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As per the answer to this question it is suggested the relative tidal pull of objects of equal angular area is equal to their relative density. Which lead me to the click-baity question:

Does a ball bearing held to eclipse the Moon have more gravitational effect on my eye than the Moon?

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  • $\begingroup$ No, because you need to measure the size from the radius of the earth, not your eyeball. (or the barycenter, I think it's the barycenter), but certainly not your eyeball). A ball bearing looks pretty tiny from 1 earth radius away. (putting as a comment because I don't know the precise formula). Now if the ball bearing was a black hole, that would be different because density needs to be added to the angular area rule. Moon tides are bigger than Sun tides cause the Moon has greater average density. $\endgroup$
    – userLTK
    Commented May 22, 2016 at 15:18
  • $\begingroup$ I'm sorry, I mis-read. greater tidal pull on your eye than the moon? Yes, I think it does, but both would be teeny tiny. $\endgroup$
    – userLTK
    Commented May 22, 2016 at 15:21
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    $\begingroup$ It would also imply that a ball of about 60km radius in orbit will have the same tidal effect on earth than the moon (if densities of ball and Moon are equal). Unintuitive but I could not find anything wrong in the answer. $\endgroup$
    – user83548
    Commented May 22, 2016 at 17:36

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