The Value of g on Earth (Cambridge International A-Level Physics)

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

$R \text{much} greater-than h$

• 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 × 10⁵ m) 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

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:

$g=\frac{GM}{r^{2}}$

• 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

$g = \frac{ G M}{( R + h )^{2}} \text{asymptotically} \text{equal} \text{to} \frac{G M}{R^{2}}$

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

$g=\frac{GM}{r^{2}}$

Step 2: Determine the value of r

$r=\mathrm{radius} \mathrm{of} \mathrm{Earth}+\mathrm{height} \mathrm{of} \mathrm{Mount} \mathrm{Everest}$

$r=6370+8.85=6378.85\mathrm{km}=6378.85\times10^{3}m$

Step 3: Substitute the known values to calculate

• At the Earth's surface:

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

• At the height of Mt Everest:

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

Step 4: Calculate the percentage decrease

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

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