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 Earths 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

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

Orbital radius, r is roughly equal to the Earths radius for an object within the Earths atmosphere

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

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

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

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

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 0.7%.

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

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

$r = 6400 + 8.800 = 6408.8 km = 6408.8 \times 10^{3} m$

Step 3: Substitute the known values to calculate

$g = \frac{(6.67 \times 10^{- 11}) \times (6.0 \times 10^{24})}{{(6408.8 \times 10^{3})}^{2}}$

$g = 9.74 N {kg}^{- 1} (3 s . f .)$

Step 4: Calculate he percentage decrease

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

$% decrease = \frac{9.81 - 9.74}{9.74} \times 100 = 0.7 %$

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