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Chapter 9: Problem 14

\(n\) -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

Short Answer

Expert verified
(a) \\(-\frac{G n m_{0}}{r_{0}}\\)

Step by step solution

01

Understanding Gravitational Potential

The gravitational potential at a point due to a single mass is given by the formula \( V = -\frac{Gm}{r} \), where \( G \) is the gravitational constant, \( m \) is the mass, and \( r \) is the distance from the mass to the point in question.
02

Determine Distance from Mass to Center

In this problem, the distance from each corner (where the mass \( m_0 \) is located) to the center of the polygon is given as \( r_0 \). Thus, this is the value of \( r \) in our potential formula.
03

Sum Up Potentials from All Masses

Since the polygon has \( n \) vertices, and each vertex has a mass \( m_0 \) placed on it, the total gravitational potential at the center due to all masses is just \( n \) times the potential due to one mass. So: \[ V_{ ext{total}} = n \cdot \left(-\frac{G m_0}{r_0}\right) = -\frac{G n m_0}{r_0} \]
04

Compare with Given Options

After calculating, we see that the potential \( V_{\text{total}} = -\frac{G n m_0}{r_0} \) matches option (a). Therefore, option (a) is the correct choice.

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Key Concepts

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

Gravitational Constant
The gravitational constant, often denoted by the symbol \( G \), is a fundamental constant in physics that appears in Newton's law of universal gravitation. This constant describes the strength of gravity between two objects, and its value is approximately \( 6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2} \).
Understanding \( G \) is crucial because it helps determine the gravitational force and potential between masses, allowing for calculations like those required in this problem. Without \( G \), the attractive force between objects due to gravity couldn't be quantified in a standardized way.
  • It acts uniformly in space and is crucial for calculating gravitational interactions.
  • In the context of this exercise, \( G \) is necessary to find the gravitational potential at the center of the polygon due to the masses at the vertices.
Regular Polygon
A regular polygon is a geometric shape where all sides are of equal length, and all angles are equal. These shapes are symmetric and often have interesting properties that are useful in physics and mathematics.
When dealing with gravitational problems, the symmetry of a regular polygon simplifies calculations, as is the case when determining the gravitational potential at its center.
  • The regular polygon's symmetry allows easier computation of potential energy from its vertices to its center.
  • Each corner or vertex can be treated equivalently due to this symmetry.
For this problem, an \( n \)-sided regular polygon is used, highlighting its usefulness in simplifying the determination of gravitational potential at specific points, like the center.
Mass Distribution
Understanding mass distribution is vital when calculating gravitational potential since the arrangement of mass impacts the resulting gravitational effect.
In this exercise, masses \( m_0 \) are evenly distributed at the vertices of a regular polygon. This uniform distribution is key because:
  • The even placement of mass means each contributes equally to the potential at the center.
  • The cumulative effect of all \( n \) masses is determined by their uniform distance \( r_0 \) from the center.
  • Any non-uniform distribution would complicate the potential calculation significantly.
Mass distribution in this case helps maintain symmetry and simplify calculations, ensuring the effect of each mass can be added directly.
Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, often expressed as \( V = -\frac{Gm}{r} \) for a point mass. In problems like this exercise, potential energy at a point, such as the center of a polygon, is due to multiple masses exerting gravitational forces.
Several important points:
  • The negative sign in the potential energy formula reflects the attractive nature of gravity, showing that potential energy decreases as objects come closer.
  • Total gravitational potential at a point is the sum of potentials due to each mass. Here, it is simple to multiply the potential from one vertex mass by \( n \), the number of masses, due to symmetry.
Understanding potential energy allows for calculating the impact of multiple gravitational sources, critical for finding the right answer in this exercise.

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Most popular questions from this chapter

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