# How long would an occultation by a TNO last?

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.

• Did you try simple geometry - angular extent of Sol and Saturn from the POV of your spacecraft? – Carl Witthoft May 1 '19 at 13:48
• Right a transit, not an occultation. Thanks – Bob516 May 1 '19 at 14:12
• @MikeG if the near object has greater angle, then it's an eclipse and hence an occultation. – Carl Witthoft May 1 '19 at 19:32
• @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 May 1 '19 at 19:33
• @CarlWitthoft I don't know how to replace linear with angular. – Bob516 May 1 '19 at 19:38

## 1 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.

• Looks much nicer, thank you! :-) – uhoh May 5 '19 at 2:04