/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 If earth were to rotate faster t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If earth were to rotate faster than its present speed, the weight of an object will : (a) increase at the equator but remain unchanged at poles (b) decrease at the equator but remain unchanged at the poles (c) remain unchanged at the equator but decrease at the poles (d) remain unchanged at the equator but increase at the poles

Short Answer

Expert verified
(b) decrease at the equator but remain unchanged at the poles.

Step by step solution

01

Understanding the Earth's Rotation Effect

The Earth's rotation creates what is known as the centrifugal force, which is directed outward from the axis of rotation. This force affects the apparent weight of objects on the surface of the Earth, particularly at the equator where rotational speed is highest.
02

Effect of Increased Rotation Speed on Centrifugal Force

If the Earth's rotational speed increases, the centrifugal force experienced at the equator will also increase. This is because centrifugal force is directly proportional to the square of the rotational speed. As a result, objects will experience a greater outward force at the equator.
03

Analyzing the Impact at the Equator and Poles

At the equator, the increased centrifugal force counteracts the gravitational pull, effectively reducing the apparent weight of objects. At the poles, the rotation has no effect on centrifugal force since the poles are the Earth's axis of rotation, thus the weight of an object remains unchanged.
04

Choosing the Correct Answer

Based on the analysis, the option that reflects a decrease in weight due to increased centrifugal force at the equator while remaining unchanged at the poles is: (b) decrease at the equator but remain unchanged at the poles.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Earth's rotation
The Earth rotates around its axis, which is an imaginary line that passes through the North and South Poles. This rotation is responsible for the cycle of day and night we experience each day. The rotation of Earth is a critical factor in various physical phenomena that occur on our planet.
The speed of Earth's rotation varies depending on where you are located. It is fastest at the equator and effectively zero at the poles. As a result, this rotation has different effects on objects located at different latitudes. Because the rotation creates a centrifugal effect, it influences the way we perceive forces such as gravity, depending on our position on Earth.
This rotational dynamic helps us understand how changes in the Earth's rotation could affect the weight of objects.
Centrifugal force
Centrifugal force is what you would feel when you're on a merry-go-round. It is an outward component due to an object's circular motion, specifically when moving along a curved path. On Earth, centrifugal force is produced as a result of its rotation. This force acts perpendicular to the axis of rotation and grows stronger as you move from the poles towards the equator.
This force impacts objects differently based on their latitude. At the equator, it can be calculated by the formula:\[ F = m \cdot \omega^2 \cdot r \]where \( m \) is mass, \( \omega \) is rotational velocity, and \( r \) is the radius of Earth at the equator.
A faster rotation suggests a higher \( \omega \), leading to greater centrifugal force exerted on objects, decreasing their apparent weight at the equator.
Gravitational force
Gravitational force is a natural force of attraction between two masses. All objects on Earth experience this force, which is directed towards the center of the planet. It is what gives us weight and keeps us grounded on the surface.
The force of gravity is calculated using the formula: \[ F_g = G \frac{m_1 \cdot m_2}{r^2} \]where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses involved, and \( r \) is the distance between the centers of the two masses.
Despite centrifugal forces, the gravitational force remains unchanged for objects at a constant distance from Earth's center. However, the apparent weight can feel less when centrifugal forces counteract gravitational pull, especially at the equator.
Apparent weight
Apparent weight is the force exerted on an object by the surface it's on, accounting for all acting forces. It can differ from actual weight because of the presence of other forces, like centrifugal force.
In terms of Earth's rotation, apparent weight is influenced by the balance between the gravitational force pulling us down and the centrifugal force pushing us outward. This balance changes depending on where you are on the Earth's surface:
  • At the equator, centrifugal force is maximal, reducing apparent weight.
  • At the poles, centrifugal force doesn't affect apparent weight, keeping it as the actual gravitational pull.
Thus, apparent weight can vary quite significantly between the equator and the poles due to different rotational effects.
Equator and poles
The equator and poles highlight Earth's extreme points regarding rotational effects. The equator is the region around the Earth where it is widest, and it experiences the maximum rotational speed. As a result, rotational effects like centrifugal force are most significant here.
Completing one rotation every 24 hours, the equatorial speed is why any increase in Earth's rotational speed profoundly impacts measurements like apparent weight in these areas. Increased speed enhances centrifugal force, causing the apparent weight to decrease, as mentioned earlier.
Conversely, the poles represent areas on Earth's axis where rotational speed is zero. Thus, changes in Earth's rotation have no centrifugal impact here. Therefore, the apparent weight remains consistent, reflecting only the gravitational force without any reduction by rotational effects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A satellite \(S\) is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth: (a) the acceleration of \(S\) is always directed towards the centre of the earth (b) the angular momentum of \(S\) about the centre of the earth changes in direction but its magnitude remains constant (c) the total mechanical energy of \(S\) varies periodically with time (d) the linear momentum of \(S\) remains constant in the magnitude

A projectile is fired vertically upwards from the surface of the earth with a velocity \(k v_{e}\), where \(v_{e}\) is the escape velocity and \(k<1\). If \(R\) is the radius of the earth, the maximum height to which it will rise measured from the centre of earth will be : (neglect air resistance) (a) \(\frac{1-k^{2}}{R}\) (b) \(\frac{R}{1-k^{2}}\) (c) \(R\left(1-k^{2}\right)\) (d) \(\frac{R}{1+k^{2}}\)

Suppose the gravitational force varies inversely as the \(n^{\text {th }}\) power of the distance. Then the time period of a planet in circular orbit of radius \(R\) around the sun will be proportional to : (a) \(R^{n}\) (b) \(R^{(n+1) / 2}\) (c) \(R^{(n-1) / 2}\) (d) \(R^{-n}\)

The period of a satellite in a circular orbit of radius \(R\) is \(T\). What is the period of another satellite in a circular orbit of radius \(4 R\) ? (a) \(4 T\) (b) \(\frac{T}{8}\) (c) \(\frac{T}{4}\) (d) \(8 T\)

\(n\) -particlos each of mass \(m_{0}\) are placed on different comers of a regular polygon of edge length \(a\). The distance between vertex and centre of polygon is \(r_{0}\). The gravitational potential at the centre of the polygon is: (a) \(-\frac{G n m_{0}}{r_{0}}\) (b) \(-\frac{G m_{0}}{r_{0}}\) (c) \(\frac{n G m_{0}}{r_{0}}\) (d) none of these

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.