An aeroplane flies from Rome (41° N, 12° E) towards Port Clarence (65° N, 168° W). The altitude of its path is 15 km above the sea level. Calculate the shortest distance that can b... An aeroplane flies from Rome (41° N, 12° E) towards Port Clarence (65° N, 168° W). The altitude of its path is 15 km above the sea level. Calculate the shortest distance that can be covered by the aeroplane for the journey from Rome to Port Clarence.
Understand the Problem
The question asks us to calculate the shortest distance an aeroplane can cover when flying from Rome (41° N, 12° E) to Port Clarence (65° N, 168° W), given that the aeroplane flies at an altitude of 15 km above sea level. This involves calculating the great-circle distance between the two locations, taking into account the Earth's curvature and the altitude of the flight.
Answer
$8258.6 \text{ km}$
Answer for screen readers
$8258.6 \text{ km}$
Steps to Solve
- Convert coordinates to radians
Convert the latitudes and longitudes of Rome and Port Clarence from degrees to radians. $$ \text{Latitude of Rome} = 41^\circ = 41 \times \frac{\pi}{180} \approx 0.7156 \text{ radians} $$ $$ \text{Longitude of Rome} = 12^\circ = 12 \times \frac{\pi}{180} \approx 0.2094 \text{ radians} $$ $$ \text{Latitude of Port Clarence} = 65^\circ = 65 \times \frac{\pi}{180} \approx 1.1345 \text{ radians} $$ Since Port Clarence is at $168^\circ$ W, we convert it to East by subtracting from $360^\circ$: $360^\circ - 168^\circ = 192^\circ$ $$ \text{Longitude of Port Clarence} = 192^\circ = 192 \times \frac{\pi}{180} \approx 3.3510 \text{ radians} $$
- Calculate the great-circle distance using the Haversine formula
The Haversine formula calculates the great-circle distance between two points on a sphere given their latitudes and longitudes. $$ d = 2r \arcsin\left(\sqrt{\sin^2\left(\frac{\phi_2 - \phi_1}{2}\right) + \cos(\phi_1)\cos(\phi_2)\sin^2\left(\frac{\lambda_2 - \lambda_1}{2}\right)}\right) $$ where: $ \phi_1 $ is the latitude of Rome $ \phi_2 $ is the latitude of Port Clarence $ \lambda_1 $ is the longitude of Rome $ \lambda_2 $ is the longitude of Port Clarence $ r $ is the radius of the Earth plus the altitude of the airplane
- Plug in the known values
The Earth's radius is approximately 6371 km. The airplane's altitude is 15 km, so we use $ r = 6371 + 15 = 6386 $ km. $$ d = 2 \times 6386 \times \arcsin\left(\sqrt{\sin^2\left(\frac{1.1345 - 0.7156}{2}\right) + \cos(0.7156)\cos(1.1345)\sin^2\left(\frac{3.3510 - 0.2094}{2}\right)}\right) $$
- Simplify the equation
$$ d = 12772 \times \arcsin\left(\sqrt{\sin^2(0.20945) + \cos(0.7156)\cos(1.1345)\sin^2(1.5708)}\right) $$ $$ d = 12772 \times \arcsin\left(\sqrt{(0.2079)^2 + (0.7547)(0.423)(1)^2}\right) $$ $$ d = 12772 \times \arcsin\left(\sqrt{0.0432 + 0.3192}\right) $$ $$ d = 12772 \times \arcsin\left(\sqrt{0.3624}\right) $$ $$ d = 12772 \times \arcsin(0.602) $$ $$ d = 12772 \times 0.6465 $$ $$ d \approx 8258.6 \text{ km} $$
$8258.6 \text{ km}$
More Information
The shortest distance between two points on a sphere is a segment of a great circle. Since the airplane is flying at an altitude of 15 km above sea level, we must consider this altitude when calculating the radius.
Tips
A common mistake is forgetting to add the altitude to the Earth's radius, which would result in an inaccurate calculation of the distance. Also, forgetting to convert the coordinates to radians before using the Haversine formula.
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