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.

  • 1
    $\begingroup$ Did you try simple geometry - angular extent of Sol and Saturn from the POV of your spacecraft? $\endgroup$ May 1, 2019 at 13:48
  • $\begingroup$ Right a transit, not an occultation. Thanks $\endgroup$
    – Bob516
    May 1, 2019 at 14:12
  • $\begingroup$ @MikeG if the near object has greater angle, then it's an eclipse and hence an occultation. $\endgroup$ May 1, 2019 at 19:32
  • $\begingroup$ @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" . $\endgroup$ May 1, 2019 at 19:33
  • $\begingroup$ @CarlWitthoft I don't know how to replace linear with angular. $\endgroup$
    – Bob516
    May 1, 2019 at 19:38

1 Answer 1


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.

  • $\begingroup$ Looks much nicer, thank you! :-) $\endgroup$
    – uhoh
    May 5, 2019 at 2:04

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