The Value of g on Earth

Gravitational field strength g is approximately constant for small changes in height near the Earth’s surface (9.81 m s^-2)
This is because from the inverse square law relationship:

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

The value of g depends on the distance from the centre of Earth r
On the Earth’s surface, r is equal to the radius of the Earth = 6400 km
Since this is much larger than the distance between the surface of the earth and centre of mass of an object on it, the small changes in height near the Earth’s surface make very little difference to the value of g
If we take a position h above the Earth’s surface, where it is reasonable to assume h is much smaller than the radius of the Earth (h << R):

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

This means g remains approximately constant until a significant distance away from the Earth’s surface

The following worked example proves that g decreases by very little even on the highest point on Earth

The highest point above the Earth's surface is at the peak of Mount Everest.

Given that this is 8800 m above the Earth's surface, show that g decreases by 0.6%.

Mass of the Earth = 6.0 × 10²⁴ kg.

Radius of the Earth = 6400 km.

Answer:

Step 1: Gravitational field strength equation

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

Step 2: Determine the value of r

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

$r = 6400 + 8.800 = 6408.8 km$

Step 3: Substitute the known values to calculate

$g = \frac{(6.67 \times 10^{- 11}) \times (6.0 \times 10^{24})}{{(6408.8 \times 10^{3})}^{3}}$

$g = 9.75 m s^{- 2} (3 s . f .)$

Step 4: Calculate he percentage decrease

$% decrease = \frac{decrease}{original} \times 100$

$% decrease = \frac{9.81 - 9.75}{9.75} \times 100 = 0.6 %$

The Value of g on Earth (CIE A Level Physics)