Prove that the ratio of the rate of change of g at a height equal to the earth's radius from the surface of the earth of the value of g at the surface of the earth is equal to (-1)/(4R_e)

Prove that the ratio of the rate of change of g at a height equal to t

1.	Prove that the ratio of the rate of change of g at a height equal to the earth's radius from the surface of the earth of the value of g at the surface of the earth is equal to (-1)/(4R_e)
Answer» <html><body><p></p>Solution :`implies` Gravitational acceleration at .r. `(r <a href="https://interviewquestions.tuteehub.com/tag/ge-1003146" style="font-weight:bold;" target="_blank" title="Click to know more about GE">GE</a> R_e)`from the <a href="https://interviewquestions.tuteehub.com/tag/centre-912170" style="font-weight:bold;" target="_blank" title="Click to know more about CENTRE">CENTRE</a> of <a href="https://interviewquestions.tuteehub.com/tag/earth-13129" style="font-weight:bold;" target="_blank" title="Click to know more about EARTH">EARTH</a> . <br/> `g(r) = (GM_e)/r^2` <br/> Differentiating w.r.t. to <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> r, <br/> `(dg(r))/(<a href="https://interviewquestions.tuteehub.com/tag/dr-959219" style="font-weight:bold;" target="_blank" title="Click to know more about DR">DR</a>) = (-2GM_e)/r^3` <br/> `= -(2GM_e)/((8R_e)R_(e)^2)[ :. (GM_e)/R_e^2="ge"]` <br/> `:. [(dg(r))/(dr)]_(2R_e) =(-"ge")/(4R_e)` <br/> =`[(dg(r))/(dr)]_(2Re)/("ge")=-1/(4R_e)` .</body></html>