Applications of Linear Approximation In this unit we’re trying to learn about applications of the derivative to real problems. Here is one such example that involves math as well as physics. Example 4: Planet Quirk satellite (with velocity v) me Figure 1: Illustration of Example 4: a satellite with velocity v speeding past “me” on planet Quirk. Let’s say I am on Planet Quirk, and that a satellite is whizzing overhead with a velocity v. The satellite has a clock on it that reports a time T . I have a watch that reports a time Tm . We want to calculate the time dilation (a concept from special relativity) that describes the diﬀerence between T and Tm . We borrow the following equation from special relativity: T Tm = � 1− v2 c2 Again, Tm is the time I measure on my wristwatch, and T is the time measured on board the satellite. How diﬀerent are Tm and T ? To avoid dividing by a square root, we’ll once again use linear approximation: � � T T 1 � ≈T 1+ u , =√ 2 2 1−u 1 − vc2 2 where u = vc2 . (If you’re wondering why we got (1 + 12 u) and not (1 − 12 u), try applying the change of variables v = −u before approximating.) And so, we ﬁnd that: � � �� T 1 v2 � Tm = ≈T 1+ 2 2 c2 1 − vc2 How does this aﬀect you in the real world? GPS transmitters are mounted on satellites, the satellites are moving, and you might be moving too. According to special relativity, there will be a diﬀerence between the time on your GPS device and the time on the satellite. This time diﬀerence will aﬀect the GPS device’s estimate of your position. The engineers who set up the GPS satellite system knew this and had to decide if they needed to take this into account when designing the system. It turns out that GPS satellites move at about v = 4 kilometers per second (km/s) 2 and c = 3 × 105 km/sec, and so vc2 ≈ 10−10 and Tm ≈ T (1.00000000005). There’s hardly any diﬀerence between the times measured on the ground and in the satellite. 2 Because vc2 is very close to zero, our linear approximation should be quite close to the actual value of Tm . Another good reason for using linear approxi mation here is that if the answer is “the diﬀerence is too small to matter”, the person doing the calculation has no use for a more precise answer which may be more diﬃcult to calculate. Nevertheless, engineers used this very approximation (along with several other such approximations) to calibrate the radio transmitters on GPS satellites. The satellites transmit at a slightly oﬀset frequency.