Step-by-Step Solution
Step 1
We have the equation of force which is given by, $F = G\frac{{{m_1}{m_2}}}{{{r^2}}}$
We have the mass of each sphere, that is ${m_1} = {m_2} = 200\;{\rm{kg}}$.
The radius of each sphere is: ${r_s} = 300\;{\rm{mm}}$
We are required to show that the given equation is dimensionally homogeneous. Also, we are asked to determine the value of gravitational force in SI units in three significant figures.
Step 2
To show the dimensional homogeneity, we should write the SI units of all the physical variables. The SI unit of force $\left( F \right)$ is Newton denoted by $\left( {\rm{N}} \right)$, masses ${m_1}$ and ${m_2}$ have the unit kilogram $\left( {{\rm{kg}}} \right)$, $r$ is the distance between two masses having the SI unit meter$\left( {\rm{m}} \right)$. The constant G is known as universal gravitational constant, whose SI unit is $\left( {{{{{\rm{m}}^3}} \mathord{\left/ {\vphantom {{{{\rm{m}}^3}} {{\rm{kg}} \cdot {{\rm{s}}^2}}}} \right. } {{\rm{kg}} \cdot {{\rm{s}}^2}}}} \right)$.
Step 3
Substitute all the SI units in equation\[\begin{array}{c} F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\\ {\rm{N}} = \left( {\frac{{{{\rm{m}}^3}}}{{{\rm{kg}} \cdot {{\rm{s}}^2}}}} \right) \times \left( {\frac{{{\rm{kg}} \times {\rm{kg}}}}{{{{\rm{m}}^2}}}} \right)\\ {\rm{N}} = \left( {\frac{{{\rm{kg}} \cdot {\rm{m}}}}{{{{\rm{s}}^2}}}} \right) \end{array}\]
Since $1\;{\rm{N}} = 1\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$, therefore the given equation is dimensionally homogeneous.
Step 4
The distance between the two spheres is the sum of the radius of both the spheres. It can be calculated as,\[r = {r_s} + {r_s}\]
Step 5
Substitute the numeric value of radius of spheres in the above equation.\[\begin{array}{c} r = 300\;{\rm{mm}} + 300\;{\rm{mm}}\\ = {\rm{600}}\;{\rm{mm}} \times \left( {\frac{{{{10}^{ – 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ = 0.600\;{\rm{m}} \end{array}\]
Step 6
The value of universal gravitational constant is $6.67 \times {10^{ – 11}}\;{{{{\rm{m}}^3}} \mathord{\left/ {\vphantom {{{{\rm{m}}^3}} {{\rm{kg}} \cdot {{\rm{s}}^2}}}} \right. } {{\rm{kg}} \cdot {{\rm{s}}^2}}}$. Substitute the values of universal gravitational constant, both masses and distance between them.\[\begin{array}{c} F = 6.67 \times {10^{ – 11}}\;\frac{{{{\rm{m}}^3}}}{{\left( {{\rm{kg}} \cdot {{\rm{s}}^2}} \right)}} \times \frac{{\left( {200\;{\rm{kg}}} \right) \times \left( {{\rm{200}}\;{\rm{kg}}} \right)}}{{{{\left( {0.6\;{\rm{m}}} \right)}^2}}}\\ = 7.41 \times {10^6}\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}} \end{array}\]