Chapter 14: Problem 3
Find the spherical distance between two points on a sphere whose radius is 10 units if the chord joining the points is 10 units.
Short Answer
Expert verified
The spherical distance is approximately 10.47 units.
Step by step solution
01
Understanding the Spherical Distance Formula
The spherical distance can be calculated when the chord length and the radius of the sphere are known. The relationship between the chord distance, the spherical distance, and the radius involves the equation \( c = 2r \sin\left(\frac{d}{2r}\right) \), where \( c \) is the chord length, \( r \) is the radius, and \( d \) is the spherical distance.
02
Substituting Known Values
We know that the radius \( r = 10 \) units and the chord \( c = 10 \) units. Substituting these into the equation gives: \( 10 = 2 \times 10 \times \sin\left(\frac{d}{20}\right) \). Simplifying this equation, we get \( 10 = 20 \sin\left(\frac{d}{20}\right) \).
03
Solving for the Sine Function
To isolate the sine term, divide both sides of the equation by 20: \( \sin\left(\frac{d}{20}\right) = \frac{10}{20} = 0.5 \). This implies that \( \frac{d}{20} = \sin^{-1}(0.5) \).
04
Calculating the Inverse Sine Value
The inverse sine value for 0.5 is \( 30^\circ \) or \( \frac{\pi}{6} \) radians. Therefore, \( \frac{d}{20} = \frac{\pi}{6} \).
05
Calculating the Spherical Distance
Multiplying both sides by 20 to solve for \( d \): \( d = 20 \times \frac{\pi}{6} = \frac{20\pi}{6} = \frac{10\pi}{3} \). Therefore, the spherical distance \( d \) is \( \frac{10\pi}{3} \approx 10.47 \) units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Spherical Geometry
Spherical geometry is a type of non-Euclidean geometry that takes place on the surface of a sphere. Unlike the flat surfaces you might be used to, such as a piece of paper, spherical geometry has some unique properties. For instance, in spherical geometry, the concepts of straight lines are quite different. Here, the shortest path between two points is not a straight line but a segment of a great circle. A great circle is the largest possible circle that can be drawn on a sphere and is defined by the intersection of the sphere with a plane that passes through the center of the sphere.
Spherical geometry finds numerous applications in areas such as astronomy, navigation, and engineering. It helps describe curved surfaces and is essential for understanding planetary movements and global navigation systems. It's important to grasp these concepts fully to solve problems involving distances on spheres.
Spherical geometry finds numerous applications in areas such as astronomy, navigation, and engineering. It helps describe curved surfaces and is essential for understanding planetary movements and global navigation systems. It's important to grasp these concepts fully to solve problems involving distances on spheres.
The Concept of Chord Length
A chord of a circle or a sphere is a straight line segment whose endpoints lie on the circle or sphere, respectively. When we talk about chord length in the context of spherical geometry, we're referring to the distance between these two points along the interior of the sphere.
In spherical geometry, the relationship between the chord length, the radius of the sphere, and the spherical distance is crucial for many calculations.
In spherical geometry, the relationship between the chord length, the radius of the sphere, and the spherical distance is crucial for many calculations.
- The chord is shorter than any great circle arc connecting the same two points.
- The familiar chord equation relates the chord length and spherical distance through the sine function: \( c = 2r \sin\left(\frac{d}{2r}\right) \).
Knowing Inverse Trigonometric Functions
Inverse trigonometric functions are powerful tools in mathematics, permitting calculations to go 'backward' from an angle's sine, cosine, or tangent values to the angle itself. These are particularly essential when dealing with spherical problems and angles in radians or degrees.
When the value of a trigonometric function like sine, cosine, or tangent is known, these inverse functions help determine the corresponding angle. For example, when solving for the spherical distance, where we encountered \( \sin\left(\frac{d}{20}\right) = 0.5 \), we used the inverse sine (or arcsin) function to find the angle, which was \( 30^\circ \) or \( \frac{\pi}{6} \) radians.
When the value of a trigonometric function like sine, cosine, or tangent is known, these inverse functions help determine the corresponding angle. For example, when solving for the spherical distance, where we encountered \( \sin\left(\frac{d}{20}\right) = 0.5 \), we used the inverse sine (or arcsin) function to find the angle, which was \( 30^\circ \) or \( \frac{\pi}{6} \) radians.
- \( \sin^{-1}(0.5) = \frac{\pi}{6} \)
Spherical Distance Formula Fundamentals
The spherical distance formula provides a way to calculate distances over the curved surface of a sphere, far more realistic for celestial bodies or Earth-based navigation than the linear approximation offered by plane geometry. This formula takes into account the constraints of spherical geometry and provides a robust method for solving such real-world problems.
The formula is derived from the relationship involving chord length, radius, and angle subtended at the sphere's center. Specifically, it's expressed as \( c = 2r \sin\left(\frac{d}{2r}\right) \), where \( c \) is the chord length, \( r \) is the sphere's radius, and \( d \) is the spherical distance. Solving for \( d \), we rearrange and use inverse trigonometric functions to determine our spherical distance.
By understanding and applying this formula efficiently, we can solve for the arc distances, which reflect more accurately the true path over the sphere than straight-line estimations would.
The formula is derived from the relationship involving chord length, radius, and angle subtended at the sphere's center. Specifically, it's expressed as \( c = 2r \sin\left(\frac{d}{2r}\right) \), where \( c \) is the chord length, \( r \) is the sphere's radius, and \( d \) is the spherical distance. Solving for \( d \), we rearrange and use inverse trigonometric functions to determine our spherical distance.
By understanding and applying this formula efficiently, we can solve for the arc distances, which reflect more accurately the true path over the sphere than straight-line estimations would.