Problem — Gravitational redshift on Everest

Problem

How much does a clock at sea level slow down compared to one at the top of Everest ($H=8848$ m)? Express the difference in nanoseconds per day and state which of the two clocks runs faster.

Solution

Gravitational redshift formula. In a uniform gravitational field with acceleration $g$, two clocks separated in altitude by $H$ keep different rates. The fractional ratio between the rates is $\frac{\Delta t}{t} = \frac{g\,H}{c^2}$

Numerical substitution. With $g = 9{,}81$ m/s², $H = 8848$ m and $c^2 = 9\cdot 10^{16}$ m²/s²: $\frac{\Delta t}{t} = \frac{9{,}81\cdot 8848}{9\cdot 10^{16}} = \frac{8{,}68\cdot 10^{4}}{9\cdot 10^{16}} \approx 9{,}64\cdot 10^{-13}$

Accumulated offset in one day. One day lasts $t = 86\,400$ s, so $\Delta t = 9{,}64\cdot 10^{-13}\cdot 86\,400\ \text{s} \approx 8{,}3\cdot 10^{-8}\ \text{s}$

$\ev{\Delta t \approx 83\ \text{ns/day}}$

Which one runs faster. The further one gets from the mass (less negative gravitational potential, greater altitude), the faster time passes. The clock at the top of Everest runs faster than the one at sea level, gaining about $83$ nanoseconds per day. This is the same effect — on a much larger scale — that must be corrected for in GPS satellites.

Links

Topics: Special relativity Concepts: Time dilation Skills: Symbolic setup