# The Sun as seen from 120 AU.

#### beginner49

Hi all

Currently, Voyager 1 is about 120 AU from the Sun. I wonder how big (or small) and bright would the Sun be seen from aboard this spacecraft. What approximate magnitude?.

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Hi all

Currently, Voyager 1 is about 120 AU from the Sun. I wonder how big (or small) and bright would the Sun be seen from aboard this spacecraft. What approximate magnitude?.

Here, thanks to Caltech, is an artist's conception of the Sun from the vicinity of Sedna at 8 billion miles' (about 86 AU) distance:

#### Chronos

Gold Member
It's still very bright, even at 120 au, at about magnitude -16.3 [the full moon from earth is about -12.7. So you would easily be able to read a newspaper. It would, however, be a virtual point source at that distance.

#### beginner49

Thanks a lot for your replies. That of being able to read a newspaper is a very interesting detail.

thanks again.

#### Feodalherren

On a related topic, I heard that voyager was still accelerating, why? What causes it to accelerate further?

#### Dickfore

The angular size θ of the Sun's disk is given by the formula:
$$\sin \left( \frac{\theta}{2} \right) = \frac{R_S}{d}$$
where RS is the radius of the Sun, and d is the distance from it.

Because the distance is much larger than the Sun's radius, the sine is very small. Therefore, to a sufficient precision we may substitute:
$$\sin \left( \frac{\theta}{2} \right) \approx \frac{\theta}{2}$$
provided that we measure the angle in radians. Nevertheless, we see that:
$$\theta \approx \frac{2 R_S}{d} \propto \frac{1}{d}$$
the angular size is approximately inversely proportional to the distance. At 1 A.U. (the Earth), the angular size of the Sun is about 31' (arc minutes). Therefore, at 120 A.U. it is:
$$\theta = \frac{31 '}{120} \times \frac{60 ''}{1 '} = 15.5 ''$$
that is about 15 arc seconds.

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