# The tallest spot on Earth is Mt. Everest, which is 8857 m above sea level. If the radius of the Earth to sea level is 6369 km, how much does the magnitude of g change between sea level and the top of Mt. Everest?

Oct 11, 2016

$\text{Decrease in magnitude of g} \approx 0.0273 \frac{m}{s} ^ 2$

#### Explanation:

Let

$R \to \text{Radius of the Earth to sea level} = 6369 k m = 6369000 m$

$M \to \text{the mass of the Earth}$

$h \to \text{the height of the tallest spot of}$
$\text{ Mt Everest from sea level} = 8857 m$

$g \to \text{Acceleration due to gravity of the Earth}$
$\text{ to sea level} = 9.8 \frac{m}{s} ^ 2$

$g ' \to \text{Acceleration due to gravity to tallest}$
$\text{ "" spot on Earth}$

$G \to \text{Gravitational constant}$

$m \to \text{mass of a body}$

When the body of mass m is at sea level, we can write

$m g = G \frac{m M}{R} ^ 2. \ldots \ldots . \left(1\right)$

When the body of mass m is at the tallest spot on Everst, we can write

$m g ' = G \frac{m M}{R + h} ^ 2. \ldots . . \left(2\right)$

Dividing (2) by (1) we get

$\frac{g '}{g} = {\left(\frac{R}{R + h}\right)}^{2} = {\left(\frac{1}{1 + \frac{h}{R}}\right)}^{2}$

$= {\left(1 + \frac{h}{R}\right)}^{- 2} \approx 1 - \frac{2 h}{R}$
(Neglecting higher power terms of $\frac{h}{R}$ as $\frac{h}{R} \text{<<} 1$)

Now $g ' = g \left(1 - \frac{2 h}{R}\right)$

So change (decrease) in magnitude of g

$\Delta g = g - g ' = \frac{2 h g}{R} = \frac{2 \times 8857 \times 9.8}{6369000} \approx 0.0273 \frac{m}{s} ^ 2$

Oct 11, 2016

$\approx - .027 m {s}^{- 2}$

#### Explanation:

Newton's Law for Gravitation

$F = \frac{G M m}{{r}^{2}}$

And $g$ is computed at the earth's surface ${r}_{e}$ as follows:

$m {g}_{e} = \frac{G M m}{{r}_{e}^{2}}$

So ${g}_{e} = \frac{G M}{{r}_{e}^{2}}$

if we were to compute different $g$'s we would get

${g}_{e v e r e s t} - {g}_{s e a} = G M \left(\frac{1}{{r}_{e v e r e s t}^{2}} - \frac{1}{{r}_{s e a}^{2}}\right)$

$G M = 3.986005 \times {10}^{14} {m}^{3} {s}^{- 2}$

approx 3.986005 times 10^14 * ( 1/(6369000 + 8857)^2) - 1/(6369000^2))

$\approx - .027 m {s}^{- 2}$

Using differentials to double check:

${g}_{e} = \frac{G M}{{r}_{e}^{2}}$

$\implies \ln \left({g}_{e}\right) = \ln \left(\frac{G M}{{r}_{e}^{2}}\right) = \ln \left(G M\right) - 2 \ln \left({r}_{e}\right)$

$\frac{{\mathrm{dg}}_{e}}{{g}_{e}} = - 2 \frac{{\mathrm{dr}}_{e}}{{r}_{e}}$

${\mathrm{dg}}_{e} = - 2 \frac{{\mathrm{dr}}_{e}}{{r}_{e}} {g}_{e} = - 2 \cdot \frac{8857}{6369000} \cdot 9.81 = - 0.027 m {s}^{- 2}$