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# At what height, is the value of g half that on the surface of earth? (R = radius of the earth)A.) 0.414RB.) RC.) 2RD.) 3.5R  Hint: In this question, g is referred to as the acceleration of gravity. Its value is $9.8{\text{ }}m/{s^2}$ on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is $9.8{\text{ }}m/{s^2}$.

We know, $\dfrac{g}{{g'}} = \dfrac{{{r^2}}}{{{R^2}}}$, where g is gravitational force on earth, g' is gravitational force at a height.

Given g' = $\dfrac{g}{2}$

g’ = $\dfrac{{g{R^2}}}{{{r^2}}} = \dfrac{g}{2}$
$\Rightarrow {{\text{r}}^{\text{2}}}{\text{ = 2}}{{\text{R}}^{\text{2}}}$
$\Rightarrow {\text{r = R}}\sqrt {\text{2}} {\text{ = R + h}}$
$\Rightarrow {\text{h = R (}}\sqrt {\text{2}} {\text{ - 1)}}$
= R (1.414 – 1)
= 0.414R.

Hence the option A is the right answer.

Distance from the earth’s center of mass is the metric that truly matters. On or above the surface, gravity decreases as the square of this distance, so it would be half at the square root of twice Earth’s radius (assuming the center of mass coincides with the geometric center). If we use 6,371 km as Earth’s radius, the distance above the surface where gravitational acceleration is halved would be 2,639 km.

Note: Gravitational acceleration is an object's free fall motion under space — with no friction. It is the steady increase of momentum induced purely by the gravitational attraction factor. Both objects move in space at the same pace at specified GPS coordinates on the Earth's surface at a given height. This distinction is valid independent of the weights or concentrations of the objects. The gravitational force is fictitious. This is no tidal acceleration, so in fact so free fall the normal acceleration and hence the four-acceleration of objects is negligible. Because of experiencing an acceleration, artifacts on the bent spacetime fly in free fall in straight lines.
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