2. The rotation of the earth causes a centrifugal effect on a mass near the surface of the earth. This slightly changes the effective acceleration due to gravity. We can call the old acceleration due to gravity go and the effective acceleration geff: a. Show that the angle between g, and geff can be expressed as 2R sin(e) cos(e) $ = tan-1 Go - Ns/s, R sin² (e) where 0 is the angle between the angular velocity of earth and the position of the mass near the surface of the earth. b. How small that o if 0 is 45° and R is the radius of the earth.
2. The rotation of the earth causes a centrifugal effect on a mass near the surface of the earth. This slightly changes the effective acceleration due to gravity. We can call the old acceleration due to gravity go and the effective acceleration geff: a. Show that the angle between g, and geff can be expressed as 2R sin(e) cos(e) $ = tan-1 Go - Ns/s, R sin² (e) where 0 is the angle between the angular velocity of earth and the position of the mass near the surface of the earth. b. How small that o if 0 is 45° and R is the radius of the earth.
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![### The Effect of Earth's Rotation on Effective Gravity
The rotation of the Earth causes a centrifugal effect on a mass near the surface of the Earth. This slight change affects the effective acceleration due to gravity. We can denote the traditional acceleration due to gravity as \( g_0 \) and the effective acceleration as \( g_{eff} \).
#### a. Derivation of the Angle Between \( g_0 \) and \( g_{eff} \)
We aim to show that the angle \( \phi \) between \( g_0 \) and \( g_{eff} \) can be expressed as:
\[
\phi = \tan^{-1} \left( \frac{\Omega_{S}/S_{0}^2 R \sin(\theta) \cos(\theta)}{g_0 - \Omega_{S}/S_{0}^2 R \sin^2(\theta)} \right)
\]
Here:
- \( \theta \) represents the angle between the angular velocity vector of the Earth and the position of the mass near the Earth’s surface.
- \( R \) is the radius of the Earth.
- \( \Omega_{S}/S_{0} \) is the given constant of proportionality for the angular velocity of Earth’s rotation.
#### b. Calculation of \( \phi \) for \( \theta = 45^\circ \)
To determine how small \( \phi \) is when \( \theta = 45^\circ \) and considering the radius \( R \) of the Earth, we substitute these values into the derived formula.
#### Diagram Explanation
Below the text, there is a diagram depicting:
- A circle representing the Earth.
- An angle \( \theta \) is marked between the radius \( r_{m/S} \) and the vertical axis of rotation \( \Omega_{S}/S_{0} \).
- A point labeled \( m \) on the surface represents the mass's position.
This diagram helps visualize the relationship between Earth’s rotation and the effective gravity at different points on the Earth's surface.
By understanding these relationships, students can grasp how centrifugal forces impact the effective gravity we experience on Earth.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F203e1487-6eea-4eb4-9d2f-d1cb93589e65%2F65769be6-c56b-4357-ae57-d190205e0415%2F09xzknl.jpeg&w=3840&q=75)
Transcribed Image Text:### The Effect of Earth's Rotation on Effective Gravity
The rotation of the Earth causes a centrifugal effect on a mass near the surface of the Earth. This slight change affects the effective acceleration due to gravity. We can denote the traditional acceleration due to gravity as \( g_0 \) and the effective acceleration as \( g_{eff} \).
#### a. Derivation of the Angle Between \( g_0 \) and \( g_{eff} \)
We aim to show that the angle \( \phi \) between \( g_0 \) and \( g_{eff} \) can be expressed as:
\[
\phi = \tan^{-1} \left( \frac{\Omega_{S}/S_{0}^2 R \sin(\theta) \cos(\theta)}{g_0 - \Omega_{S}/S_{0}^2 R \sin^2(\theta)} \right)
\]
Here:
- \( \theta \) represents the angle between the angular velocity vector of the Earth and the position of the mass near the Earth’s surface.
- \( R \) is the radius of the Earth.
- \( \Omega_{S}/S_{0} \) is the given constant of proportionality for the angular velocity of Earth’s rotation.
#### b. Calculation of \( \phi \) for \( \theta = 45^\circ \)
To determine how small \( \phi \) is when \( \theta = 45^\circ \) and considering the radius \( R \) of the Earth, we substitute these values into the derived formula.
#### Diagram Explanation
Below the text, there is a diagram depicting:
- A circle representing the Earth.
- An angle \( \theta \) is marked between the radius \( r_{m/S} \) and the vertical axis of rotation \( \Omega_{S}/S_{0} \).
- A point labeled \( m \) on the surface represents the mass's position.
This diagram helps visualize the relationship between Earth’s rotation and the effective gravity at different points on the Earth's surface.
By understanding these relationships, students can grasp how centrifugal forces impact the effective gravity we experience on Earth.
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