Chapter 8, Problem 10P

(a)

To determine

The linear speed of a point on the equator.

Answer to Problem 10P

The linear speed of a point on the equator is $\text{ 465}\text{.4 m/s}$

Explanation of Solution

Given:

Earth rotates by 360 degrees, $\text{ (Δθ = 2π radians)}$

Time taken for this rotation is 1 day.

Radius of earth is 6400 km.

Formula used:

Linear speed of a point on the equator is determined by the following formula,

  $\text{Linear Speed (v) = r}\times \text{ω}$

  $\text{ω = }\frac{\text{Δθ }}{\text{Δt}}$

Calculation:

Substitute the given values, to find the angular velocity.

  $\begin{array}{c}\text{ω= }\frac{\text{2π radians }}{\text{1 day}}\\ =\frac{\text{2π radians }}{\text{1}\times \text{24}\times \text{60}\times \text{60 seconds}}\\ =\frac{\text{2π radians }}{\text{8}\text{.64}\times {\text{10}}^4\text{seconds}}\\ =7.2722\times {10}^{-5}\text{ rad/sec}\end{array}$

Linear speed is,

  $\begin{array}{c}\text{v = r}\times \text{ω}\\ \text{ = 6400 km}\times \text{(7}\text{.2722}\times {\text{10}}^{\text{-5}}\text{ rad/sec)}\\ \text{ = 6400}\times {\text{10}}^3\times \text{7}\text{.2722}\times {\text{10}}^{-5}\text{ rad/sec }\\ \text{= 465}\text{.4 m/s}\end{array}$

Conclusion:

The linear speed of a point on the equator is $\text{ 465}\text{.4 m/s}$

(b)

To determine

Linear speed of a point on the Arctic circle at a given altitude.

Answer to Problem 10P

Linear speed of a point on the Arctic circle of latitude 66.5⁰ N is $\text{185}\text{.6 m/s}$

Explanation of Solution

Given:

Earth rotates by 360 degrees, $\text{ (Δθ = 2π radians)}$

Time taken for this rotation is 1 day.

Radius of earth is 6400 km.

Latitude is $\text{(θ) = 66}{\text{.5}}^{\text{0}}\text{ N}$

Formula used:

Linear speed of a point on the Arctic circle is determined by the following formula,

  $\text{Linear Speed (v) = (r}\times \text{cosθ)}\times \text{ω}$

Angular velocity is,

  $\text{ω = }\frac{\text{Δθ }}{\text{Δt}}$

Calculation:

Substitute the given values, to find the angular velocity.

  $\begin{array}{c}\text{ω = }\frac{\text{2π radians }}{\text{1 day}}\\ =\frac{\text{2π radians }}{\text{1}\times \text{24}\times \text{60}\times \text{60 seconds}}\\ =\frac{\text{2π radians }}{\text{8}\text{.64}\times {\text{10}}^4\text{seconds}}\\ =7.2722\times {10}^{-5}\text{ rad/sec}\end{array}$

Linear speed is,

  $\begin{array}{c}\text{v = (r}\times \text{cosθ }\times \text{ ω)}\\ \text{= 6400 km}\times \text{cos 66}{\text{.5}}^0\times \text{(7}\text{.2722}\times {\text{10}}^{\text{-5}}\text{ rad/sec) }\\ \text{= 6400}\times {\text{10}}^3\times \text{0}\text{.3987}\times \text{7}\text{.2722}\times {\text{10}}^{-5}\text{ rad/sec}\\ \text{= 185}\text{.6 m/s}\end{array}$

Conclusion:

Linear speed of a point on the Arctic circle of latitude 66.5⁰ N is $\text{185}\text{.6 m/s}$

(c)

To determine

Linear speed of a point at a given latitude.

Answer to Problem 10P

Linear speed of a point at a latitude of 45⁰ N is $\text{329 m/s}$

Explanation of Solution

Given:

Earth rotates by 360 degrees, $\text{ (Δθ = 2π radians)}$

Time taken for this rotation is 1 day.

Radius of earth is 6400 km.

Latitude is ${\text{(θ) = 45}}^{\text{0}}\text{ N}$

Formula used:

Linear speed of a point at a latitude of 45⁰ is determined by the following formula,

  $\text{Linear Speed (v) = (r}\times \text{cosθ)}\times \text{ω}$

Angular velocity is,

  $\text{ω = }\frac{\text{Δθ }}{\text{Δt}}$

Calculation:

Substitute the given values, to find the angular velocity.

  $\begin{array}{c}\text{ω = }\frac{\text{2π radians }}{\text{1 day}}\\ =\frac{\text{2π radians }}{\text{1}\times \text{24}\times \text{60}\times \text{60 seconds}}\\ =\frac{\text{2π radians }}{\text{8}\text{.64}\times {\text{10}}^4\text{seconds}}\\ =7.2722\times {10}^{-5}\text{ rad/sec}\end{array}$

Linear speed is,

  $\begin{array}{c}\text{v = (r}\times \text{cosθ}\times \text{ω) }\\ \text{= 6400 km}\times {\text{cos 45}}^0\times \text{(7}\text{.2722}\times {\text{10}}^{\text{-5}}\text{ rad/sec) }\\ \text{= 6400}\times {\text{10}}^3\times \text{0}\text{.707}\times \text{7}\text{.2722}\times {\text{10}}^{-5}\text{ rad/sec}\\ \text{= 329}\text{.1 m/s}\end{array}$

  $$

Conclusion:

Linear speed of a point at a latitude of 45⁰ N is $\text{329}\text{.1 m/s}$