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Home Forums General Discussion Farthest object in Solar system

Viewing 6 posts - 1 through 6 (of 6 total)
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  • #12300
    Brian Blake
    Participant
    Topics: 188
    Replies: 409
    #12302
    Andy Sawers
    Moderator
    Topics: 134
    Replies: 604

    So I’m wondering, what’s the angular diameter of an object the size of ‘Land’s End to John O’Groats’ when it’s 15.5 billion kilometres/103AU away – and what’s its apparent magnitude??

    #12307
    Mike Meynell
    Moderator
    Topics: 119
    Replies: 756

    So I’m wondering, what’s the angular diameter of an object the size of ‘Land’s End to John O’Groats’ when it’s 15.5 billion kilometres/103AU away

    Tiny 😉

    Seriously though, the formula is α = 2*arctan(g/(2r)), where α is the angle, g is the real size and r is the distance [it doesn’t matter what units you use for g and r, as long as they are consistent].

    So, in this case, if g = 1000km and r = 15.5bn km, then α is about 3.7 x 10-6 degrees.

    and what’s its apparent magnitude

    Dunno that one… it would depend on the albedo of the object.

    #12308
    Andy Sawers
    Moderator
    Topics: 134
    Replies: 604

    So basically, what you’re saying is, it’s not very easy to see.

    On the basis of the xkcd cartoon below, unless I’ve slipped a zero this new object has an angular diameter on the surface of the earth of about 40cm…

    xkcd.com

    • This reply was modified 4 years, 6 months ago by Andy Sawers. Reason: image reload
    • This reply was modified 4 years, 6 months ago by Andy Sawers. Reason: image
    #12311
    Mike Meynell
    Moderator
    Topics: 119
    Replies: 756

    So basically, what you’re saying is, it’s not very easy to see

    At 0.01 arseconds! No 🙂

    Neptune is 2.3 arcseconds… Pluto is 0.1… so this is 10 times smaller than Pluto!!

    #12312
    Andy Sawers
    Moderator
    Topics: 134
    Replies: 604

    They must be using one of those cool telescopes from the John Lewis ads, then 😉

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