A LevelPhysicsCambridge (CIE)Revision Notes13. Gravitational FieldsGravitational Field of a Point MassThe Value of g on Earth

The Value of g on Earth (Cambridge (CIE) A Level Physics): Revision Note

Exam code: 9702

Written by: Leander Oates

Reviewed by: Caroline Carroll

Updated on 2 July 2026

The value of g on earth

Gravitational field strength g is approximately constant for relatively small changes in height near the Earth’s surface
- Close to the Earth's surface, within the Earth's atmosphere, g = 9.81 N kg⁻¹
g can be approximated as constant because the Earth's radius, R is much larger than the distance between the Earth's surface and the position of an object in the Earth's atmosphere, h

$R ≫ h$

Where small changes in height do not affect the total height of an object, h
- Consider an object orbiting at a height, h of 150 km (1.5 × 10⁵m) above the Earth's surface
- The radius of the Earth, R is 6400 km = 6.4 × 10⁶ m
- So the total orbital radius, r is (6.4 × 10⁶) + (1.5 × 10⁵) = 6.55 × 10⁶ m
- Hence 6.4 × 10⁶ ≅ 6.55 × 10⁶

Diagram of orbital height

6-1-2-worked-example-solution-cie-igcse-23-rn

Orbital radius, r is roughly equal to the Earth's radius for an object within the Earth's atmosphere

From the equation for Gravitational field strength due to a point mass:

$g = \frac{G M}{r^{2}}$

Gravitational field strength, g, and orbital radius, r, have an inverse square law relationship:

$g \propto \frac{1}{r^{2}}$

Therefore, small changes in r result in small changes in g

$g = \frac{G M}{{(R + h)}^{2}} ≃ \frac{G M}{R^{2}}$

Worked Example

The highest point above the Earth's surface is at the peak of Mount Everest, a distance of 8850 m above the Earth's surface.

Show that the value of g at the top of Everest is about 0.3% less than the value of g at the Earth's surface.

Mass of the Earth = 5.97 × 10²⁴ kg

Radius of the Earth = 6370 km

Answer:

Step 1: Gravitational field strength equation

$g = \frac{G M}{r^{2}}$

Step 2: Determine the value of r

$r = radius of Earth + height of Mount Everest$

$r = 6370 + 8.85 = 6378.85 km = 6378.85 \times 10^{3} m$

Step 3: Substitute the known values to calculate

At the Earth's surface:

$g_{0} = \frac{(6.67 \times 10^{- 11}) \times (5.97 \times 10^{24})}{{(6370 \times 10^{3})}^{2}} = 9.8134 N {kg}^{- 1}$

At the height of Mt Everest:

$g' = \frac{(6.67 \times 10^{- 11}) \times (5.97 \times 10^{24})}{{(6378.85 \times 10^{3})}^{2}} = 9.7862 N {kg}^{- 1}$

Step 4: Calculate the percentage decrease

$% g = \frac{g_{0} - g'}{g_{0}} \times 100 %$

$% g = \frac{9.8134 - 9.7862}{9.8134} \times 100 = 0.28 % \approx 0.3 %$

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