Weight in Terms of Distance From Earth

[bfr]

[SOLVED] Weight in Terms of Distance From Earth

Homework Statement

At what height above the Earth's surface will an astronaut have a true weight that is one half his weight on earth?

Homework Equations

F=G(m1*m2)/r^2
(GM)/(4*pi^2)=(R^3)/t^2

The Attempt at a Solution

I did:
F/2=(G*m1*m2)/r^2
r=sqrt((2*G*m1*m2)/F)

My teacher got sqrt(2) times the radius of the earth. Where do I go from there?

 

[tiny-tim]

F=G(m1*m2)/r^2
…
I did:
F/2=(G*m1*m2)/r^2
r=sqrt((2*G*m1*m2)/F)

Hi bfr!

You're confusing yourself by using the letter r for two different distances.

Hint: call the Earth's radius r, and call the radius you're looking for s.

Then write two equations, one with r and one with s.

Then … does that help?

 

[mysqlpress]

g=GM/R^2

so g is directly proportional to 1/R^2
g'/g = R^2/R'^2
1/2=R^2/R'^2
let R = 6400*1000 m
R'^2 = 2R^2
R' = sqrt2 R = 9050km
atitude = 9050-6400 =2651km above the Earth's surface

of coz , u may realize that
GM =gRe^2 , where Re is the radius of the Earth
then
g'= gRe^2/R'^2
which is the same.

 

[tiny-tim]

Yes! But a bit long-winded.

You could just say:

F =G.m1.m2/r^2
F´ =G.m1.m2/r´^2
so F/F´= r´^2/r^2 = (r´/r)^2.

Do you see the advantage now of being careful to use different letters?!

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