## 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.

1. anonymous

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

2. highschoolmom2010

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

3. 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. $F _{earth}=\frac{ GMm }{ r^2_{earth} }$ $F _{orbit}=\frac{ GMm }{ r^2 }$ $\frac{ F _{orbit} }{ F _{earth} }=\frac{ 1 }{ 16 }$

4. anonymous

Does that make sense?

5. anonymous

|dw:1435680564384:dw|

6. highschoolmom2010

ok i think i follow you

7. anonymous

ok. were you able to solve for r?

8. anonymous

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

9. Michele_Laino

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

10. highschoolmom2010

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

11. Michele_Laino

so we can write this: $\begin{gathered} W = G\frac{{{M_T}m}}{{R_T^2}} \hfill \\ {W_1} = G\frac{{{M_T}m}}{{{{\left( {{R_T} + h} \right)}^2}}} \hfill \\ \end{gathered}$

12. 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

13. 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

14. Michele_Laino

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

15. highschoolmom2010

im actually confused now :/

16. Michele_Laino

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

17. anonymous

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

18. anonymous

you can cross multiply

19. highschoolmom2010

i was understanding ok until after we simplified for r

20. highschoolmom2010

at @Michele_Laino first reply threw me off

21. anonymous

|dw:1435682627087:dw|

22. 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

23. highschoolmom2010

oh ok

24. anonymous

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

25. highschoolmom2010

|dw:1435682797670:dw|

26. anonymous

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

27. 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: $W = G\frac{{{M_T}m}}{{R_T^2}}$

28. Michele_Laino

now, using the data from your problem , we have to solve this equation: $\frac{{{W_1}}}{W} = \frac{1}{{16}} = {\left( {\frac{{{R_T}}}{{{R_T} + h}}} \right)^2}$

29. Michele_Laino

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

30. Michele_Laino

or to this one: $4{R_T} = {R_T} + h$

31. Michele_Laino

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

32. highschoolmom2010

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

33. highschoolmom2010

i hope i did it right that time

34. 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.

35. highschoolmom2010

so 25600-6400?

36. anonymous

yes

37. highschoolmom2010

19200

38. anonymous

39. highschoolmom2010

thank you :)

40. anonymous

you're welcome