The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 x 106 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 77.0° north of the equator.
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
![**Title:** Calculating Speed and Centripetal Acceleration on Earth's Surface
**Description:**
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of \(6.38 \times 10^6\) meters, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of \(77.0^\circ\) north of the equator.
**Illustration:**
The provided image includes a diagram illustrating the Earth, segmented to show its spherical shape and rotation axis. Key features in the diagram:
- **\(R_e\)** represents the Earth's radius.
- **\(r\)** is the distance from the rotation axis to a point on the Earth's surface at latitude \( \theta \).
Two points on the Earth's surface are highlighted for spherical coordinates:
- One on the equator, \( \theta = 0^\circ \).
- Another at latitude \(77.0^\circ\).
**Equations and Variables:**
- Centripetal speed: \( v = \omega \cdot r \)
- Centripetal acceleration: \( a_c = \frac{v^2}{r} = \omega^2 \cdot r \)
where \( \omega = \frac{2 \pi}{T} \) with \( T = 86400 \) seconds (24 hours).
### Calculation Fields:
**a) At the Equator:**
- **Speed \(v\):**
\[ \text{Units:} \quad \text{m/s} \]
- **Centripetal Acceleration \(a_c\):**
\[ \text{Units:} \quad \text{m/s}^2 \]
**b) At a Latitude of \(77.0^\circ\) North:**
- **Speed \(v\):**
\[ \text{Units:} \quad \text{m/s} \]
- **Centripetal Acceleration \(a_c\):**
\[ \text{Units:} \quad \text{m/s}^2 \]
---
This resource allows students to input values and select units to compute the speed and centripetal acceleration at given points on Earth's surface, facilitating a deeper understanding of rotational dynamics](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb1b37c1-d10c-4bcc-8f78-f20c960fb27f%2F4ebac1c9-0720-49dd-9a59-9385c65c6dbe%2F19ss4cq_processed.png&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images
