Simple Physics of Malaysia Airlines MH370
Figure 1 shows variables in spherical and geodetic coordinates. They are different in defining the angle along the longitude line. In the spherical coordinate, the northern pole, the southern pole, and the equator are assigned to \( \theta = \) = 0o, 180o, and 90o, respectively. To the contrary, in the geodetic coordinate, the northern pole, the southern pole, and the equator are assigned to \( \lambda \) = 90o, latitude, \( \lambda \) = -90o latitude and \( \lambda \) = 0o latitude. These two angles are related by \( \theta = \pi / 2 – \lambda \). The unit vectors in the spherical coordinates are
\( \hat{r} = \hat{i} \sin \theta \cos \varphi +\hat{j} \sin \theta \sin \varphi + \hat{k} \cos \theta \)\( \hat{\theta} = \hat{i} \sin \theta \cos \varphi +\hat{j} \sin \theta \sin \varphi + \hat{k} \sin \theta \)\( \hat{\varphi} = -\hat{i} \sin \varphi +\hat{j} \cos \varphi \)
Expressed in latitude angle, the first and the second equations can be rewritten as
\( \hat{r} = \hat{i} \cos \lambda \cos \varphi +\hat{j} \cos \lambda \sin \varphi + \hat{k} \sin \lambda \)\( \hat{\theta} =\hat{i} \sin \lambda \cos \varphi +\hat{j} \sin \lambda \sin \varphi + \hat{k} \cos \lambda \)Assume the aircraft is moving at cruising altitude (constant altitude), usually at about 35,000-40,000 feet. The aircraft has only \( \hat{\theta} \) and \( \hat{\varphi}\) components of velocity, and its velocity relative to the ground can be written as
\( \vec{v}_{ag} = v_\theta \hat{\theta}+v_\varphi \hat{\varphi}\)\( =(v_\theta \sin \lambda \cos \varphi – v_\varphi \sin \varphi) \hat{i}+ (v_\theta \sin \lambda \sin \varphi – v_\varphi \cos \varphi ) \hat{j} + (v_\theta \cos \lambda) \hat {k} \)Because the Earth rotates once a day, the Earth’s surface (ground) is moving toward \( \hat{\varphi} \) direction relative to the fixed frame of reference (fixed relative to distant stars). This velocity depends on the latitude angle as
\( \vec{v}_g = \omega R \sin \theta \hat{\varphi} = \omega R \cos \lambda (-\hat{i} \sin \varphi +\hat{j} \cos \varphi) \)The velocity of the aircraft relative to the fixed frame of reference is obtained from the last two equations as
\( \vec{v}_a=\vec{v}_{ag}+\vec{v}_g = (v_\theta \sin \lambda \cos \varphi – v_\varphi \sin \varphi – \omega R \cos \lambda \sin \varphi) \hat{i} \)\(+ (v_\theta \sin \lambda \sin \varphi + v_\varphi \cos \varphi + \omega R \cos \lambda \cos \varphi) \hat{j}+ v_theta \cos \lambda \hat{k} \)To determine the Doppler effect on the electromagnetic wave emitted by the aircraft and detected by the satellite, we need to find the velocity of the aircraft relative to the satellite. For this purpose firstly we derive the velocity of the satellite relative to the fixed frame of reference as
\( \vec{V}_s = V_{s \theta} \hat{\theta}’+ V_{s \varphi} \hat{\varphi}’ \)with
\( \hat{\theta}’ = \hat{i} \cos {\theta}’ \cos {\varphi}’+ \hat{j} \cos {\theta}’ \sin {\varphi}’+ \hat{k} \sin {\theta}’ \)\( = \hat{i} \sin {\lambda}’ \cos {\varphi}’+ \hat{j} \sin {\lambda}’ \sin {\varphi}’+ \hat{k} \cos {\theta}’ \)and
\( \hat{\varphi}’ = -\hat{i} \sin {\varphi}’+ \hat{j} \cos {\varphi}’ \)Substifuting the corresponding unit vectors we obtain
\( \vec{V}_s = (V_{s \theta} \sin {\lambda}’ \cos {\varphi}’ – V_{s \phi} \sin {\varphi}’) \hat{i} + (V_{s \theta} \sin {\lambda}’ \sin {\varphi}’- V_{s \phi} \cos {\varphi}’) \hat{j} + V_{s \theta} \cos {\theta}’ \hat{k} \)Finally we obtain the velocity of the aircraft relative to satellite as
\( \vec{v}_{as} = \vec{v}_a – \vec{V}_s \)If \(r \) is the radius of satellite orbit, one has
\( {{G M_E m_s} \over r^2} = {{m V_s^2} \over r} \)or
\(V_s = \sqrt{{G M_E} \over r} \)The next query is how we understand the aircraft is approaching or leaving the satellite. This is important because it affects the frequency detected by the satellite: decreases or increases relative to the original frequency (emitted by the aircraft). For this purpose, we do the following procedures. The vector position of the aircraft and satellite are
\(\vec{r}_a = (R+h) \hat{r}(\theta, \varphi) \approx R \hat{r}(\theta, \varphi) = R (\sin \theta \cos \varphi + \hat{j} \sin \theta \sin \varphi + \hat{k} \cos \theta) \)\(= R (\cos \lambda \cos \varphi + \hat{j} \cos \lambda \sin \varphi + \hat{k} \sin \lambda) \)and
\( \vec{r}_s = r (\cos \lambda’ \cos \varphi’ + \hat{j} \cos \lambda’ \sin \varphi’ + \hat{k} \sin \lambda’ ) \)The position of aircraft relative to satellite is
\( \vec{r}_{as} = \vec{r}_a – \vec{r}_s \)We can conclude whether the aircraft approaching or leaving the satellite based on the scalar product of \( \vec{r}_{as} \) and \( \vec{v}_{as} \). If \( \vec{r}_{as} . \vec{v}_{as} < 0 \) the aircraft is approaching the satellite and if \( \vec{r}_{as} . \vec{v}_{as} > 0 \) the leaving the satellite.
The aircraft started to move at the latitude and longitude angles of \( (\lambda_1,\varphi_1) \). Having traveled by a distance \( d \) the latitude and longitude angles became \( (\lambda_2,\varphi_2) \). The distance traveled by the aircraft satisfies
\(d = (R + h) \Delta \sigma \)with
\( \Delta \sigma = \cos^{-1} (\sin \lambda_1 \sin \lambda_2 + \cos \lambda_1 \cos \lambda_2 \cos (\varphi_2 – \varphi_1)) \)From this equation we obtain
\( \varphi_2 = \varphi_1 \pm \cos^{-1} {{\cos (d/(R+h)) – \sin \lambda_1 \sin \lambda 2} \over {\cos \lambda_1 \cos \lambda_2}} \)Figure 2 shows the relationship between \( \varphi \) and \( \lambda_2 \).
The positive sign is used when the aircraft moves to the east direction (longitude angle increases) while the minus sign is used when the aircraft moves to the west (longitude angle decreases). By assuming the aircraft traveled at a constant speed, we can write
\( d = v_a t \)
To what direction is the aircraft moving? We assumed the direction of aircraft velocity is parallel to the direction of the spherical arc connecting the initial and the current positions of the aircraft as illustrated in Fig. 3. This direction can be determined by the following procedure. The difference in vector positions of the aircraft at
\( \vec{r}_{a,21} = \vec{r}_a (\lambda_2,\varphi_2) – \vec{r}_a (\lambda_1,\varphi_1) \)The component of this vector parallel to the Earth surface at position of \( (\lambda_2,\varphi_2) \) is
\(\vec{r}_{a,21,//}=r_{a,21,//,\theta} \hat{\theta} (\lambda_2, \varphi_2) + r_{a,21,//,\varphi} \hat{\varphi} (\lambda_2, \varphi_2) \)with
\(r_{a,21,//,\theta} = \vec{r}_{a,21,//}. \hat{\theta} (\lambda_2, \varphi_2) \)and
\(r_{a,21,//,\varphi} = \vec{r}_{a,21,//}. \hat{\varphi} (\lambda_2, \varphi_2) \)The direction of aircraft velocity is assumed to be the same as the vector \( \vec{r}_{a,21,//} \). Therefore, we can write the component of velocity of the aircraft is
\( v_\theta = v_a { r_{a,21,//,\theta} \over r_{a,21,//}} \)and
\( v_\varphi = v_a { r_{a,21,//,\varphi} \over r_{a,21,//}} \)The aircraft emitted a ping wave with a wavelength \( \lambda_0 \). Due to Doppler’s effect on the electromagnetic wave, the wavelength detected by the satellite is
\( \lambda = \left ({1 \pm {v_{as} \over c}} \right ) \lambda_0 \)The change in wavelength due to Doppler effect is
\( \Delta \lambda = \pm {v_{as} \over c} \lambda_0 \)RESULTS
The maximum and minimum latitude angles reached by the aircraft is satisfied if it traveled due north or south. The maximum latitude angle (the aircraft traveled due north) satisfies
\(\lambda_{maks} – \lambda_1 = {d \over {R+h}} \)or
\(\lambda_{maks} = \lambda_1 + {d \over {R+h}} \)and the minimum latitude (the aircraft traveled due south) satisfies
\( \lambda_{min} = \lambda_1 – {d \over {R+h}} \)The allowed latitude angle,
\( \lambda_{min}< \lambda_2 < \lambda_{maks} \)Application to MH370
Let us apply the above equation to MH370 aircraft that have missed at March 8, 2014. Final position of the aircraft detected by radar is around 200 km west of Penang Island, Malaysia. Let us used the reported data. The aircraft was assumed to start travelled north or south at a location around Penang Island, Malaysia. The coordinate of this position is 5.4o north and 100.24o east. This means that \( \lambda_1 = 5,4^o = 0,094 \) rad. The distance of 200 km corresponds to an angle of around \( 200/(6.400 \sin (\pi /2 – \lambda_1) = 0,0314 \) rad = \( 1,8^o \). Therefore, the initial longitude of the aircraft is \( \varphi_1 = 100,24^o – 1,8^o = 98,44^o = 1,718 \) rad.
After leaving this point, we assumed the aircraft traveled at normal cruising speed of \(v_a = 471 \) knot = 872 km/h = 242 m/s. We also assumed the aircraft travelled at normal cruising altitude of around 35,000 feet or h = 11.000 [/latex] m. The Earth radius is \( R = 6,400 \) km. The angular velocity of Earth rotation is \( \omega = 2 \pi /T = 7,27 \times 10^{-5} \) rad/s.
The satellite detecting the ping signal of the aircraft is the Inmarsat satellite. The radius of this geostationary satellite satisfies \( r = 42.000 \) km and its orbiting speed is \( V_a = 3,1 \) km/s. The longitude of the Inmarsat-3 F1 satellite is 64.5o east. Therefore we have \( \varphi’ = 64,5^o = 1,125 \) rad.
Figure 4 is the calculated dependence of the Doppler shift shift on \( \lambda_2 \) for MH370.
The aircraft was last detected by military radar at around Penang Island at 02.15 Malaysian time. The ping of the aircraft was received by Inmarsat-3 F1 satellite at 08.11 Malaysian time. Therefore we have \( t = 08.11 – 02.15 \) = 5 hours 56 minutes = 21,360 s. The distance travelled by the aircraft is
\( d = v_a t = 242 \times 21.360 = 5,7 \times 10^6 \) m
dan kita dapatkan
\( \lambda_{min} = \lambda_1 – {d \over {R+h}} = -0,714 \) rad
\( \lambda_{maks} = \lambda_1 + {d \over {R+h}} = 0,902 \) rad
Based on the measurement \( \Delta \lambda \) we can calculate \( \lambda_2 \) so that we can determine to which direction the aircraft has moved.
Feature image source: en.wikipedia.org
