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    At what altitude h above the North Pole is the weight of an object reduced to one half its value on the Earth’s surface? Assume the Earth to be a sphere radius R and express h as a fraction of R.
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    (Original post by runny4)
    At what altitude h above the North Pole is the weight of an object reduced to one half its value on the Earth’s surface? Assume the Earth to be a sphere radius R and express h as a fraction of R.
    The weight of an object is inversely proportional to the square of the distance between that object and the centre of the Earth. Use ratios therefore to find (R+h) in terms of R.
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    (Original post by runny4)
    At what altitude h above the North Pole is the weight of an object reduced to one half its value on the Earth’s surface? Assume the Earth to be a sphere radius R and express h as a fraction of R.
    When you say above the North Pole, is it to take account of the effect of angular velocity of the earth on the weight of an object when placed at the equator? There has to be a place mentioned for comparison.
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    (Original post by 16Characters....)
    The weight of an object is inversely proportional to the square of the distance between that object and the centre of the Earth. Use ratios therefore to find (R+h) in terms of R.
    so ive got that the height must be root 2 times r but then what do u do?
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    (Original post by runny4)
    so ive got that the height must be root 2 times r but then what do u do?
    The height and the distance from the centre of the Earth are not the same since we measure height from the Earth's surface, not the centre
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    that's a really hard question i have to admit- this is probably no help to u at all
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    (Original post by 16Characters....)
    The height and the distance from the centre of the Earth are not the same since we measure height from the Earth's surface, not the centre
    so what is h and why?
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    (Original post by runny4)
    so what is h and why?
    You already had height_{from center} = \sqrt{2} R so height_{from surface} = \sqrt{2} R -R =(\sqrt{2}-1)R
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    (Original post by EricPiphany)
    You already had height_{from center} = \sqrt{2} R so height_{from surface} = \sqrt{2} R -R =(\sqrt{2}-1)R
    ok thanks
 
 
 
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