Question

Two spheres each of mass 70 kg are 0.14 m apart. Find the force of attraction between them.

Given $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer 1

Step 1: Given data

Given,

$$\begin{gathered} Massofboththespheres,M_1=M_2=70kg \\ Dis\tan cebetweenboththespheres,R=0.14m \\ G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2} \\ Forceofattraction,F=? \end{gathered}$$

Step 2: Formula for gravitational force (attraction) between both the spheres

$F=G\times \frac{M_1\times M_2}{R^2}$

Step 3: Substituting the given values in the formula

$$\begin{gathered} F=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}\times \frac{70kg\times 70kg}{{\left(0.14m\right)}^2} \\ =166.75\times {10}^{-7}N \end{gathered}$$

Hence, the force between the two spheres will be $166.75\times {10}^{-7}N$.