Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 13-57, identical blocks with identical masses m=2.00 kg hang from strings of different lengths on a balance at Earth’s surface.The strings have negligible mass and differ in length by h=5.00 cm. Assume Earth is spherical with a uniform density r=5.5. g/cm³. What is the difference in the weight of the blocks due to one being closer to Earth than the other?

The difference in weight of the blocks $= 3.1 \times 10^{- 5} N$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Listing the given quantities

The mass of each of the two identical blocks $= 2.00 kg$

The difference in the length of the string $= 5.00 cm$

Earth is sphere and has uniform density $= 5.50 g / {cm}^{2}$ .

Step 2: Understanding the concept of gravitational force

The gravitational force (and acceleration) is inversely proportional to the square of the distance between the particle and the earth.

Formula:

$\vec{F} = \frac{{GM}_{E} m}{R_{E}^{2}} a = \frac{{GM}_{E}}{R_{E}^{2}}$

Step 3: Explanation

The earth exerts gravitational force on a particle of mass m, given as

$\vec{F} = \frac{{GM}_{E} m}{R_{E}^{2}}$

i.e., the acceleration due to gravity is

$a = \frac{{GM}_{E}}{R_{E}^{2}}$

The weight difference between the two object is

$\begin{array}{rcl} w & = & m (g - a_{g}) = \frac{GMm}{R_{E}^{2}} - \frac{GMm}{{(R_{E} + h)}^{2}} \\ w & = & \frac{GMm}{R_{E}^{2}} [1 - {(1 + \frac{h}{R_{E}})}^{- 2}] \\ w & \approx & \frac{GMm}{R_{E}^{2}} \frac{2 h}{R_{E}} \\ w & = & \frac{2 GMmh}{R_{E}^{3}} \end{array}$

We know

$M = \frac{4}{3} {πR}_{E}^{3} ρ$

Putting the value of M in the above equation

$w = \frac{2 Gmh}{R_{E}^{3}} \frac{4}{3} {πR}_{E}^{3} ρ w = \frac{8 πGmhρ}{3}$

Substituting the values, we have

$\begin{array}{rcl} w & = & \frac{8 π}{3} (5.5 \times 10^{3}) (6.67 \times 10^{- 11}) (2.00) (0.50) \\ = & 3.07 \times 10^{- 7} N \end{array}$

The difference in weight of the blocks $= 3.1 \times 10^{- 5} N$ .

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