Variation In Acceleration Due To Gravity

VARIATION IN ACCELERATION DUE TO GRAVITY

(a) Variation of g with altitude

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

for h <<R,

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

At a height of 100 km from the surface, the acceleration due to gravity decreases by a fraction of $\frac{2h}{R}=\frac{2\times 100}{6400}\cong$ 3%

(b) Variation of g with depth

g¢ = $g\left(1-\frac{x}{R}\right)$ ( since R = R_S + x)

Therefore, at a depth of 10 km below the earth's surface, the fractional decrease in the acceleration due to gravity is

approximately, $\frac{10}{6400}\cong$ 0.16%

(c) Variation with latitude

Therefore, g'= g – $\omega$² R

or, g'= $g\left(1-\frac{{\omega }^{2}R}{g}{\cos }^2\theta \right)$ …(1)

Now, $\omega$= 2$\pi$/86400 s^-1, R$\cong$ 6400 km, g = 9.8 m/s²

So, $\omega$² R/g $\cong$ 0.34%, which is indeed a very small effect

At poles $\theta$ = 90° $\Rightarrow$ g' = g

$\Rightarrow$ the rotation of the earth has no effect on the gravity at poles.

At equator:

$\theta$ = 0°

$\Rightarrow$ $g^{\prime }=g\left(1-\frac{{\omega }^{2}R}{g}\right)$

(d) Effect of the surface of Earth

The equatorial is about 21 km longer than its polar radius. We know, $g=\frac{G{M}_e}{R_e^2}Hence{g}_{pole}>g_{equator}$. The weight of the body increase as the body taken from the equator to the pole.

Illustration 1:

Determine the speed with which the earth has to rotate on its axis so that a person on the equator would weigh 20% as much as present. Take the equatorial radius as 6400 km.

Solution: The apparent weight of a person at a place on the surface of the earth due to rotation of the earth is given by

W' = W - mR$\omega$²cos²$\lambda$ (where $\lambda$ is the latitude of the place)

For equator $\lambda$ = 0

So, W' = W - mR$\omega$²

Since W = mg, W' = 20% of mg = (20/100)mg =$\frac{1}{5}$ mg

$\frac{1}{5}$mg = mg - mR$\omega$²

$\Rightarrow$ mR$\omega$² = $\frac{4}{5}mg$

$\omega$ = $\sqrt{\left(\frac{4}{5}\frac{g}{R}\right)}$, g = 9.8 m/s², R = 6400 × 10³ m

$\omega$ = $\sqrt{\left(\frac{4}{5}\times \frac{9.8}{6400\times {10}^3}\right)}$ radian/sec. = 1.106 × 10^-3 rad/sec