Results 1 to 11 of 11

Thread: Diameter of the Moon

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    42

    Diameter of the Moon

    The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello somanyquestions
    Quote Originally Posted by somanyquestions View Post
    The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.
    The circumference of a circle of radius 240,000 miles is 2\pi\times 240000 miles.

    The circumference subtends an angle of 360^o at the centre of the circle. So what fraction of the circumference subtends an angle of 1^o at the centre? And so what fraction of the circumference subtends an angle of \tfrac12^o?

    Well, to calculate the diameter of the moon, you need this fraction of the whole circumference,
    2\pi\times 240000.

    Can you complete it now?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by somanyquestions View Post
    The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.
    Draw a diagram, then look at the triangle formed by the observer, centre of the moon and a point on the visible edge of the moon, then:

    let R be the radius of the Moon:

    \tan(0.5^{\circ}/2)=\frac{R}{240000}

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,108
    Thanks
    3648
    Quote Originally Posted by somanyquestions View Post
    The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.
    cosine law ...

     <br />
d^2 = 2(240000)^2 - 2(240000)^2 \cdot \cos(0.5^\circ)<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by somanyquestions View Post
    The full moon subtends an angle of 1/2 degrees. The moon is 240,000 miles away. Find the diameter of the moon.
    If the shortest distance from the observer to the moon is 240,000 miles

    sin\left(\frac{0.5^o}{2}\right)=\frac{R}{240,000+R  }

    R=(240,000+R)Sin\left(\frac{0.5^o}{2}\right)=240,0  00Sin0.25^o+Rsin0.25^o

    R\left(1-sin0.25^o\right)=240,000sin0.25^o

    R=\frac{240,000sin0.25^0}{1-sin0.25^o}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello somanyquestions

    I'm not sure why so many different solutions are being offered to this question. Some are good alternatives to the solution I gave you. Some - like this one -
    Quote Originally Posted by Archie Meade View Post
    If the shortest distance from the observer to the moon is 240,000 miles

    sin\left(\frac{0.5^o}{2}\right)=\frac{R}{240,000+R  }

    R=(240,000+R)Sin\left(\frac{0.5^o}{2}\right)=240,0  00Sin0.25^o+Rsin0.25^o

    R\left(1-sin0.25^o\right)=240,000sin0.25^o

    R=\frac{240,000sin0.25^0}{1-sin0.25^o}
    are not!

    I suggest that you look carefully at the method I suggested, since that is similar to another question you have recently posted - the latitude of Toronto - and don't worry about the others unless you really want to.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    See attachment.
    Attached Thumbnails Attached Thumbnails Diameter of the Moon-moon-radius.jpg  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello Archie Meade -

    Sorry to pursue this further, but what you (and others) don't seem to be taking into account is that the distance of the earth from the moon - given as 240,000 miles - is:
    (a) very large compared to the diameter of the moon, and, for that matter, the earth;

    (b) only given to 2 significant figures, anyway. (The mean distance is given on WikiAnswers as 238,857 miles.)
    Thus, taking the moon's radius into account is a waste of time and effort. You might just as well consider the radius of the earth as well.

    The fact that the distance between the earth and the moon is large compared to their radii is the reason that I was able to offer the solution that I did. Because of this fact, I was able to treat the visible surface of the moon as an arc of a circle, whose centre is the observer on the surface of the earth.

    Captain Black's solution (using the tangent of half the subtended angle) assumes that the distance of 240,000 miles is the distance from the observer to the centre of the moon. But equally, you could take it (with no reasonable loss of accuracy) as the distance from the observer to the visible edge of the moon, in which case the tangent of the angle is replaced by its sine.

    skeeter's solution, using the Cosine Rule, assumes the same thing.

    The fact is that for small values of \theta,\; \sin\theta \approx \tan\theta \approx \theta (in radians). So the difference between the arc length (which was the method I used) and the length of the chord is totally irrelevant compared to the other factors I have listed above.

    So rather than offer more complicated solutions that add nothing of value, please let's keep it simple!

    Grandad
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Given that the 240,000 miles and the 0.5 degree angles are approximate,
    all solutions will be approximate anyway.
    There are many ways to obtain an approximate answer,
    by deciding certain values are insignificant.

    If we were dealing with "exact" values,
    then certain calculations give exact results.

    I was simply offering a solution,
    I wasn't rebuking anyone.

    Many threads contain numerous approaches.
    To me, it's lack of clarity that causes confusion.

    In any case, the poster can decide what's appropriate or not.

    Unconstructive criticism isn't something I place much value on.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello everyone

    It's time this discussion was drawn to a close, so I'm closing this thread now.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    I think the moral of the story here is to state any assumptions made at the start of a solution.

    I think everyone has made a valuable contribution to this thread and I think, in hindsight, everyone would agree. The OP can decide which of the many approaches best appeals to him/her.

    Critique of solutions is important and often clarifies them, diversity of approach can be important too. I'm as guilty as the next person of offering a solution that is too complicated, and I'm also guilty of disagreeing with solutions.

    Let's all have a nice day (or night as the case may be).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Feb 23rd 2011, 12:11 AM
  2. Miles From Moon to Earth
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Sep 13th 2010, 08:14 AM
  3. Replies: 3
    Last Post: Jul 12th 2010, 03:27 AM
  4. image of the moon
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 3rd 2008, 07:35 AM
  5. Effect of moon..
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Aug 16th 2006, 11:46 AM

Search tags for this page

The full Moon viewed from the Earth is about 1⁄2 degree. If the average distance to the Moon from the Earth is 238,900 miles, what is the diameter of the Moon?
,

moon diameter question trigonimetry

,

assuming that the moon diameter subtents an angle of 30' at the eye of an observer, find how far from the eye coin of one cm diameter must be held so as just to hide the moon

,

diameter of a full moon

,

moon's distance from earth is 360000 & diameter subtends an angle of 42° at eye of observer. the diamter of moon is

,

calculate diameter of moon if moon's distance from earth 360000km at an angle 42minutes

,

radius of earth is 6400 km and that it subtends an angle of 57' at the centre of the moon find the distance of the centre of the moon from the earth

,

earth radius is 6400 km and that subtended angle of 57 degree at the center of moon

,

distance between moon and earth is 360000km. if moon's diameter make elevation angle 30'.find the diameter of moon?

,

the moon's distance from the earth is 360000 km nd its diameter subtends an angle of 42' at the eye of observer. What is the diameter of moon?

,

if the moon disk subtends a maximum angle of 0.52 degrees at the surface of the earth, calculate the moon's radius

,

If the Moon disk subtends a maximum angle of 0°31'8'' at the surface of the Earth , what is the Moon's radius?

,

the radius of earth is 6300km if it substend an angle of 57' at the center of the moon find the distance of moon from the earth surface

,

what does the moon subtends at the earth mean

,

The moon's diameter subtends an angle of about 31.1 degrees from the center of the earth. Taking the distance from the earth to the moon as 239 000 miles. Find the diameter of the moon.

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum