 2023-01-01)

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?

**Tags:** the-sun, earth, distances, radiation, size
> **Comments:**
> **username_1**: "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](https://en.wikipedia.org/wiki/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.  

---

## 2 answers

### Answer by username_5 on 2023-01-01. Score: 13 | Accepted answer

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](https://astronomy.stackexchange.com/a/47731/16685), 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](https://en.m.wikipedia.org/wiki/Solar_luminosity) 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](https://en.wikipedia.org/wiki/World_energy_supply_and_consumption) 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.

> **Comments:**
> **username_6**: Good answer. However, the units of luminosity are W, not W/m2.  Also, can you add a link to wikipedia: https://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.  
> **username_7**: You almost state this, but consider explicitly answering the question "What fraction of the Sun's sky does the Earth take up?" -- it's $4.5 \times 10^{-10}$.  
> **username_8**: For comparison, the angular diameter of Mars as viewed from Earth ranges from $3''$—$25''$, and that of Saturn ranges from $15''$–$20''$.  

---

### Answer by username_2 on 2023-01-02. Score: 8

According to [space.com](https://www.space.com/17001-how-big-is-the-sun-size-of-the-sun.html) 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](https://earthsky.org/space/coincidence-that-sun-and-moon-seem-same-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.

> **Comments:**
> **username_3**: You could replace the Sun by the Moon.  It is visually the same size as the Sun but the size is easier to grasp.  
> **username_2**: Thankyou - I have modified my answer.  
> **username_4**: It's pretty small, we might not notice earth if standing on the sun.  

---

