Question

(a) Show that the rate of change of the free-fall acceleration with vertical position near the Earth’s surface is

$\frac{\mathbf{dg}}{\mathbf{dr}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}{\mathbf{GM}}_{\mathbf{E}}}{{\mathbf{R}}_{\mathbf{E}}^{\mathbf{3}}}$

This rate of change with position is called a gradient.

(b) Assuming h is small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is

$\left|\Delta g\right|\mathbf{=}\frac{\mathbf{2}\mathbf{G}{\mathbf{M}}_{\mathbf{E}}\mathbf{h}}{{\mathbf{R}}_{\mathbf{E}}^{\mathbf{3}}}$

(c) Evaluate this difference for$\mathit{h}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathit{m}$ , a typical height for a two-story building.

Short answer

(a)$\frac{dg}{dr}=-\frac{2G{M}_E}{R_E^3}$

(b) $\left|\Delta g\right|=\frac{2G{M}_{E}h}{R_E^3}$

(c)$\left|∆g\right|=1.85\times {10}^{-5}m/s^2$

Step-by-step solution

Gravitational acceleration

Gravitational acceleration is directly proportional to the mass of the object and inversely proportional to the square of the distance between the object and earth.

Given

h=6.00 m

(a) Acceleration

The free fall that will be get produce by the earth will be,

$\begin{array}{rcl}g& =& \frac{G{M}_E}{r^2}\\ \frac{dg}{dr}& =& -\frac{2G{M}_E}{r^3}\\ & & \end{array}$

Near the earth surface the negative sign is indicating that g is decreasing with increasing height, so at the earth surface it will be,

$\frac{dg}{dr}=-\frac{2G{M}_E}{R_E^3}$

(b) Differences in free fall accelerations

With the height the difference in the acceleration will be,

$\begin{array}{rcl}\frac{\left|\Delta g\right|}{\Delta r}& =& \frac{2G{M}_E}{R_E^3}\\ \frac{\left|\Delta g\right|}{\Delta r}& =& \frac{\left|\Delta g\right|}{h}\\ \frac{\left|\Delta g\right|}{h}& =& \frac{2G{M}_E}{R_E^3}\\ \left|\Delta g\right|& =& \frac{2G{M}_{E}h}{R_E^3}\\ & & \end{array}$

(c) Evaluation

When ewe will put the given values we will be the evalution,

$\begin{array}{rcl}\left|∆g\right|& =& \frac{2\times 6.67\times {10}^{-11}N·m^2/kg\times 5.98\times {10}^{24}kg\times 6.00m}{{\left(6.37\times {10}^{6}m\right)}^3}\\ \left|∆g\right|& =& 1.85\times {10}^{-5}m/s^2\end{array}$