Home
Class 11
PHYSICS
Five point masses m each are kept at fiv...

Five point masses `m` each are kept at five vertices of a regular pentagon. Distance of center of pentagon from any one of the vertice is 'a'. Find gravitational potential and filed strength at center.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the gravitational potential and field strength at the center of a regular pentagon with five point masses \( m \) at its vertices, we can follow these steps: ### Step 1: Understand the Geometry We have a regular pentagon with five vertices, each occupied by a mass \( m \). The distance from the center of the pentagon to any vertex is given as \( a \). ### Step 2: Calculate the Gravitational Potential The gravitational potential \( V \) due to a single mass \( m \) at a distance \( r \) is given by the formula: \[ V = -\frac{Gm}{r} \] where \( G \) is the gravitational constant. Since there are five identical masses \( m \) located at the vertices of the pentagon, the total gravitational potential \( V_{total} \) at the center \( O \) of the pentagon is the sum of the potentials due to each mass. Since potential is a scalar quantity, we can simply add them: \[ V_{total} = V_1 + V_2 + V_3 + V_4 + V_5 = -\frac{Gm}{a} - \frac{Gm}{a} - \frac{Gm}{a} - \frac{Gm}{a} - \frac{Gm}{a} \] This simplifies to: \[ V_{total} = -5 \frac{Gm}{a} \] ### Step 3: Calculate the Gravitational Field Strength The gravitational field strength \( E \) at a point is defined as the force per unit mass experienced by a small test mass placed at that point. The gravitational field \( E \) due to a single mass \( m \) at a distance \( r \) is given by: \[ E = -\frac{Gm}{r^2} \] However, in this case, we need to consider the contributions from all five masses. When we place a test mass at the center, the forces due to each of the five masses will act towards their respective vertices. Due to the symmetry of the pentagon, the forces exerted by the masses will cancel each other out. Therefore, the net gravitational field \( E_{total} \) at the center \( O \) will be: \[ E_{total} = 0 \] ### Final Results 1. The gravitational potential at the center of the pentagon is: \[ V_{total} = -\frac{5Gm}{a} \] 2. The gravitational field strength at the center of the pentagon is: \[ E_{total} = 0 \]

To solve the problem of finding the gravitational potential and field strength at the center of a regular pentagon with five point masses \( m \) at its vertices, we can follow these steps: ### Step 1: Understand the Geometry We have a regular pentagon with five vertices, each occupied by a mass \( m \). The distance from the center of the pentagon to any vertex is given as \( a \). ### Step 2: Calculate the Gravitational Potential The gravitational potential \( V \) due to a single mass \( m \) at a distance \( r \) is given by the formula: \[ ...
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    DC PANDEY ENGLISH|Exercise Exercise 13.4|4 Videos
  • GRAVITATION

    DC PANDEY ENGLISH|Exercise Exercise 13.5|6 Videos
  • GRAVITATION

    DC PANDEY ENGLISH|Exercise Exercise 13.2|9 Videos
  • GENERAL PHYSICS

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|2 Videos
  • KINEMATICS

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|10 Videos