• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

100P

Expert-verified
Found in: Page 1

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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$=3.1×{10}^{-5}\mathrm{N}$.

See the step by step solution

## Step 1: Listing the given quantities

The mass of each of the two identical blocks $=2.00\mathrm{kg}$

The difference in the length of the string $=5.00\mathrm{cm}$

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

## 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:

$\stackrel{\to }{\mathrm{F}}=\frac{{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{E}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{{\mathrm{GM}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{E}}^{2}}$

## Step 3: Explanation

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

$\stackrel{\to }{\mathrm{F}}=\frac{{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{E}}^{2}}$

i.e., the acceleration due to gravity is

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

The weight difference between the two object is

$\begin{array}{rcl}\mathrm{w}& =& \mathrm{m}\left(\mathrm{g}-{\mathrm{a}}_{\mathrm{g}}\right)=\frac{\mathrm{GMm}}{{\mathrm{R}}_{\mathrm{E}}^{2}}-\frac{\mathrm{GMm}}{{\left({\mathrm{R}}_{\mathrm{E}}+\mathrm{h}\right)}^{2}}\\ \mathrm{w}& =& \frac{\mathrm{GMm}}{{\mathrm{R}}_{\mathrm{E}}^{2}}\left[1-{\left(1+\frac{\mathrm{h}}{{\mathrm{R}}_{\mathrm{E}}}\right)}^{-2}\right]\\ \mathrm{w}& \approx & \frac{\mathrm{GMm}}{{\mathrm{R}}_{\mathrm{E}}^{2}}\frac{2\mathrm{h}}{{\mathrm{R}}_{\mathrm{E}}}\\ \mathrm{w}& =& \frac{2\mathrm{GMmh}}{{\mathrm{R}}_{\mathrm{E}}^{3}}\end{array}$

We know

$\mathrm{M}=\frac{4}{3}{\mathrm{\pi R}}_{\mathrm{E}}^{3}\mathrm{\rho }$

Putting the value of M in the above equation

$\mathrm{w}=\frac{2\mathrm{Gmh}}{{\mathrm{R}}_{\mathrm{E}}^{3}}\frac{4}{3}{\mathrm{\pi R}}_{\mathrm{E}}^{3}\mathrm{\rho }\phantom{\rule{0ex}{0ex}}\mathrm{w}=\frac{8\mathrm{\pi Gmh\rho }}{3}$

Substituting the values, we have

$\begin{array}{rcl}\mathrm{w}& =& \frac{8\mathrm{\pi }}{3}\left(5.5×{10}^{3}\right)\left(6.67×{10}^{-11}\right)\left(2.00\right)\left(0.50\right)\\ & =& 3.07×{10}^{-7}\mathrm{N}\end{array}$

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