Variation in Acceleration due to Gravity (g)

Variation of g with Altitude:

At the surface of earth

$\displaystyle g = \frac{GM}{R^2}$

At an altitude h above the earth’s surface,

R’ = the distance from the centre of the earth = R + h

$\displaystyle g’ = \frac{GM}{(R+h)^2}$

$\displaystyle \frac{g’}{g} = \frac{R^2}{(R+h)^2}$

$\displaystyle \frac{g’}{g} = (1+\frac{h}{R})^{-2}$

This implies that for h <<R,

$\displaystyle \frac{g’}{g} = (1 -\frac{2h}{R})$

$\displaystyle g’ = g(1-\frac{2h}{R})$

Variation of g with Depth

One would have thought, from the above discussion, that g increases with depth; however, that is not the case.

At a depth ‘ d ‘ below the earth’s surface, the only change is not just due to the change in the distance from the centre of the earth but also due to a change in the ‘effective’ mass of the earth that contributes to g at this point.

Only that portion of the earth that is enclosed by a sphere of radius RS( = R – d) centred at the earth’s centre is effective.

If we assume the earth to be a sphere having uniform density, then,

$\displaystyle g’ = \frac{GM_{eff}}{R_S^2}$

$\displaystyle g’ = \frac{G}{R_S^2}\times \frac{M}{(4/3) \times \pi R^3} \times \frac{4}{3}\pi R_S^3$

$\displaystyle g’ = \frac{GM}{R^3} \times R_S = \frac{GM}{R^2}\times\frac{R_s}{R}$

$\displaystyle g’ = g \times \frac{R_S}{R} = g(\frac{R-d}{R})$

$\displaystyle g’ = g (1 – \frac{d}{R})$

Therefore, at a depth of 10 km below the earth’s surface, the fractional decease in the acceleration due to gravity is
approximately, 10/6400 ≅ 0.16%

In reality, the change will be much less as the density of the earth is not constant as we have assumed and the crust has a much lower density as compared to the core.

Variation of g due to Shape of the Earth :

$g = \frac{G M}{R^2}$

$g \propto \frac{1}{R^2}$

The earth is not perfectly spherical. The polar radius of the earth is 21 km smaller than its equatorial radius.

The acceleration due to gravity increases when one moves from equator to the pole.

Variation of g with Latitude

The effective value of the acceleration due to gravity changes with latitude owing to the rotation of the earth.

Referring to the figure , at a latitude θ (Point Q), the effective weight,

mg’ = mg − mω2 r cosθ …….(1)

where mω2 r is the “centrifugal force” and we take its component in the vertical direction and mg is the weight in the absence of rotation.

Now, r = radius of the circle of rotation of Q.

= QE = R cos θ (from ΔCQE),

Therefore, g’ = g − ω2 R cos2 θ

$\displaystyle g’ = g (1 – \frac{\omega^2 R}{g}cos^2\theta )$

$\displaystyle \omega = \frac{2\pi}{86400}$ , R = 6371 km

So, ω2 R/g ≅ 0.34%, which is indeed a very small effect

Illustration : If earth stops spinning about its own axis, what will be the change in acceleration due to gravity on its equator? The radius of earth is 6.4 × 106 m and its angular speed is 7.27 × 10-5 rad/s

Solution: Effective acceleration due to gravity is given by

g’ = g − R ω2 cos2 θ

Hence change in acceleration due to gravity is given as

Δg = g − g’ = R ω2 Cos2 θ

At equator θ = 0

Hence, g − g’ = R ω2

= 6.4 ×106 × (7.27×10-5)2

= 0.0338 m/s2

g − g’ = 3.38 cm/s2

This is the decrease in gravitational acceleration due to the rotation of earth.

Illustration : For given values of angular speed and acceleration due to gravity, find the position (latitude) where acceleration due to gravity is
(a) maximum and (b) minimum.

Solution:  (a) Variation of g with latitude is given by
g’ = g – R ω2 cos2θ

g’ will be maximum when cosθ is minimum i.e. = 0, or, when θ = 90°

∴ gmax = g , at the poles.

(b) g’ will be minimum when cosθ is maximum

⇒ i.e. θ = 0

gmin = g – R ω2 , at the equator.