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Chapter 8: Problem 29

The gravitational potential in a region is given by \(V=(3 x+4 y+12 z) J / k g\). The modulus of the gravitational field at \((x=1, y=0, z=3)\) is (a) \(20 \mathrm{~N} \mathrm{~kg}^{-1}\) (b) \(13 \mathrm{~N} \mathrm{~kg}^{-1}\) (c) \(12 \mathrm{~N} \mathrm{~kg}^{-1}\) (d) \(5 \mathrm{~N} \mathrm{~kg}^{-1}\)

Short Answer

Expert verified
The modulus of the gravitational field at the given point is approximately 36 N/kg.

Step by step solution

01

Calculate the Gradient of the Potential Function

The gravitational field vector \( \mathbf{g} \) is the negative gradient of the potential function: \( \mathbf{g} = - \nabla V \). Therefore, we have to calculate the gradients of V in terms of x, y and z respectively. These are: \(\frac{\partial V}{\partial x} = 3 N/kg\), \( \frac{\partial V}{\partial y} = 4 N/kg\), and \( \frac{\partial V}{\partial z} = 12 N/kg \).
02

Substitute the Given Point into the Gradients

The next step is to substitute the coordinates of the given point (x=1, y=0, z=3) into the gradients. The result is: \(\frac{\partial V}{\partial x}(1,0,3) = 3 N/kg\), \( \frac{\partial V}{\partial y}(1,0,3) =0 N/kg \), and \( \frac{\partial V}{\partial z}(1,0,3)= 36 N/kg \).
03

Calculate the Modulus of the Field Vector

With the components of the field vector \( -\mathbf{g} \): (3,0,36), we can calculate its modulus using the formula for the magnitude of a vector: \( g = |\mathbf{g}|=\sqrt{(g_x)^2+(g_y)^2+(g_z)^2} =\sqrt{(3)^2+(0)^2+(36)^2} = 36.0555 \mathrm{N} \mathrm{kg^{-1}} \).
04

Choose the Closest Option

If we compare our calculated result to the given answer options, we can see that the nearest (in modulus) option to our result is \(36 \mathrm{N} \mathrm{kg^{-1}}\)

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Most popular questions from this chapter

A projectile is projected with velocity \(k v_{e}\) in vertically upward direction from the ground into the space. \(\left(v_{e}\right.\) is escape velocity and \(k<1\) ). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be ( \(R=\) radius of earth) (a) \(\frac{R}{k^{2}+1}\) (b) \(\frac{R}{k^{2}-1}\) (c) \(\frac{R}{1-k^{2}}\) (d) \(\frac{R}{k+1}\)

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