Since Pluto and Charon have distinct orbits around their barycenter (located approximately 960 km above Pluto's surface), the centripedal acceleration must be different on the outer hemisphere of Pluto (in relation to the barycenter).

Does that mean I would weigh less standing on the top of the outer hemisphere than I would standing on top of the inner hemisphere (facing the barycenter)?

Or is that cancelled out by the pull of Charon?

Pluto-Charon System

  • $\begingroup$ Should that read "I would weight less" rather than more? $\endgroup$ – James Thorpe Jan 21 '16 at 13:44
  • $\begingroup$ @JamesThorpe - yep, corrected ;-) $\endgroup$ – mhbuur Jan 21 '16 at 13:53
  • 2
    $\begingroup$ I was halfway through writing a comment about Charon adding to the weight rather than cancelling it out when I realised that that was the wrong way round, which led me back to the more->less issue :) $\endgroup$ – James Thorpe Jan 21 '16 at 13:53

Let's try and do some math with this. Will use Wikipedia for numbers.

  • Gravity at Pluto's surface: 0.61711215789 $m/s^2$
  • Gravity from Charon- Pluto's near side- 0.0002724276 $m/s^2$
  • Gravity from Charon- Pluto's far side- 0.00027242753 $m/s^2$
  • Centripetal Acceleration- Near side- 0.00001749536 $m/s^2$
  • Centripetal Acceleration- Far side- 0.00006647414 $m/s^2$

For reference, the formulas to calculate these are as follows:

  • Gravity- $a_g=G* m/d^2$ where $G=6.673×10^{-11} N m^2 kg^{-2}$. As $F=m*a$, this is basically the same as the gravitational equation, assuming a much smaller body than the primary mass.
  • Centripetal- $a_c=v_t^2/r$. $v_t=2*\pi*r_{orbit}/orbital Period$

Bottom line, Charon's gravity has more of an effect than the Centripetal acceleration, but both are at least 3 orders of magnitude smaller than the base gravity. This is less than the difference of weight in a person from the equator to the poles.

  • 2
    $\begingroup$ So how many bags of space Doritos I eat will have more of an impact than Charon? $\endgroup$ – corsiKa Jan 21 '16 at 16:31
  • 8
    $\begingroup$ @corsiKa: The single-serving bags of Doritos are 1 oz or 28 grams. If you are the average weight (62 kg) then it would take about two to cancel out the effect of Charon on one side vs. Charon on the other. $\endgroup$ – Charles Jan 21 '16 at 18:48
  • $\begingroup$ Anothr aspect: On Earth, the tidal wave facing the Moon is the same height (I suppose) as the one opposite it, which indicates that the effect on the directly facing side should be the same as on the directly opposite side - but perpendicular to it, things are different $\endgroup$ – Hagen von Eitzen Jan 21 '16 at 21:29
  • 1
    $\begingroup$ Interesting. Despite the larger mass of Earth's moon its much greater distance dominates. Its hard to visualise two points together - that the relative acceleration due to the moon is a tenth of a Dorito by the earlier measure and yet it is also responsible for up to 15m tides. $\endgroup$ – Puffin Jan 22 '16 at 13:37
  • 1
    $\begingroup$ The 15m tide is essentially a funnel effect. 1m is a more normal value. A part of this is because of the Sun, and a second part due to the repeatability of the motion, which has some resonance effects. And there are other effects too. $\endgroup$ – PearsonArtPhoto Jan 22 '16 at 13:41

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.