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What would be the size of Sun as seen from surface of Pluto? It obviously depends on the distance and one would expect it to be very small. However, due to different optical phenomenon, the size of Sun seen on surface of Pluto could be larger. Can it can be estimated from the annular eclipse Sun-Pluto eclipse picture or other pictures taken by New Horizons probe ? Thanks for your insight.

The picture is also posted on this forum question.

Edit: From the answer given by @Conrad Turner below, Sun should be just pin point at that distance. However, when seen by New Horizons probe from behind Pluto, its size is seen to be almost as large as Pluto. The atmosphere of Pluto must be bending the light so much to make this happen.

Update: This picture was taken from a distance of 2 million km from Pluto. At this distance Sun will have an angular size of 0'49" (distance 5804550000 km, diameter 1392000 km), while Pluto has the angular size of 4'5" (distance 2000000 km, diameter 2372 km). However, the picture is showing Sun to have an angular size slightly larger than Pluto (or nearly the same size). Hence, high refraction or scattering or some other phenomenon are at work here.

Considering sizes and distances, if atmosphere of Pluto does nothing to light, the ring of Sun should have been visible around Pluto once probe was 10 million km beyond Pluto. However, the ring of Sun is seen in this picture when probe is only 2 million km beyond Pluto. The scientists must have known the optical characteristics of Pluto's atmosphere and hence they got the probe early in proper position to take this picture.

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  • $\begingroup$ Obviously you can position yourself at a distance from Pluto where the Sun and Pluto have the same angular size. The moon is much smaller than the Sun but we still get total eclipses... $\endgroup$ – Rob Jeffries Oct 16 '15 at 11:29
  • $\begingroup$ You are right about equal sizes. But it can happen only when size of Pluto becomes small as the distance increases. The angular size of Sun cannot increase more than a pin point at that distance! Here we are seeing Sun of a size much more than 1' estimated using distance alone. Some other factor must be at work here. $\endgroup$ – rnso Oct 16 '15 at 11:31
  • $\begingroup$ The Sun is not present in the picture - it is completely hidden. Even if this was just refraction - it is small. Scattering is almost certainly what is going on here. Refraction causes an object to appear smaller from the surface of Pluto. $\endgroup$ – Rob Jeffries Oct 16 '15 at 13:51
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    $\begingroup$ This question has led to discussion about very interesting optical phenomenon related to atmosphere of Pluto. I cannot understand the downvotes. $\endgroup$ – rnso Oct 17 '15 at 2:04
  • $\begingroup$ If you find yourself on Pluto, it's still not a good idea to stare at the sun, even at that distance it could still damage your retina. $\endgroup$ – userLTK Oct 22 '15 at 21:33
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New Horizons encountered Pluto when it was about 33.5 au from the Sun.

The solar diameter is $1.39\times 10^{9}$ m.

An astronomical unit is $1.496\times 10^{11}$ m.

The angle subtended by the Sun at Pluto (in radians) is the ratio of the diameter of the Sun to its distance from Pluto. Converting to degrees, minutes, seconds, gives 0.01589 degrees, or 0.95 arcminutes. The atmosphere of Pluto is way too thin (much thinner than the Earth's) to make any significant difference to this at all.

The diameter of Pluto is $2.37\times 10^6$ m. In order for Pluto and the Sun to subtend a similar angle in an image, you just need to move far enough away from Pluto to make the same angle.

Thus $$ D_P/R_P \simeq (D_P + D_{PS})/R_S,$$ where $D_P$ would be the distance from New Horizons to Pluto, $D_{PS}$ the distance from Pluto to the Sun, $R_P$ the radius of Pluto and $R_S$ the radius of the Sun. Because I will show that $D_P \ll D_{PS}$ we can say that $$ D_P \simeq \frac{R_P}{R_S} D_{PS} \simeq 0.057\ au$$.

New Horizons is travelling at about 14.5 km/s. So a simple bit of Maths tells us it would takes about 6.5 days to get into this position.

The picture I think you are talking about was taken on July 15th 2015 only 7 hours after the flyby. The Sun and Pluto do not share the same angular size in this photo. You are not seeing the Sun at all in this picture. It is eclipsed by Pluto completely. All that is being seen is sunlight refracted (or more likely scattered) through Pluto's atmosphere. The Sun acts more-or-less like a point source of light at more-or-less infinite distance.

The angle that the light needs to deviate in order to be seen through the atmosphere is about $0.19$ degrees (the half-angle subtended by Pluto at the Sun [negligible] plus the angle subtended by Pluto at New Horizons after 7 hours). If we were to model the atmosphere as a glass prism. The deviation angle is $\delta \simeq (n-1) \alpha$, where $n$ is the refractive index and $\alpha$ is the opening angle of the prism. If you let $\alpha$ be quite large (let's say 1 radian) to simulate rays coming through the atmosphere, then $n = 1 + \delta$, where $\delta$ is in radians. So $n\simeq 1.003$. NB: This assumes there is no scattering in Pluto's atmosphere, but I suspect that is not true and that my estimate of the refractive index is a large overestimate. In fact I'm sure it is an overestimate because this is larger than the refractive index of air on Earth. i.e. I am sure that what we are seeing is scattered light, not refracted light.

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  • $\begingroup$ Your answer seems to agree with my hypothesis. $\endgroup$ – rnso Oct 16 '15 at 12:26
  • $\begingroup$ Why are you assuming refractive index there cannot be larger that on Earth? Also, can you estimate the angular size of Sun from surface of Pluto with the refractive index that you obtained? $\endgroup$ – rnso Oct 16 '15 at 12:34
  • $\begingroup$ The refractive index I have calculated is an upper limit. The refractive index cannot be larger than Earth because the refractive index of a gas depends mostly on its pressure. The Sun's apparent diameter would be smaller by a factor of the refractive index. $\endgroup$ – Rob Jeffries Oct 16 '15 at 13:45
  • $\begingroup$ You're off by a factor of two, @RobJeffries. It looks like you use the solar diameter instead of the solar radius in your calculation of $D_P$. Your argument still stands. $\endgroup$ – David Hammen Oct 17 '15 at 11:41
  • $\begingroup$ @RobJeffries - WA and I get 0.057 AU . $\endgroup$ – David Hammen Oct 17 '15 at 20:37
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The atmoshere on Pluto will have very little effect on the apparent size of the Sun, It is negligable compared to that on Earth ($\sim 0.3-1.0 \mbox{ Pa}$ compared to $\sim 10^5 \mbox{ Pa}$, approximately the air pressure at 100 km on the Earth).

Now angular size is proportional to the ratio of the objects diameter to its range. At the Earth's distance ($1 \mbox{ AU}$) the Sun subtends about $30'$ of arc. So at Pluto (as it is at a distance from the Sun of between $26.7 \mbox{AU}$ and $49.3 \mbox{AU}$ depending on where in its orbit it is) the Sun will subtend between $30/26.7\approx 1.1'$ and $30/49.3\approx 0.6'$ of arc. Which is close to the resolution of the human eye corresponding to 20/20 vision.

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  • $\begingroup$ Even though the atmosphere may be thin, its constitution can affect the refractive index. Hence, theoretically it is still possible that the size seen is larger. $\endgroup$ – rnso Oct 16 '15 at 6:12
  • $\begingroup$ Other mechanisms/phenomenon like scatter can also make apparent size larger. $\endgroup$ – rnso Oct 16 '15 at 6:22

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