Question

If g is acceleration due to gravity at a height equal to radius of the earth, its corresponding value at the surface of the earth is

Answer 1

Answer:

g/4

Explanation:

g'=g/(1+h/Re)

g'=g/(1+Re/Re) as(h=Re)

g'=g/(1+1)

g'=g/(2)

g'=g/4

$g'=g/(1+h/Re) {}^{2} </p><p>g'=g/(1+Re/Re) {}^{2} as(h=Re)</p><p>g'=g/(1+1) {}^{2} </p><p>g'=g/(2) {}^{2} \\ g'=g/(2) {}^{2} </p><p>g'=g/4$