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Math Help - Radius of Earth

  1. #1
    Member purplec16's Avatar
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    Radius of Earth

    A spacelab circles Earth at an altitude of 380 miles. When an astronaut views the horizon of Earth the angle theta is 65.8 degrees. Use this information to estimate the radius of Earth.


    What do I do?
    (I know how it looks so you dont have to draw it but I dont know what am finding, hence I cant find it)
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    Quote Originally Posted by purplec16 View Post
    A spacelab circles Earth at an altitude of 380 miles. When an astronaut views the horizon of Earth the angle theta is 65.8 degrees. Use this information to estimate the radius of Earth.


    What do I do?
    (I know how it looks so you dont have to draw it but I dont know what am finding, hence I cant find it)
    \sin{\theta} = \frac{r}{r+h}
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  3. #3
    Member purplec16's Avatar
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    Why are we using sin if we have adjacent?
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    Quote Originally Posted by purplec16 View Post
    Why are we using sin if we have adjacent?
    better take another look at your sketch.

    the adjacent side is the distance from the astronaut to the point where the horizon touches the earth's surface.

    the hypotenuse is (r+h)

    the opposite side is r
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  5. #5
    Member purplec16's Avatar
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    So what am I solving for h or r?, r right? so dont i have to find h? whats h?
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    Quote Originally Posted by purplec16 View Post
    So what am I solving for h or r?, r right? so dont i have to find h? whats h?
    the height of the spacecraft above the earth's surface.
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  7. #7
    Member purplec16's Avatar
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    I still dont understand what to do
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  8. #8
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    h = 380 miles

    \theta = 65.8^\circ

    solve this equation for r ...

    \sin{\theta} = \frac{r}{r+h}

    check if your solution is reasonable by looking up the value of the earth's radius in miles.
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  9. #9
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    hi purplec,

    is your sketch like the attachment?

    then, as skeeter showed

    Sin(65.8^o)=\frac{opposite\ to\ angle}{hypotenuse}=\frac{R}{R+380}

    multiply both sides by the denominator

    (R+380)Sin(65.8^o)=\frac{R(R+380)}{R+380}=R

    multiply out the left

    RSin(65.8^o)+380Sin(65.8^o)=R

    subtract R from both sides to bring them together

    RSin(65.8^o)-R+380Sin(65.8^o)=R-R=0

    subtract the 380Sin part from both sides to have the R terms alone on the left

    RSin(65.8^o)-R+380Sin(65.8^o)-380Sin(65.8^o)=-380Sin(65.8^o)

    RSin(65.8^o)-R=-380Sin(65.8^o)

    factorise the R parts

    R[Sin(65.8^o)-1]=-380Sin(65.8^o)

    divide both sides by the multiplier of R

    R=\frac{-380Sin(65.8^0)}{Sin(65.8^o)-1}=380\frac{(-1)Sin(65.8^o)}{(-1)[1-Sin(65.8^o)]}=380\frac{Sin(65.8^o)}{1-Sin(65.8^o)}
    Attached Thumbnails Attached Thumbnails Radius of Earth-spacelab.jpg  
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Archie Meade View Post
    hi purplec,

    is your sketch like the attachment?

    then, as skeeter showed

    Sin(65.8^o)=\frac{opposite\ to\ angle}{hypotenuse}=\frac{R}{R+380}

    multiply both sides by the denominator

    (R+380)Sin(65.8^o)=\frac{R(R+380)}{R+380}=R

    multiply out the left

    RSin(65.8^o)+380Sin(65.8^o)=R

    subtract R from both sides to bring them together

    RSin(65.8^o)-R+380Sin(65.8^o)=R-R=0

    subtract the 380Sin part from both sides to have the R terms alone on the left

    RSin(65.8^o)-R+380Sin(65.8^o)-380Sin(65.8^o)=-380Sin(65.8^o)

    RSin(65.8^o)-R=-380Sin(65.8^o)

    factorise the R parts

    R[Sin(65.8^o)-1]=-380Sin(65.8^o)

    divide both sides by the multiplier of R

    R=\frac{-380Sin(65.8^0)}{Sin(65.8^o)-1}=380\frac{(-1)Sin(65.8^o)}{(-1)[1-Sin(65.8^o)]}=380\frac{Sin(65.8^o)}{1-Sin(65.8^o)}
    Just a note [tex]\sin\left(65.8^{\circ}\right)[/tex] gives \sin\left(95.8^{\circ}\right)
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  11. #11
    Member purplec16's Avatar
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    no its not I tried to upload it but it wont
    its the earth then its this triangle forming at the edge of it, and it bottom length is 360 mi and it tips the earth then the line extends and to the center of the earth and thats what their tellin me to find and the angle touching it is 68.5 degrees
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  12. #12
    Member purplec16's Avatar
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    I mean Archie Meade btw
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  13. #13
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    Quote Originally Posted by purplec16 View Post
    no its not I tried to upload it but it wont
    its the earth then its this triangle forming at the edge of it, and it bottom length is 360 mi and it tips the earth then the line extends and to the center of the earth and thats what their tellin me to find and the angle touching it is 68.5 degrees
    your original post ...

    A spacelab circles Earth at an altitude of 380 miles. When an astronaut views the horizon of Earth the angle theta is 65.8 degrees. Use this information to estimate the radius of Earth.
    so ... which is correct?
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  14. #14
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    hi purplec,

    if it's similar to this attachment,
    the calculations will be the same,
    as the geometry is still the same
    Attached Thumbnails Attached Thumbnails Radius of Earth-spacelab2.jpg  
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  15. #15
    Member purplec16's Avatar
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    Yeah your right its like how you did it the 2nd time with out the two ends of the triangle sticking out
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