Gravity at different altitudes on Everest

Problem

Explain why the acceleration due to gravity $g$ is greater at sea level than at the top of Everest, and why the difference is negligible for everyday life. Estimate the ratio $g_\text{Everest}/g_0$.

Solution

The acceleration due to gravity at distance $r$ from the Earth’s centre is: $g(r) = \frac{G M_T}{r^2}$ At altitude $h$ above sea level the distance from the centre is $r = R_T + h$, so: $g(h) = \frac{G M_T}{(R_T+h)^2}$ As we climb, $r$ increases and $g$ decreases (the $1/r^2$ dependence): hence $g$ is greater at sea level than at the top of Everest. The ratio between the two accelerations is: $\frac{g_\text{Everest}}{g_0} = \left(\frac{R_T}{R_T+h}\right)^2$ With $R_T \approx 6371$ km and $h \approx 8848$ m $= 8{,}848$ km: $\frac{g_\text{Everest}}{g_0} = \left(\frac{6371}{6371+8{,}85}\right)^2 = \left(\frac{6371}{6379{,}85}\right)^2 \approx 0{,}997$ The altitude of the summit is barely $0{,}14\%$ of Earth’s radius, so $g$ drops by only about $0{,}3\%$: a completely imperceptible change in everyday life. $\ev{\dfrac{g_\text{Everest}}{g_0} \approx 0{,}997}$

Links

Topics: Gravitation Concepts: Law of universal gravitation · Weight force Skills: Fermi estimates