<p>Due to the <a href="https://en.wikipedia.org/wiki/Intercept_theorem" rel="noreferrer">basic proportionality theorem</a>, the width of the boundary of the shadow is</p>

<p><span class="math-container">$$
w=\frac{\ell D}L\;,
$$</span></p>

<p>where <span class="math-container">$\ell$</span> is the distance from the moon to the planet&rsquo;s surface, <span class="math-container">$L$</span> is the distance from the planet to the sun, and <span class="math-container">$D$</span> is the diameter of the sun. Here</p>

<p><span class="math-container">\begin{eqnarray}
D&amp;\approx&amp;1.4\cdot10^9\mathrm m\;,\\
L_♁&amp;\approx&amp;1.5\cdot10^{11}\mathrm m\;,\\
\ell_♁&amp;\approx&amp;3.8\cdot10^8\mathrm m\;,\\
L_♃&amp;\approx&amp;7.8\cdot10^{11}\mathrm m\;,\\
\ell_♃&amp;\approx&amp;3.5\cdot10^8\mathrm m\;.
\end{eqnarray}</span></p>

<p>(Note that <span class="math-container">$\ell$</span> is the radius of the moon&rsquo;s orbit minus the radius of the planet; subtracting the radius of the planet makes an appreciable difference in the case of Jupiter and Io but not in the case of Earth and the Moon.) Thus the width of the boundary of Io&rsquo;s shadow on Jupiter is</p>

<p><span class="math-container">$$
w_♃=\frac{\ell_♃ D}{L_♃}\approx\frac{3.5\cdot10^8\mathrm m\cdot1.4\cdot10^9\mathrm m}{7.8\cdot10^{11}\mathrm m}\approx6.3\cdot10^5\mathrm m\;,
$$</span></p>

<p>whereas the width of the boundary of the Moon&rsquo;s shadow on Earth is</p>

<p><span class="math-container">$$
w_♁=\frac{\ell_♁ D}{L_♁}\approx\frac{3.8\cdot10^8\mathrm m\cdot1.4\cdot10^9\mathrm m}{1.5\cdot10^{11}\mathrm m}\approx3.5\cdot10^6\mathrm m\;.
$$</span></p>

<p>Since <span class="math-container">$D$</span> is the same in both cases and the distances from the moons to the planets also happen to be roughly equal, the difference in the width of the boundary is mostly due to the difference in the distances from the sun, as stated in @christopherlovell&rsquo;s answer.</p>

<p>We can also calculate the relative widths, both relative to the radius of  the moon and relative to the radius of the planet, both of which contribute to whether the shadow appears as hard or soft on photos such as the ones you included in the question. The radius of the moon is, to a very good approximation, the radius of the shadow up to the centre of its boundary, that is, the radius of the shadow as it would be if the sun were a point source. This is because this radius is the radius of the moon, magnified by the factor <span class="math-container">$\frac{L+\ell}L=1+\frac\ell L$</span> (again due to the basic proportionality theorem), which is approximately <span class="math-container">$1$</span> since <span class="math-container">$\ell\ll L$</span>.</p>

<p>The width of the boundary of the shadow relative to the radius <span class="math-container">$r$</span> of the moon is</p>

<p><span class="math-container">$$
\frac{w_♃}{r_♃}\approx\frac{6.3\cdot10^5\mathrm m}{1.8\cdot10^6\mathrm m}\approx35\%
$$</span></p>

<p>for Io&rsquo;s shadow on Jupiter and</p>

<p><span class="math-container">$$
\frac{w_♁}{r_♁}\approx\frac{3.5\cdot10^6\mathrm m}{1.7\cdot10^6\mathrm m}\approx200\%
$$</span></p>

<p>for the Moon&rsquo;s shadow on Earth. (In fact the Moon&rsquo;s orbit is sufficiently elliptical for the relative width of the boundary to fluctuate around <span class="math-container">$200\%$</span>, which is why there are both <a href="https://en.wikipedia.org/wiki/Solar_eclipse#Types" rel="noreferrer">annular and total solar eclipses</a>.) Again, Io and the Moon happen to have very similar radii, so the difference is mostly due to the difference in the distances from the sun.</p>

<p>The width of the boundary of the shadow relative to the radius <span class="math-container">$R$</span> of the planet is</p>

<p><span class="math-container">$$
\frac{w_♃}{R_♃}\approx\frac{6.3\cdot10^5\mathrm m}{7.0\cdot10^7\mathrm m}\approx1\%
$$</span></p>

<p>for Io&rsquo;s shadow on Jupiter and</p>

<p><span class="math-container">$$
\frac{w_♁}{R_♁}\approx\frac{3.5\cdot10^6\mathrm m}{6.4\cdot10^6\mathrm m}\approx55\%
$$</span></p>

<p>for the Moon&rsquo;s shadow on Earth. Here the large difference between the radii of Jupiter and the Earth considerably enhances the difference between the boundary widths. You compared two photos in which roughly similar parts of the planets are visible, whereas the shadows/moons, which are actually of similar size, appear at a very different scale; this enhances the impression of different hardness of the shadows, since the already smaller width of the boundary of Io&rsquo;s shadow is further scaled down.</p>

CC BY-SA 4.0
2020-05-22T21:37:14.130Z
2020-05-23T07:06:13.677Z
2020-05-23T07:06:13.677Z
joriki
<p>Another factor is involved in addition to the above answers.</p>

<p>Io's umbral shadow is a bit over two million kilometers long, almost six times longer than the ~350 thousand kilometer distance between Jupiter's surface and Io. This means that most of Io's shadow on Jupiter's surface as seen from space is Io's umbral rather than its penumbral shadow. Io's umbral shadow is pitch black.</p>

<p>Compare that to the Earth-Moon system. The length of the Moon's umbral shadow is about the same as the distance between the center of the Moon and the Earth's surface. This means that almost all of the shadow created by the Moon on the surface os the Earth results from the Moon's penumbral shadow rather than from its umbral shadow. Even in the case of a total solar eclipse, it's rather difficult to see from space that there's a tiny bit of pitch black in the center of the Moon's shadow on the Earth. Most of what is seen from space is a gradual transition from fully lit to rather dark in the center of the Moon's shadow.</p>

<p>Moreover, that the distance between the Sun and Io is over five times larger than is the distance between the Sun and the Moon means that Io's penumbral shadow is rather sharp compared to the Moon's penumbral shadow.</p>

CC BY-SA 4.0
2020-05-23T13:58:29.540Z
+1 but it should be noted that this factor is not in addition to the other ones, it's just another wording of the same factor - that is, relation distances. However, +1 because it may be helpful at understanding.
Pere
CC BY-SA 4.0
2020-05-23T12:40:50.417Z
2020-05-23T14:28:12.447Z
2020-05-23T14:28:12.447Z
David Hammen
<p>The farther you go, the smaller the light source gets.</p>

<p>Imagine a tubelight and an LED diode, as in the picture below. The diffusion factor increases as you get closer. Animation artists will know what I'm talking about. </p>

<p>But to a common man: the bigger the light source, the softer will be the light's shadow. And we know Earth is nearer to the Sun than Jupiter. Hence the rays of light will be crisper around the edges.</p>

<p><a href="https://i.sstatic.net/MGJ8j.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MGJ8j.png" alt="Light, teapot and shadow"></a></p>

CC BY-SA 4.0
2020-05-23T22:23:20.803Z
2020-05-26T09:10:27.037Z
2020-05-26T09:10:27.037Z
Hello Gandu
<p>This can be seen geometrically:</p>

<p><a href="https://i.sstatic.net/xIA3R.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xIA3R.png" alt="enter image description here"></a></p>

<p>Everywhere the lines touch a circle, they are tangent.</p>

<p>In some parts of the shadow, the sun is not visible at all. This is the <a href="https://en.wikipedia.org/wiki/Umbra,_penumbra_and_antumbra" rel="nofollow noreferrer">umbra</a>. It is visible in this construction as the inner cone of the shadow.</p>

<p>Outside this, the sun is only partially obscured. This is called the penumbra.</p>

<p>If the penumbra is small relative to the umbra, the shadow looks hard. If the penumbra is relatively large, the shadow is soft.</p>

<p>Both Io and Earth's moon are about the same size. Both orbit at a similar altitude over their respective bodies, as well. But Io is approximately 5 times more distant from the sun. Accordingly that makes the Sun "smaller" for Io, which according to the geometry above it should be intuitively obvious that the penumbra becomes smaller, and thus, the shadow becomes harder.</p>

<p>If it's not immediately obvious, consider the limiting case where the Sun is just a single point: in that case there is no penumbra at all.</p>

CC BY-SA 4.0
2020-05-25T04:42:04.687Z
2020-05-25T04:42:04.687Z
Phil Frost
<p>Imagine what the shadow of our sister Moon would look like if brother Earth circled the Sun closer than Mercury. Assuming the happy tidally locked twin survives the intense radiation.</p>
<p>Mother Sun would look gigantic in size. Sister Moon would cover a small part of her mother, and no place on her twin brother would ever be in the dark, apart from the nightside.</p>

CC BY-SA 4.0
2022-06-10T15:57:00.757Z
2022-06-10T15:57:00.757Z
Felicia
<p>It's due to the larger relative apparent size of the Sun. When the source of light is a point source the shadow is <em>harder</em>, and when it is extended it is <em>softer</em>.</p>

<p>Jupiter is approximately 5 times more distant from the Sun than the Earth, so the Sun is approximately 5 times smaller in the sky.</p>

<p><a href="https://i.sstatic.net/WsilW.png" rel="noreferrer"><img src="https://i.sstatic.net/WsilW.png" alt="Hard and soft shadows"></a>
*Source: <a href="https://www.cs.unc.edu/~dm/UNC/COMP236/LECTURES/SoftShadows.pdf" rel="noreferrer">University of North Carolina CS</a></p>

CC BY-SA 4.0
2020-05-23T15:17:54.953Z
But the (relative) distance between body 2 and body 3 must also play a role (for the same distance to the sun). If they are sufficiently far apart it would all be penumbral.
Peter Mortensen
CC BY-SA 4.0
2020-05-23T19:07:39.377Z
Yes, as well as the scale of the photographs of the shadows.
Dronz
CC BY-SA 4.0
2020-05-24T02:28:29.853Z
@PeterMortensen Correct. Every dotted line on the image is defined by the position of the vertices of the orange triangle _and_ the vertices of the blue square, both their positions matter. But the distance between the moon (blue square) and planet (white square) absolutely pales in comparison to the distance between the sun (orange triangle) and the moon (blue square). More correctly, the proportional difference between Sun-Luna and Sun-Io is much larger than the difference between Luna-Earth and Io-Jupiter..
Flater
CC BY-SA 4.0
2020-05-22T12:58:58.180Z
2020-05-22T12:58:58.180Z
christopherlovell
<p>Io (<a href="https://apod.nasa.gov/apod/ap021207.html" rel="noreferrer">source</a>):</p>

<p><a href="https://i.sstatic.net/FLiI8.jpg" rel="noreferrer"><img src="https://i.sstatic.net/FLiI8.jpg" alt="enter image description here"></a></p>

<p>The Moon (<a href="https://www.nasa.gov/audience/forstudents/k-4/stories/moons-shadow" rel="noreferrer">source</a>):</p>

<p><a href="https://i.sstatic.net/oXdov.jpg" rel="noreferrer"><img src="https://i.sstatic.net/oXdov.jpg" alt="enter image description here"></a></p>

<the-moon><solar-eclipse><io>
Why does Io cast a hard shadow on Jupiter, but the Moon casts a soft shadow on Earth?
stackzebra
# Why does Io cast a hard shadow on Jupiter, but the Moon casts a soft shadow on Earth?

(Asked by: username_0 on 2020-05-22)

Io ([source](https://apod.nasa.gov/apod/ap021207.html)):

[![enter image description here](https://i.sstatic.net/FLiI8.jpg)](https://i.sstatic.net/FLiI8.jpg)

The Moon ([source](https://www.nasa.gov/audience/forstudents/k-4/stories/moons-shadow)):

[![enter image description here](https://i.sstatic.net/oXdov.jpg)](https://i.sstatic.net/oXdov.jpg)

**Tags:** the-moon, solar-eclipse, io


---

## 6 answers

### Answer by username_7 on 2020-05-22. Score: 88 | Accepted answer

It's due to the larger relative apparent size of the Sun. When the source of light is a point source the shadow is *harder*, and when it is extended it is *softer*.

Jupiter is approximately 5 times more distant from the Sun than the Earth, so the Sun is approximately 5 times smaller in the sky.

[![Hard and soft shadows](https://i.sstatic.net/WsilW.png)](https://i.sstatic.net/WsilW.png)
\*Source: [University of North Carolina CS](https://www.cs.unc.edu/~dm/UNC/COMP236/LECTURES/SoftShadows.pdf)

> **Comments:**
> **username_8**: But the (relative) distance between body 2 and body 3 must also play a role (for the same distance to the sun). If they are sufficiently far apart it would all be penumbral.  
> **username_9**: Yes, as well as the scale of the photographs of the shadows.  
> **username_10**: @username_8 Correct. Every dotted line on the image is defined by the position of the vertices of the orange triangle _and_ the vertices of the blue square, both their positions matter. But the distance between the moon (blue square) and planet (white square) absolutely pales in comparison to the distance between the sun (orange triangle) and the moon (blue square). More correctly, the proportional difference between Sun-Luna and Sun-Io is much larger than the difference between Luna-Earth and Io-Jupiter..  

---

### Answer by username_1 on 2020-05-22. Score: 32

Due to the [basic proportionality theorem](https://en.wikipedia.org/wiki/Intercept_theorem), the width of the boundary of the shadow is

$$
w=\frac{\ell D}L\;,
$$

where $\ell$ is the distance from the moon to the planet’s surface, $L$ is the distance from the planet to the sun, and $D$ is the diameter of the sun. Here

\begin{eqnarray}
D&\approx&1.4\cdot10^9\mathrm m\;,\\
L\_♁&\approx&1.5\cdot10^{11}\mathrm m\;,\\
\ell\_♁&\approx&3.8\cdot10^8\mathrm m\;,\\
L\_♃&\approx&7.8\cdot10^{11}\mathrm m\;,\\
\ell\_♃&\approx&3.5\cdot10^8\mathrm m\;.
\end{eqnarray}

(Note that $\ell$ is the radius of the moon’s orbit minus the radius of the planet; subtracting the radius of the planet makes an appreciable difference in the case of Jupiter and Io but not in the case of Earth and the Moon.) Thus the width of the boundary of Io’s shadow on Jupiter is

$$
w\_♃=\frac{\ell\_♃ D}{L\_♃}\approx\frac{3.5\cdot10^8\mathrm m\cdot1.4\cdot10^9\mathrm m}{7.8\cdot10^{11}\mathrm m}\approx6.3\cdot10^5\mathrm m\;,
$$

whereas the width of the boundary of the Moon’s shadow on Earth is

$$
w\_♁=\frac{\ell\_♁ D}{L\_♁}\approx\frac{3.8\cdot10^8\mathrm m\cdot1.4\cdot10^9\mathrm m}{1.5\cdot10^{11}\mathrm m}\approx3.5\cdot10^6\mathrm m\;.
$$

Since $D$ is the same in both cases and the distances from the moons to the planets also happen to be roughly equal, the difference in the width of the boundary is mostly due to the difference in the distances from the sun, as stated in @username_7’s answer.

We can also calculate the relative widths, both relative to the radius of the moon and relative to the radius of the planet, both of which contribute to whether the shadow appears as hard or soft on photos such as the ones you included in the question. The radius of the moon is, to a very good approximation, the radius of the shadow up to the centre of its boundary, that is, the radius of the shadow as it would be if the sun were a point source. This is because this radius is the radius of the moon, magnified by the factor $\frac{L+\ell}L=1+\frac\ell L$ (again due to the basic proportionality theorem), which is approximately $1$ since $\ell\ll L$.

The width of the boundary of the shadow relative to the radius $r$ of the moon is

$$
\frac{w\_♃}{r\_♃}\approx\frac{6.3\cdot10^5\mathrm m}{1.8\cdot10^6\mathrm m}\approx35\%
$$

for Io’s shadow on Jupiter and

$$
\frac{w\_♁}{r\_♁}\approx\frac{3.5\cdot10^6\mathrm m}{1.7\cdot10^6\mathrm m}\approx200\%
$$

for the Moon’s shadow on Earth. (In fact the Moon’s orbit is sufficiently elliptical for the relative width of the boundary to fluctuate around $200\%$, which is why there are both [annular and total solar eclipses](https://en.wikipedia.org/wiki/Solar_eclipse#Types).) Again, Io and the Moon happen to have very similar radii, so the difference is mostly due to the difference in the distances from the sun.

The width of the boundary of the shadow relative to the radius $R$ of the planet is

$$
\frac{w\_♃}{R\_♃}\approx\frac{6.3\cdot10^5\mathrm m}{7.0\cdot10^7\mathrm m}\approx1\%
$$

for Io’s shadow on Jupiter and

$$
\frac{w\_♁}{R\_♁}\approx\frac{3.5\cdot10^6\mathrm m}{6.4\cdot10^6\mathrm m}\approx55\%
$$

for the Moon’s shadow on Earth. Here the large difference between the radii of Jupiter and the Earth considerably enhances the difference between the boundary widths. You compared two photos in which roughly similar parts of the planets are visible, whereas the shadows/moons, which are actually of similar size, appear at a very different scale; this enhances the impression of different hardness of the shadows, since the already smaller width of the boundary of Io’s shadow is further scaled down.


---

### Answer by username_2 on 2020-05-23. Score: 13

Another factor is involved in addition to the above answers.

Io's umbral shadow is a bit over two million kilometers long, almost six times longer than the ~350 thousand kilometer distance between Jupiter's surface and Io. This means that most of Io's shadow on Jupiter's surface as seen from space is Io's umbral rather than its penumbral shadow. Io's umbral shadow is pitch black.

Compare that to the Earth-Moon system. The length of the Moon's umbral shadow is about the same as the distance between the center of the Moon and the Earth's surface. This means that almost all of the shadow created by the Moon on the surface os the Earth results from the Moon's penumbral shadow rather than from its umbral shadow. Even in the case of a total solar eclipse, it's rather difficult to see from space that there's a tiny bit of pitch black in the center of the Moon's shadow on the Earth. Most of what is seen from space is a gradual transition from fully lit to rather dark in the center of the Moon's shadow.

Moreover, that the distance between the Sun and Io is over five times larger than is the distance between the Sun and the Moon means that Io's penumbral shadow is rather sharp compared to the Moon's penumbral shadow.

> **Comments:**
> **username_3**: +1 but it should be noted that this factor is not in addition to the other ones, it's just another wording of the same factor - that is, relation distances. However, +1 because it may be helpful at understanding.  

---

### Answer by username_4 on 2020-05-23. Score: 8

The farther you go, the smaller the light source gets.

Imagine a tubelight and an LED diode, as in the picture below. The diffusion factor increases as you get closer. Animation artists will know what I'm talking about.

But to a common man: the bigger the light source, the softer will be the light's shadow. And we know Earth is nearer to the Sun than Jupiter. Hence the rays of light will be crisper around the edges.

[![Light, teapot and shadow](https://i.sstatic.net/MGJ8j.png)](https://i.sstatic.net/MGJ8j.png)


---

### Answer by username_5 on 2020-05-25. Score: 4

This can be seen geometrically:

[![enter image description here](https://i.sstatic.net/xIA3R.png)](https://i.sstatic.net/xIA3R.png)

Everywhere the lines touch a circle, they are tangent.

In some parts of the shadow, the sun is not visible at all. This is the [umbra](https://en.wikipedia.org/wiki/Umbra,_penumbra_and_antumbra). It is visible in this construction as the inner cone of the shadow.

Outside this, the sun is only partially obscured. This is called the penumbra.

If the penumbra is small relative to the umbra, the shadow looks hard. If the penumbra is relatively large, the shadow is soft.

Both Io and Earth's moon are about the same size. Both orbit at a similar altitude over their respective bodies, as well. But Io is approximately 5 times more distant from the sun. Accordingly that makes the Sun "smaller" for Io, which according to the geometry above it should be intuitively obvious that the penumbra becomes smaller, and thus, the shadow becomes harder.

If it's not immediately obvious, consider the limiting case where the Sun is just a single point: in that case there is no penumbra at all.


---

### Answer by username_6 on 2022-06-10. Score: 1

Imagine what the shadow of our sister Moon would look like if brother Earth circled the Sun closer than Mercury. Assuming the happy tidally locked twin survives the intense radiation.

Mother Sun would look gigantic in size. Sister Moon would cover a small part of her mother, and no place on her twin brother would ever be in the dark, apart from the nightside.


---

