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:

$F = G\frac{m_1m_2}{r^2}$

where $G$ is the gravitational constant, $m_1$ and $m_2$ 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.