highschoolmom2010
  • highschoolmom2010
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
You know this formula? \[F=\frac{ GMm }{ r^2 }\]
highschoolmom2010
  • highschoolmom2010
i know of it but not too much on how to apply it
anonymous
  • anonymous
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. \[F _{earth}=\frac{ GMm }{ r^2_{earth} }\] \[F _{orbit}=\frac{ GMm }{ r^2 }\] \[\frac{ F _{orbit} }{ F _{earth} }=\frac{ 1 }{ 16 }\]

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anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
|dw:1435680564384:dw|
highschoolmom2010
  • highschoolmom2010
ok i think i follow you
anonymous
  • anonymous
ok. were you able to solve for r?
anonymous
  • anonymous
The G, M, and m cancel so you're left with \[\frac{ 1 }{ 16 }=\frac{ 6400^2 }{ r^2 }\]
Michele_Laino
  • Michele_Laino
the situation described in your problem is: |dw:1435681788781:dw|
highschoolmom2010
  • highschoolmom2010
|dw:1435681791475:dw| well since you did it before i could figure out how to draw it YAY
Michele_Laino
  • Michele_Laino
so we can write this: \[\begin{gathered} W = G\frac{{{M_T}m}}{{R_T^2}} \hfill \\ {W_1} = G\frac{{{M_T}m}}{{{{\left( {{R_T} + h} \right)}^2}}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
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
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
hint, we have: \[\Large \frac{{{R_T}}}{{{R_T} + h}} = \frac{1}{4}\]
highschoolmom2010
  • highschoolmom2010
im actually confused now :/
Michele_Laino
  • Michele_Laino
after a simplification, we get: \[\Large 4{R_T} = {R_T} + h\] what is h?
anonymous
  • anonymous
do you know how to solve the equation you wrote above?
anonymous
  • anonymous
you can cross multiply
highschoolmom2010
  • highschoolmom2010
i was understanding ok until after we simplified for r
highschoolmom2010
  • highschoolmom2010
at @Michele_Laino first reply threw me off
anonymous
  • anonymous
|dw:1435682627087:dw|
anonymous
  • anonymous
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
highschoolmom2010
  • highschoolmom2010
oh ok
anonymous
  • anonymous
the benefit of her way is you don't have to subtract at the end
highschoolmom2010
  • highschoolmom2010
|dw:1435682797670:dw|
anonymous
  • anonymous
square 6400 first then multiply by 16 then take the square root
Michele_Laino
  • Michele_Laino
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: \[W = G\frac{{{M_T}m}}{{R_T^2}}\]
Michele_Laino
  • Michele_Laino
now, using the data from your problem , we have to solve this equation: \[\frac{{{W_1}}}{W} = \frac{1}{{16}} = {\left( {\frac{{{R_T}}}{{{R_T} + h}}} \right)^2}\]
Michele_Laino
  • Michele_Laino
which is equivalent to this one: \[\frac{{{R_T}}}{{{R_T} + h}} = \frac{1}{4}\]
Michele_Laino
  • Michele_Laino
or to this one: \[4{R_T} = {R_T} + h\]
Michele_Laino
  • Michele_Laino
from which we get the requested height as below: \[\Large h = 3{R_T} = 3 \times 6400 = ...Km\]
highschoolmom2010
  • highschoolmom2010
oh ok 6400^2=40960000 40960000*16=655360000 \[\sqrt{655360000}=25600\] @peachpi
highschoolmom2010
  • highschoolmom2010
i hope i did it right that time
anonymous
  • anonymous
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.
highschoolmom2010
  • highschoolmom2010
so 25600-6400?
anonymous
  • anonymous
yes
highschoolmom2010
  • highschoolmom2010
19200
anonymous
  • anonymous
there's your answer :)
highschoolmom2010
  • highschoolmom2010
thank you :)
anonymous
  • anonymous
you're welcome

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