/*! 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 1 The earth (mass \(=10^{24} \math... [FREE SOLUTION] | 91Ó°ÊÓ

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The earth (mass \(=10^{24} \mathrm{~kg}\) ) revolves round the Sun with an angular velocity \(2 \times 10^{-7} \mathrm{rad} \mathrm{s}^{-1}\) in a circular orbit of radius \(1.5 \times 10^{8} \mathrm{~km}\). Find the force exerted by the Sun on the earth \(\left(\mathrm{in} \times 10^{21} \mathrm{~N}\right)\)

Short Answer

Expert verified
The force exerted by the Sun on the Earth is \( 6 \times 10^{21} \text{ N} \).

Step by step solution

01

Understanding Centripetal Force Formula

The centripetal force required to keep an object moving in a circular path is given by the formula \( F = m \cdot r \cdot \omega^2 \), where \( F \) is the force, \( m \) is the mass of the object, \( r \) is the radius of the circular path, and \( \omega \) is the angular velocity.
02

Convert Units to Meters

We need to ensure all our units are consistent. The given radius is \( 1.5 \times 10^8 \) km. Convert this to meters: \( r = 1.5 \times 10^8 \times 10^3 \) m, which is \( 1.5 \times 10^{11} \) m.
03

Substitute Into the Formula

Substitute the known values into the centripetal force formula: \( F = 10^{24} \times 1.5 \times 10^{11} \times (2 \times 10^{-7})^2 \).
04

Calculate the Force

Calculate \( F = 10^{24} \times 1.5 \times 10^{11} \times 4 \times 10^{-14} = 6 \times 10^{21} \) N.
05

Convert Force to Desired Units

The problem asks for the force in terms of \( \times 10^{21} \text{ N} \). Calculate \( F = 6 \times 10^{21} \text{ N} \).

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Key Concepts

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

Circular Motion
Circular motion is a fascinating topic in physics. It describes the movement of an object along the circumference of a circle or a circular path. When an object moves in a circle, it is constantly changing direction. This requires a continuous application of force towards the center of the circle to keep the object in its path.

One important concept in circular motion is centripetal force. This is the force that acts on the object to maintain its circular path. Without this force, the object would move off in a straight line. Imagine swinging a ball on a string; the tension in the string provides the centripetal force. In our exercise, the Sun exerts a force on the Earth to keep it in orbit. The Earth moves in a circular path around the Sun due to this centripetal force.

In summary, circular motion requires two key elements:
  • A force directed towards the center (centripetal force).
  • Continuous change in the object's direction as it follows the circular path.
Understanding these fundamentals is crucial when analyzing problems involving orbits and revolving objects.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates or revolves around a central point. It tells us how much angle an object sweeps out in a certain period of time. In the exercise, the Earth's angular velocity around the Sun is given as \(2 \times 10^{-7}\, \text{rad}\, \text{s}^{-1}\).

When an object travels in a circular path, it has both linear velocity and angular velocity. Linear velocity relates to the distance traveled along the circle's edge, whereas angular velocity relates to the rotation angle. Angular velocity \(\omega\) is measured in radians per second (rad/s). One complete circle corresponds to \(2\pi\) radians. Higher angular velocity means the object covers more angular distance in a shorter time.

In physics, the relationship between the angular velocity \(\omega\), radius \(r\), and the linear velocity \(v\) is given by:
\[ v = r \cdot \omega \]
This formula shows how the angular speed affects the linear speed in circular motion. By knowing the angular velocity, we can derive other essential parameters of circular motion, such as the centripetal force needed to keep an object in its circular path.
Physics Formulas
Physics relies heavily on formulas to describe and predict the behavior of objects. Understanding the right formulas is key to solving physics problems, like the one presented in the exercise. For circular motion, one of the essential formulas is the centripetal force formula:

\[ F = m \cdot r \cdot \omega^2 \]
Where:
  • \(F\) is the centripetal force,
  • \(m\) is the mass of the object,
  • \(r\) is the radius of the circular path,
  • \(\omega\) is the angular velocity.
The formula tells us that the force needed to keep an object moving in a circular path depends on the mass, radius, and square of the angular velocity. This is why larger and faster-spinning objects require more force to maintain their circular paths.

Certain calculations also involve unit conversions, like meters for distances instead of kilometers, and use of standardized units aligns results. It's important to understand the application and manipulation of these formulas to solve different physics problems effectively.

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Most popular questions from this chapter

Two particles of mass \(m\) and \(4 m\) are at rest at an infinite separation. They move towards each other under mutual gravitational attraction. If \(G\) is the universal gravitational constant. Then at separation \(r\), (Assume zero reference potential energy at infinite separation) (1) the total energy of the two object is zero (2) their relative velocity of approach is \(\left(\frac{10 G m}{r}\right)^{1 / 2}\) in magnitude (3) the total kinetic energy of the object is \(\frac{4 G m^{2}}{r}\) (4) Net angular momentum of both the particles is zero about any point

An astronaut, inside an earth satellite, experiences weightlessness because (1) no external force is acting on him (2) he is falling freely (3) no reaction is exerted by the floor of the satellite (4) he is far away from the earth's surface

The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is \(R\) and the radius of the planet is \(R^{\prime}\). Then wat is the value of \(R / R^{\prime}\) ?

If three particles, each of mass \(M\), are placed at the three corners of an equilateral triangle of side \(a\), the forces exerted by this system on another particle of mass \(M\) placed (i) at the midpoint of a side and (ii) at the centre of the triangle are, respectively, (1) \(0, \frac{4 G M t}{3 a^{2}}\) (2) \(\frac{+G \| F}{3 a^{2}}, 0\) (3) \(3 \frac{3 G M^{2}}{a^{2}}, \frac{G M t^{2}}{a^{2}}\) (4) 0,0

The earth is assumed to be a sphere of radius \(R\). A platform is arranged at a height \(3 R\) from the surface of the earth. The escape velocity of a body from this platform is \(f v_{e}\), where \(v_{e}\) is its escape velocity from the surface of the earth. Find the value of \(f\).

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