understand why g is approximately constant for small changes in height near the Earth’s surface

Gravitational Force Between Point Masses

Objective

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.

1. Newton’s Law of Universal Gravitation

The force between two point masses \$m1\$ and \$m2\$ separated by a distance \$r\$ is given by

\$\$

F = G\frac{m1 m2}{r^{2}}

\$\$

where \$G = 6.674 \times 10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}\$ is the universal gravitational constant.

2. Gravitational Acceleration at a Distance \$r\$ from the Earth’s Centre

Taking \$m1 = 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}}.

\$\$

3. Approximation Near the Earth’s Surface

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}}.

\$\$

4. Why \$g\$ Is Approximately Constant for Small \$h\$

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}\$.

• The correction term \$2h/R_{\oplus}\$ is dimensionless.
• For \$h = 100\ \text{m}\$, \$2h/R_{\oplus} \approx 3.1\times10^{-5}\$, i.e. a change of only 0.003 %.
• Thus, for heights up to a few kilometres, the variation in \$g\$ is negligible for most practical calculations.

5. Numerical Illustration

6. Practical Consequences

1. Projectile motion problems near the surface can safely use \$g = 9.81\ \text{m s}^{-2}\$.
2. Engineering calculations for buildings, bridges, and short‑range rockets typically ignore the \$h\$‑dependence of \$g\$.
3. For high‑altitude or orbital applications, the full expression \$g(h)=G M{\oplus}/(R{\oplus}+h)^{2}\$ must be employed.

7. Summary

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.