To understand why the acceleration due to gravity, $g$, can be treated as approximately constant for small changes in height near the Earth’s surface.
The force between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by
$$ F = G\frac{m_1 m_2}{r^{2}} $$where $G = 6.674 \times 10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}$ is the universal gravitational constant.
Taking $m_1 = M_{\oplus}$ (mass of the Earth) and $m_2 = m$ (mass of a small object), the gravitational force on the object is
$$ F = G\frac{M_{\oplus} m}{r^{2}} = m\,g(r) $$Hence the local gravitational acceleration is
$$ g(r) = G\frac{M_{\oplus}}{r^{2}}. $$Let $R_{\oplus}$ be the mean radius of the Earth ($\approx 6.37\times10^{6}\,\text{m}$). For a point at height $h$ above the surface, the distance from the Earth’s centre is
$$ r = R_{\oplus}+h. $$Substituting into $g(r)$ gives
$$ g(h) = G\frac{M_{\oplus}}{(R_{\oplus}+h)^{2}}. $$When $h \ll R_{\oplus}$ we can expand $g(h)$ using a first‑order Taylor series about $h=0$:
$$ g(h) \approx g(0)\left[1 - 2\frac{h}{R_{\oplus}}\right], $$where $g(0)=G M_{\oplus}/R_{\oplus}^{2}\approx9.81\ \text{m s}^{-2}$.
| Height $h$ (m) | Exact $g(h)$ (m s⁻²) | Approx. $g(0)[1-2h/R_{\oplus}]$ (m s⁻²) | Relative error (%) |
|---|---|---|---|
| 0 | 9.80665 | 9.80665 | 0 |
| 100 | 9.80399 | 9.80399 | 0.00 |
| 1 000 | 9.77900 | 9.77902 | 0.00 |
| 10 000 | 9.72600 | 9.72612 | 0.0012 |
| 100 000 | 9.51500 | 9.51530 | 0.0031 |
The gravitational acceleration decreases with the square of the distance from the Earth’s centre. Because the Earth’s radius is very large compared with everyday height changes, the fractional change in $g$ is extremely small. This justifies treating $g$ as a constant ($\approx9.81\ \text{m s}^{-2}$) for most problems confined to the near‑surface region.