At what altitude the value of g become one fourth than on the surface of the earht?

Question

anonymous · Answer

$$ g=G\frac{m}{r^2}$$

$$ gs=G\frac{M_E}{r_e^2}$$
$$ gs=G\frac{M_E}{r_a^2}$$

You should just be able to set up a proportion

$$ \frac{g_s}{g_a}=4$$
For the gravitational field at the surface, gs,  divided by the gravitational field at some altitude, ga.  Then everything but their radii cancel, and yo should be able to solve for r_a )the radius at some altitude) directly in terms of r_e (the radius of the earth).

^_^

anonymous · Answer

derp, sorry, the third equation should be

$$ ga=G\frac{M_E}{r_a^2}$$

imtiaz7 · Answer

can you explaing the fourth equation @AllTehMaffs

anonymous · Answer

If the strength of the gravitational field of ga is just 1/4th gs, then their ratio (with the gravitational field of the surface being in the numerator) would be 1/(1/4) - the better way to set it up would be the intuitive way

$$\frac{g_a}{g_s}=1/4$$

I just flipped it for absolutely no reason..... sorry for the confusingness :P

imtiaz7 · Answer

thanks @AllTehMaffs for your help

anonymous · Answer

^_^ Welcome!

imtiaz7 · Answer

will you help in other question