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In older times, people used to think tha...

In older times, people used to think that the earth was flat. Imagine that the earth is indeed not a sphere of radius `R`, but an infinite plate of thickness `H`. What value of `H` is needed to allow the same gravitational acceleration to be experienced as on the surface of the actual Earth? (Assume that the Earth's density is uniform and equal in the tow models.)

A

`(2R)/(3)`

B

`(4R)/(3)`

C

`(8R)/(3)`

D

`(R )/(3)`

Text Solution

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The correct Answer is:
To solve the problem of finding the thickness \( H \) of an infinite plate that results in the same gravitational acceleration as that experienced on the surface of a spherical Earth, we can follow these steps: ### Step 1: Understanding the Gravitational Field of an Infinite Plate For an infinite plate of uniform density \( \rho \) and thickness \( H \), the gravitational field \( g \) at a point above the plate can be derived using Gauss's law. The gravitational field due to an infinite sheet is given by: \[ g = 2 \pi G \rho H \] where \( G \) is the universal gravitational constant. ### Step 2: Gravitational Acceleration on the Surface of a Sphere For a spherical Earth of radius \( R \) and uniform density \( \rho \), the gravitational acceleration \( g \) at the surface is given by: \[ g = \frac{GM}{R^2} \] where \( M \) is the mass of the Earth. The mass \( M \) can be expressed in terms of density and volume: \[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \] Substituting this into the equation for gravitational acceleration gives: \[ g = \frac{G \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R \] ### Step 3: Setting the Two Expressions for Gravitational Acceleration Equal To find \( H \) such that the gravitational acceleration from the infinite plate equals that from the spherical Earth, we set the two expressions for \( g \) equal: \[ 2 \pi G \rho H = \frac{4}{3} \pi G \rho R \] ### Step 4: Simplifying the Equation We can cancel \( \pi G \rho \) from both sides (assuming \( \rho \neq 0 \)): \[ 2H = \frac{4}{3} R \] ### Step 5: Solving for \( H \) Now, we can solve for \( H \): \[ H = \frac{4}{3} \cdot \frac{1}{2} R = \frac{2}{3} R \] ### Conclusion Thus, the thickness \( H \) needed for the infinite plate to provide the same gravitational acceleration as that on the surface of the actual Earth is: \[ H = \frac{2R}{3} \]
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