| Date | November 2016 | Marks available | 1 | Reference code | 16N.3.HL.TZ0.9 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Suggest | Question number | 9 | Adapted from | N/A |
Question
The global positioning system (GPS) uses satellites that orbit the Earth. The satellites transmit information to Earth using accurately known time signals derived from atomic clocks on the satellites. The time signals need to be corrected due to the gravitational redshift that occurs because the satellites are at a height of 20 Mm above the surface of the Earth.
The gravitational field strength at 20 Mm above the surface of the Earth is about 0.6 N kg–1. Estimate the time correction per day needed to the time signals, due to the gravitational redshift.
Suggest, whether your answer to (a) underestimates or overestimates the correction required to the time signal.
Markscheme
\(\frac{{\Delta f}}{f} = \frac{{gh}}{{{c^2}}}\) so \(\Delta f = \frac{{0.6 \times 20000000}}{{{{\left( {3 \times {{10}^8}} \right)}^2}}} = 1.3 \times {10^{ - 10}}\)
\(\frac{{\Delta f}}{f} = \frac{{\Delta t}}{t}\)
1.3 × 10–10 × 24 × 3600 = 1.15×10–5 «s» «running fast»
Award [3 max] if for g 0.6 OR 9.8 OR average of 0.6 and 9.8 is used.
ALTERNATIVE 1
g is not constant through ∆h so value determined should be larger
Use ECF from (a)
Accept under or overestimate for SR argument.
ALTERNATIVE 2
the satellite clock is affected by time dilation due to special relativity/its motion
