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An infinite number of particles each of ...

An infinite number of particles each of mass `m` are placed on the positive X-axis of `1m, 2m, 4m, 8m,...` from the origin. Find the magnitude of the resultant gravitational force on mass `m` kept at the origin.

A

`-8Gm`

B

`-3Gm`

C

`-4Gm`

D

`-2Gm`

Text Solution

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To find the magnitude of the resultant gravitational force on a mass \( m \) kept at the origin due to an infinite number of particles each of mass \( m \) placed at positions \( 1m, 2m, 4m, 8m, \ldots \) on the positive X-axis, we will follow these steps: ### Step 1: Understand the Gravitational Force Formula The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by: \[ F = \frac{G m_1 m_2}{r^2} \] In our case, both masses are \( m \), so the force exerted by each mass on the mass at the origin will be: \[ F = \frac{G m^2}{r^2} \] ### Step 2: Identify the Positions and Distances The particles are located at distances \( r_1 = 1m, r_2 = 2m, r_3 = 4m, r_4 = 8m, \ldots \). The general position of the \( n \)-th particle can be expressed as: \[ r_n = 2^{n-1} \text{ for } n = 1, 2, 3, \ldots \] ### Step 3: Calculate the Gravitational Force from Each Particle The gravitational force exerted by the \( n \)-th particle on the mass at the origin is: \[ F_n = \frac{G m^2}{(2^{n-1})^2} = \frac{G m^2}{4^{n-1}} \] ### Step 4: Sum the Forces from All Particles The total gravitational force \( F_0 \) on the mass at the origin is the sum of the forces from all particles: \[ F_0 = \sum_{n=1}^{\infty} F_n = \sum_{n=1}^{\infty} \frac{G m^2}{4^{n-1}} \] ### Step 5: Recognize the Series as a Geometric Series The series can be rewritten as: \[ F_0 = G m^2 \sum_{n=0}^{\infty} \frac{1}{4^n} \] This is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{4} \). ### Step 6: Calculate the Sum of the Geometric Series The sum of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \] Substituting the values: \[ S = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \] ### Step 7: Substitute Back to Find \( F_0 \) Now substituting back into the expression for \( F_0 \): \[ F_0 = G m^2 \cdot \frac{4}{3} \] ### Final Result Thus, the magnitude of the resultant gravitational force on the mass \( m \) kept at the origin is: \[ F_0 = \frac{4}{3} G m^2 \]
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Knowledge Check

  • A mass m is placed at point P lies on the axis of a ring of mass M and radius R at a distance R from its centre. The gravitational force on mass m is

    A
    `(GMm)/(sqrt(2)R^(2))`
    B
    `(GMm)/(2R^(2))`
    C
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    D
    `(GMm)/(4R^(2))`
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