<p>From the observer's point of view 1 million km away, the TNO's apparent angular motion is</p>

<p><span class="math-container">$$\mathrm{\frac{8~m/s}{10^9~m} = 8 \times 10^{-9}~rad/s = 0.00165~^\circ/h}.$$</span></p>

<p>Assuming that the observer at 18 billion km = 120 au is in a circular orbit around the Sun, the orbital period is
120<sup>3/2</sup> = 1320 years, making the Sun appear to move the other way at</p>

<p><span class="math-container">$$\mathrm{\frac{360~^\circ}{1320~y} = 0.27~^\circ/y = 3.1 \times 10^{-5~\circ}/h},$$</span></p>

<p>so the TNO's apparent motion relative to the Sun is</p>

<p><span class="math-container">$$\mathrm{0.00165~^\circ/h + 3.1 \times 10^{-5~\circ}/h = 0.00168~^\circ/h}.$$</span></p>

<p>The Sun would be totally occulted for</p>

<p><span class="math-container">$$\mathrm{\frac{0.126~^\circ - 0.004~^\circ}{0.00168~^\circ/h} = 72.6~h}$$</span></p>

<p>and partially occulted for</p>

<p><span class="math-container">$$\mathrm{\frac{0.004~^\circ}{0.00168~^\circ/h} = 2.4~h}$$</span></p>

<p>at each end.</p>

CC BY-SA 4.0
2019-05-05T02:04:45.610Z
Looks *much nicer*, thank you! :-)
uhoh
CC BY-SA 4.0
2019-05-03T17:12:39.137Z
2019-05-05T01:57:05.540Z
2019-05-05T01:57:05.540Z
Mike G
<p>A trans-Neptunian object, 1 million km from the observer, with an angular diameter of 0.126° occults the Sun (angular diameter of 0.004°) and the TNO and the observer are moving in the same direction, in the same plane.  The TNO is moving 8m/sec with respect to the observer, how long would the occultation last?</p>

<p>Assume the observer is 18 billion km from the Sun.</p>

CC BY-SA 4.0
2019-05-01T13:48:53.857Z
Did you try simple geometry - angular extent of Sol and Saturn from the POV of your spacecraft?
Carl Witthoft
CC BY-SA 4.0
2019-05-01T14:12:04.900Z
Right a transit, not an occultation.   Thanks
Bob516
CC BY-SA 4.0
2019-05-01T19:32:38.060Z
@MikeG  if the near object has greater angle, then it's an eclipse and hence an occultation.
Carl Witthoft
CC BY-SA 4.0
2019-05-01T19:33:35.373Z
@Bob516 your edit fails to note that you can't answer because you mixed angular dimensions with linear.  Replace your " 8m/sec" with  "X degrees/sec" .
Carl Witthoft
CC BY-SA 4.0
2019-05-01T19:38:57.547Z
@CarlWitthoft I don't know how to replace linear with angular.
Bob516
CC BY-SA 4.0
2019-05-01T19:43:33.550Z
Should I start the question all over again, and just use the sizes of the objects, the speeds and the distances between them?
Bob516
<the-sun><occultation><trans-neptunian-object>
How long would an occultation by a TNO last?
Bob516
# How long would an occultation by a TNO last?

(Asked by: username_0 on 2019-05-01)

A trans-Neptunian object, 1 million km from the observer, with an angular diameter of 0.126° occults the Sun (angular diameter of 0.004°) and the TNO and the observer are moving in the same direction, in the same plane. The TNO is moving 8m/sec with respect to the observer, how long would the occultation last?

Assume the observer is 18 billion km from the Sun.

**Tags:** the-sun, occultation, trans-neptunian-object
> **Comments:**
> **username_1**: Did you try simple geometry - angular extent of Sol and Saturn from the POV of your spacecraft?  
> **username_0**: Right a transit, not an occultation.   Thanks  
> **username_1**: @MikeG  if the near object has greater angle, then it's an eclipse and hence an occultation.  
> **username_1**: @Bob516 your edit fails to note that you can't answer because you mixed angular dimensions with linear.  Replace your " 8m/sec" with  "X degrees/sec" .  
> **username_0**: @CarlWitthoft I don't know how to replace linear with angular.  
> **username_0**: Should I start the question all over again, and just use the sizes of the objects, the speeds and the distances between them?  

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## 1 answers

### Answer by username_2 on 2019-05-03. Score: 6 | Accepted answer

From the observer's point of view 1 million km away, the TNO's apparent angular motion is

$$\mathrm{\frac{8~m/s}{10^9~m} = 8 \times 10^{-9}~rad/s = 0.00165~^\circ/h}.$$

Assuming that the observer at 18 billion km = 120 au is in a circular orbit around the Sun, the orbital period is
1203/2 = 1320 years, making the Sun appear to move the other way at

$$\mathrm{\frac{360~^\circ}{1320~y} = 0.27~^\circ/y = 3.1 \times 10^{-5~\circ}/h},$$

so the TNO's apparent motion relative to the Sun is

$$\mathrm{0.00165~^\circ/h + 3.1 \times 10^{-5~\circ}/h = 0.00168~^\circ/h}.$$

The Sun would be totally occulted for

$$\mathrm{\frac{0.126~^\circ - 0.004~^\circ}{0.00168~^\circ/h} = 72.6~h}$$

and partially occulted for

$$\mathrm{\frac{0.004~^\circ}{0.00168~^\circ/h} = 2.4~h}$$

at each end.

> **Comments:**
> **username_3**: Looks *much nicer*, thank you! :-)  

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