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» Solution :`implies` Gravitational acceleration at .r. `(r GE R_e)`from the CENTRE of EARTH . `g(r) = (GM_e)/r^2` Differentiating w.r.t. to DISTANCE r, `(dg(r))/(DR) = (-2GM_e)/r^3` `= -(2GM_e)/((8R_e)R_(e)^2)[ :. (GM_e)/R_e^2="ge"]` `:. [(dg(r))/(dr)]_(2R_e) =(-"ge")/(4R_e)` =`[(dg(r))/(dr)]_(2Re)/("ge")=-1/(4R_e)` .