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

Key Concepts

Newton’s law of universal gravitation tells us that any two point masses attract each other with a force:

\$F = G\frac{m1m2}{r^2}\$

where \$G\$ is the gravitational constant, \$m1\$ and \$m2\$ are the masses, and \$r\$ is the distance between their centres.

Near the Earth’s surface the force on a mass \$m\$ is usually written as \$F = mg\$, where \$g\$ is the acceleration due to gravity.

Why is \$g\$ Approximately Constant?

For a point mass at the Earth’s centre, the gravitational acceleration is:

\$g = \frac{GM{\text{Earth}}}{R{\text{Earth}}^2}\$

When we move a small height \$h\$ above the surface, the distance to the centre becomes \$R_{\text{Earth}} + h\$. The new acceleration is:

\$g(h) = \frac{GM{\text{Earth}}}{(R{\text{Earth}} + h)^2}\$

Because \$h \ll R_{\text{Earth}}\$ (even a 10 m climb is tiny compared to 6 300 km), we can expand using a binomial approximation:

\$g(h) \approx g \left(1 - \frac{2h}{R_{\text{Earth}}}\right)\$

So the change in \$g\$ is tiny: for \$h = 10\$ m, \$\Delta g \approx 0.0003\,\text{m/s}^2\$, far below the precision of most experiments.

Examples & Analogies

• 🏔️ Mountain Climb Analogy: Imagine walking up a 100 m hill. The change in gravity is like a whisper compared to the roar of the whole Earth.
• 📏 Scale Example: A 1 kg mass weighs 9.81 N at sea level. At 100 m up, it weighs 9.809 N – a difference of only 0.01 N.
• 🌍 Planetary Scale: If you were on the Moon, \$g\$ would be 1.6 m/s², but even a 10 m jump changes it by \$<0.0001\$ m/s².

Exam Tips

When tackling questions about \$g\$ near the surface:

1. Show the formula: \$g = \frac{GM}{R^2}\$.
2. Explain why \$h \ll R\$ allows the approximation \$g(h) \approx g\$.
3. Use the binomial expansion if the question asks for the change in \$g\$ over a small height.
4. Remember to keep units consistent (m, kg, s).
5. Include a quick sanity check: \$\Delta g\$ should be \$<0.01\$ m/s² for typical heights.