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.

- highschoolmom2010

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

You know this formula?
\[F=\frac{ GMm }{ r^2 }\]

- highschoolmom2010

i know of it but not too much on how to apply it

- 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.
\[F _{earth}=\frac{ GMm }{ r^2_{earth} }\]
\[F _{orbit}=\frac{ GMm }{ r^2 }\]
\[\frac{ F _{orbit} }{ F _{earth} }=\frac{ 1 }{ 16 }\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Does that make sense?

- anonymous

|dw:1435680564384:dw|

- highschoolmom2010

ok i think i follow you

- anonymous

ok. were you able to solve for r?

- anonymous

The G, M, and m cancel so you're left with
\[\frac{ 1 }{ 16 }=\frac{ 6400^2 }{ r^2 }\]

- Michele_Laino

the situation described in your problem is:
|dw:1435681788781:dw|

- highschoolmom2010

|dw:1435681791475:dw|
well since you did it before i could figure out how to draw it YAY

- 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

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

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

hint, we have:
\[\Large \frac{{{R_T}}}{{{R_T} + h}} = \frac{1}{4}\]

- highschoolmom2010

im actually confused now :/

- Michele_Laino

after a simplification, we get:
\[\Large 4{R_T} = {R_T} + h\]
what is h?

- anonymous

do you know how to solve the equation you wrote above?

- anonymous

you can cross multiply

- highschoolmom2010

i was understanding ok until after we simplified for r

- highschoolmom2010

at @Michele_Laino first reply threw me off

- anonymous

|dw:1435682627087:dw|

- 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

oh ok

- anonymous

the benefit of her way is you don't have to subtract at the end

- highschoolmom2010

|dw:1435682797670:dw|

- anonymous

square 6400 first
then multiply by 16
then take the square root

- 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

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

which is equivalent to this one:
\[\frac{{{R_T}}}{{{R_T} + h}} = \frac{1}{4}\]

- Michele_Laino

or to this one:
\[4{R_T} = {R_T} + h\]

- Michele_Laino

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

- highschoolmom2010

oh ok
6400^2=40960000
40960000*16=655360000
\[\sqrt{655360000}=25600\]
@peachpi

- highschoolmom2010

i hope i did it right that time

- 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

so 25600-6400?

- anonymous

yes

- highschoolmom2010

19200

- anonymous

there's your answer :)

- highschoolmom2010

thank you :)

- anonymous

you're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.