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? $\endgroup$ – Carl Witthoft May 1 '19 at 13:48
  • $\begingroup$ Right a transit, not an occultation. Thanks $\endgroup$ – Bob516 May 1 '19 at 14:12
  • $\begingroup$ @MikeG if the near object has greater angle, then it's an eclipse and hence an occultation. $\endgroup$ – Carl Witthoft May 1 '19 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" . $\endgroup$ – Carl Witthoft May 1 '19 at 19:33
  • $\begingroup$ @CarlWitthoft I don't know how to replace linear with angular. $\endgroup$ – Bob516 May 1 '19 at 19:38

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.

| improve this answer | |
  • $\begingroup$ Looks much nicer, thank you! :-) $\endgroup$ – uhoh May 5 '19 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.