Variation of g with altitude, depth, and latitude on Earth

At the surface: $g = \frac{GM}{R^2}$

At height $h$ above surface, the distance from Earth’s centre becomes $(R + h)$:

For small $h$ (where $h \ll R$), we use the binomial approximation:

Key point: $g$ decreases with altitude. At $h = R$, gravity drops to $g/4$.

At depth $d$, only the mass of the sphere of radius $(R - d)$ contributes to gravity (by the shell theorem, the outer shell exerts zero net force).

Mass of inner sphere: $M' = \frac{4}{3}\pi(R-d)^3\rho$

Since $g = \frac{4}{3}\pi G\rho R$, we get:

Key point: $g$ decreases linearly with depth. At the centre ($d = R$), $g = 0$.

Earth rotates, so a body on the surface experiences a centrifugal effect. At latitude $\lambda$, the effective gravity is:

• At the equator ($\lambda = 0°$): $g_{eq} = g - R\omega^2$ — minimum value
• At the poles ($\lambda = 90°$): $g_{pole} = g$ — maximum value

The difference: $g_{pole} - g_{eq} = R\omega^2 \approx 0.034$ m/s²

Additionally, Earth is not a perfect sphere — it’s an oblate spheroid (flattened at poles). The polar radius is less than the equatorial radius, which further increases $g$ at the poles.