100P - Fundamentals Of Physics

Question

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?

Step-by-Step Solution

Answer

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

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:

$$\begin{gathered} \overrightarrow{\mathrm{F}}=\frac{{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{E}}^2} \\ \mathrm{a}=\frac{{\mathrm{GM}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{E}}^2} \end{gathered}$$

Step 3: Explanation

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

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

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

$$\begin{gathered} \mathrm{w}=\frac{2\mathrm{Gmh}}{{\mathrm{R}}_{\mathrm{E}}^3}\frac{4}{3}{\mathrm{\pi R}}_{\mathrm{E}}^3\mathrm{\rho } \\ \mathrm{w}=\frac{8\mathrm{\pi Gmh\rho }}{3} \end{gathered}$$

Substituting the values, we have

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

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