# The supergiant star Betelgeuse has a measured angular diameter of 0.044 arcsecond. Its distance has been measured to be 427 light-years. What is the actual diameter of Betelgeuse?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2
Zack M. Share
Dec 13, 2015

$861 , 000 , 000 \text{ km}$

#### Explanation:

This is a pretty straight forward trigonometry problem. We can set up a diagram showing that the distance to Betelgeuse and the radius of Betelgeuse make a right angle.

Therefore, we can use the $\sin$ function to find the radius of Betelgeuse. Since $\theta$ is very small, we can use the small angle approximation, $\sin \left(\theta\right) \approx \theta$ if we convert $\theta$ to radians.

$.044 \text{ arc seconds" = 2.13 xx 10^-7 " radians}$

Since $\theta$ is the total diameter of Betelgeuse, we want to use $\sin \left(\theta \text{/} 2\right)$ to calculate the radius.

$r = d \sin \left(\frac{\theta}{2}\right) \approx d \frac{\theta}{2}$

But the radius is $1 \text{/} 2$ of the diameter, $D$, so we have;

$\frac{D}{2} = d \frac{\theta}{2}$

Canceling the $2$s leaves;

$D = d \theta$

Now we have an expression for the diameter, we can plug in what we know.

$D = \left(427 \text{ light years}\right) \left(2.13 \times {10}^{-} 7\right)$

$D = 9.10 \times {10}^{-} 5 \text{ light years}$

Light years are not the most practical units for measuring the diameter of a star, however, so lets convert to $\text{km}$ instead.

$D = \left(9.10 \times {10}^{-} 5 \text{ light years")(9.46 xx 10^12 " km / light year}\right)$

$D = 8.61 \times {10}^{8} \text{ km}$

This is about 600 times the diameter of the sun!

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2
Trippy Share
Dec 13, 2015

It has a diameter of 861744026,4 km, so $\approx$862 million km.

#### Explanation:

In astronomy, the angular diameter is given by the following formula:

$\delta = \frac{206265 d}{D}$.

$\delta$ - Angular Diameter
$d$ - diameter of the object
$D$ - distance between the observer and the object

206265 is simply the number of arcsecs in a radian.

So before we chuck in the values we have to rearrange the equation in a way that we isolate $d$, which is what we want, so let's do just so:

$\delta = \frac{206265 d}{D}$$\Leftrightarrow$$\delta \cdot D = 206265 d$ (I moved $D$ to the other side here).

$\delta \cdot D = 206265 d$$\Leftrightarrow$$\frac{\delta \cdot D}{206265} = d$, by moving 206265 to the left hand side, we get $d$ isolated. So now we chuck in the values:

$\frac{0.044 \cdot 428}{206265} = d$$\Leftrightarrow$$d = 9 , 1 \cdot {10}^{-} 5$ light years, because we were using light years in our $D$, we get the result in light years.

Now to turn our $d$ into km, we need to know how many km's there is in a light year. A light year is equal to $9.4607 \cdot {10}^{12}$ km.

$9.4607 \cdot {10}^{12} \cdot 9.1 \cdot {10}^{-} 15 = 861744026 , 37367578$ which is $\approx$$861744026 , 4$ km

• 10 minutes ago
• 11 minutes ago
• 11 minutes ago
• 12 minutes ago
• 3 minutes ago
• 5 minutes ago
• 6 minutes ago
• 8 minutes ago
• 10 minutes ago
• 10 minutes ago
• 10 minutes ago
• 11 minutes ago
• 11 minutes ago
• 12 minutes ago