0
$\begingroup$

I have read that a star is chosen from the left hand side of pluto and as pluto moves it blocks the starlight and when the star appears on the right hand side again, meaning it reappears we get the diameter of pluto. Can somebody explain this to me in layman's terms?

$\endgroup$
2
  • $\begingroup$ Aside: My first thought was: this will never happen in my lifetime. But it has, more than once. In 2011 both Pluto and Charon occulted the same star. Four days later, Pluto and Hydra occulted a different star. "Anything that can happen, will happen." $\endgroup$
    – garyp
    Jan 25, 2017 at 15:29
  • $\begingroup$ While the occultation method you're asking about was historically important for Pluto, and still important for asteroids, note that since 2015 we have a much more direct measurement of Pluto's size and many other properties. $\endgroup$
    – rob
    Feb 2, 2017 at 4:05

2 Answers 2

1
$\begingroup$

We time it. We ideally want a bright star. We time when the light from the star goes out and when it reappears. We know how fast Pluto is moving in its orbit, the star is treated as fixed in position, so that is one way we measure it, but we will only get a true diameter if the equator of Pluto passes across the star.

$\endgroup$
2
  • $\begingroup$ Lets take an example: If pluto's speed is 5 km/sec and the star takes 20 seconds to reappear then how do we calculate the diameter of pluto from this info? $\endgroup$
    – avito009
    Jan 26, 2017 at 7:13
  • $\begingroup$ $5\times 20\,\mathrm{km}$, or in fact at least this since it may not have passed in front of the star centrally. Dimensional analysiscwould have told you this. $\endgroup$
    – user107153
    Jan 27, 2017 at 11:11
0
$\begingroup$

I think I have the answer. Distance is speed into time. So when we know plutos speed and time it takes for pluto to move which in the above example is 20 seconds. The answer to the above hence will be 5x20=100 (COrrect me if I am wrong and also give me the units). This distance will be the diameter of pluto.

$\endgroup$

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