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highschoolmom2010

  • one year ago

13. Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of your current weight.

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  1. anonymous
    • one year ago
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    You know this formula? \[F=\frac{ GMm }{ r^2 }\]

  2. highschoolmom2010
    • one year ago
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    i know of it but not too much on how to apply it

  3. anonymous
    • one year ago
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    ok. in G is a gravitation constant, M is the mass of the bigger body (Earth) and m is the mass of the smaller one (you). This application is comparing your weight on the Earth to somewhere in space so you can make a ratio. \[F _{earth}=\frac{ GMm }{ r^2_{earth} }\] \[F _{orbit}=\frac{ GMm }{ r^2 }\] \[\frac{ F _{orbit} }{ F _{earth} }=\frac{ 1 }{ 16 }\]

  4. anonymous
    • one year ago
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    Does that make sense?

  5. anonymous
    • one year ago
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    |dw:1435680564384:dw|

  6. highschoolmom2010
    • one year ago
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    ok i think i follow you

  7. anonymous
    • one year ago
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    ok. were you able to solve for r?

  8. anonymous
    • one year ago
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    The G, M, and m cancel so you're left with \[\frac{ 1 }{ 16 }=\frac{ 6400^2 }{ r^2 }\]

  9. Michele_Laino
    • one year ago
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    the situation described in your problem is: |dw:1435681788781:dw|

  10. highschoolmom2010
    • one year ago
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    |dw:1435681791475:dw| well since you did it before i could figure out how to draw it YAY

  11. Michele_Laino
    • one year ago
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    so we can write this: \[\begin{gathered} W = G\frac{{{M_T}m}}{{R_T^2}} \hfill \\ {W_1} = G\frac{{{M_T}m}}{{{{\left( {{R_T} + h} \right)}^2}}} \hfill \\ \end{gathered} \]

  12. anonymous
    • one year ago
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    lol :) so once you have that r, that's the distance from the center of the earth. You have to subtract 6400 from that to get your final answer

  13. Michele_Laino
    • one year ago
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    now using your data we have to solve this equation: \[\frac{{{W_1}}}{W} = \frac{1}{{16}} = {\left( {\frac{{{R_T}}}{{{R_T} + h}}} \right)^2}\] please solve that equation for h, and you will get your answer, where: R_T is the radius of the Earth

  14. Michele_Laino
    • one year ago
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    hint, we have: \[\Large \frac{{{R_T}}}{{{R_T} + h}} = \frac{1}{4}\]

  15. highschoolmom2010
    • one year ago
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    im actually confused now :/

  16. Michele_Laino
    • one year ago
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    after a simplification, we get: \[\Large 4{R_T} = {R_T} + h\] what is h?

  17. anonymous
    • one year ago
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    do you know how to solve the equation you wrote above?

  18. anonymous
    • one year ago
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    you can cross multiply

  19. highschoolmom2010
    • one year ago
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    i was understanding ok until after we simplified for r

  20. highschoolmom2010
    • one year ago
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    at @Michele_Laino first reply threw me off

  21. anonymous
    • one year ago
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    |dw:1435682627087:dw|

  22. anonymous
    • one year ago
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    that's just a slightly different way of doing it. she separated the earth's radius and the height, and I just kept it all as one

  23. highschoolmom2010
    • one year ago
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    oh ok

  24. anonymous
    • one year ago
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    the benefit of her way is you don't have to subtract at the end

  25. highschoolmom2010
    • one year ago
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    |dw:1435682797670:dw|

  26. anonymous
    • one year ago
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    square 6400 first then multiply by 16 then take the square root

  27. Michele_Laino
    • one year ago
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    reassuming: the weight of a mass m which is locate h meters above the Earth's surface is: \[{W_1} = G\frac{{{M_T}m}}{{{{\left( {{R_T} + h} \right)}^2}}}\] whereas the weight of our mass m when it is located upon the Eart's surface is: \[W = G\frac{{{M_T}m}}{{R_T^2}}\]

  28. Michele_Laino
    • one year ago
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    now, using the data from your problem , we have to solve this equation: \[\frac{{{W_1}}}{W} = \frac{1}{{16}} = {\left( {\frac{{{R_T}}}{{{R_T} + h}}} \right)^2}\]

  29. Michele_Laino
    • one year ago
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    which is equivalent to this one: \[\frac{{{R_T}}}{{{R_T} + h}} = \frac{1}{4}\]

  30. Michele_Laino
    • one year ago
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    or to this one: \[4{R_T} = {R_T} + h\]

  31. Michele_Laino
    • one year ago
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    from which we get the requested height as below: \[\Large h = 3{R_T} = 3 \times 6400 = ...Km\]

  32. highschoolmom2010
    • one year ago
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    oh ok 6400^2=40960000 40960000*16=655360000 \[\sqrt{655360000}=25600\] @peachpi

  33. highschoolmom2010
    • one year ago
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    i hope i did it right that time

  34. anonymous
    • one year ago
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    yes that's right. that's the distance from the center of the earth, so you have to subtract the radius of earth (6400) to get the distance from the surface.

  35. highschoolmom2010
    • one year ago
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    so 25600-6400?

  36. anonymous
    • one year ago
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    yes

  37. highschoolmom2010
    • one year ago
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    19200

  38. anonymous
    • one year ago
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    there's your answer :)

  39. highschoolmom2010
    • one year ago
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    thank you :)

  40. anonymous
    • one year ago
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    you're welcome

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