Chapter 7, Problem 67P

One of the first things you learn in physics class is the law of universal gravitation. $F=G\frac{m_1{m}_2}{r^2}$ , where F is the attractive force between two bodies, $m_1$ and $m_2$ are the masses of the two bodies, r is the distance between the two bodies, and G is the universal gravitational constant equal to $\left(6.67428\pm 0.00067\right)\times {10}^{-11}$ [the units of G are not given here], (a) Calculate the SI units of G. For consistency, give your answer in terms of kg. m. and s. (b) Suppose you don't remember the law of universal gravitation, but you are clever enough to know that F is a function of $G,m_1,m_2$, and r. Use dimensional analysis and the method of repeating variables (show all your work) to generate a nondimensional expression for $F=F\left(G,m_1,m_2,r\right)$. Give your answer as ${\Pi }_1=$function of $\left({\Pi }_2={\Pi }_3,\mathrm{...}\right)$ . (c) Dimensional analysis cannot yield the exact form of the function. However, compare your result to the law of universal gravitation to find the form of the function (e.g., $\Pi ={\Pi }_2^2$ or some other functional form).

To determine

(a)

S.I. unit of $G$.

Answer to Problem 67P

S.I. unit of $G$ is $\frac{{\text{m}}^{\text{3}}}{\text{kg}\cdot {\text{s}}^2}$.

Explanation of Solution

Expression for gravitational force,
$F=\frac{G{m}_1{m}_2}{r^2}$...... (I)
Here, the gravitational constant is $G$, the mass of the first body is $m_1$, the mass of the second body is $m_2$ and the distance between the two bodies is $r$.

Dimensional formula for $F$,
$F=\left[M^1{L}^1{T}^{-2}\right]$

Here, dimension of mass is $\left[M\right]$, dimension of length is $\left[L\right]$ and the dimension of time is $\left[T\right]$.

Dimensional formula for mass,
$m=\left[M^1\right]$

Dimensional formula for $r$,
$r=\left[L^1\right]$

Calculation:

Rearrange the Equation (I) for $G$.

$G=\frac{F{r}^2}{m_1{m}_2}$...... (II)
Substitute $[M^1{L}^1{T}^{-2}]$ for dimensional formula of $F$. $[M^0{L}^1{T}^0]$ for dimensional formula of $r$, $[M^1{L}^0{T}^0]$ for the dimensional formula of $m_1$ and $[M^1{L}^0{T}^0]$ for the dimensional formula of $m_2$ in Equation (II).

$\begin{array}{c}G=\frac{[M^1{L}^1{T}^{-2}]{[M^0{L}^1{T}^0]}^2}{[M^1{L}^0{T}^0][M^1{L}^0{T}^0]}\\ =\frac{[M^1{L}^1{T}^{-2}]{[M^0{L}^1{T}^0]}^2}{{[M^1{L}^0{T}^0]}^2}\\ =\frac{[M^1{L}^1{T}^{-2}][L^2]}{[M^2]}\\ =[M^{-1}{L}^3{T}^{-2}]\end{array}$

Conclusion:

Therefore, the S.I unit of $G$ is $\frac{{\text{m}}^{\text{3}}}{\text{kg}\cdot {\text{s}}^2}$.

To determine

(b)

The non dimensional expression for $F$ using dimensional analysis and method of repeating variables.

Answer to Problem 67P

The required expression is $F=\frac{\frac{m^{2}G}{r^{2}F}}{\varphi \left(\frac{m_2}{m_1}\right)}$.

Explanation of Solution

Write the expression for number of $\Pi$ terms.

$k=n-j$...... (III)
Here, the number of variables are $n$ and the primary dimensions are $j$.

Substitute $5$ for $n$ and $3$ for $j$ in Equation (III).

$\begin{array}{c}k=5-3\\ =2\end{array}$

Therefore, the number of $\Pi$ terms is $2$.

Write the expression for first $\Pi$ term.

${\Pi }_1=r^{a_1}{m}_1^{b_1}{F}^{c_1}G$...... (IV)
Write the expression for the second $\Pi$ term.

${\Pi }_2=r^{a_2}{m}_1^{b_2}{F}^{c_2}{m}_2$...... (V)
Calculation:

The dimensional formula for $\Pi$ term is $[M^0{L}^0{T}^0]$.

Substitute $[M^0{L}^0{T}^0]$ for $\Pi$, $[M^1{L}^1{T}^{-2}]$ for dimensional formula of $F$. $[M^0{L}^1{T}^0]$ for dimensional formula of $r$, $[M^1{L}^0{T}^0]$ for the dimensional formula of $m$ and $[M^{-1}{L}^3{T}^{-2}]$ for the dimensional formula of $G$ in Equation (IV) for first $\Pi$ term.

$\begin{array}{l}[M^0{L}^0{T}^0]=\left[{[M^0{L}^1{T}^0]}^{a_1}{[M^1{L}^0{T}^0]}^{b_1}{[M^1{L}^1{T}^{-2}]}^{c_1}[M^{-1}{L}^3{T}^{-2}]\right]\\ [M^0{L}^0{T}^0]=\left[{[M]}^{b_1+c_1-1}{[L]}^{a_1+c_1+3}{[T]}^{-2c_1-2}\right]\end{array}$...... (VI)
Compare the exponential coefficient of $T$ in Equation (V).

$\begin{array}{l}-2c_1-2=0\\ c_1=\frac{2}{-2}\\ c_1=-1\end{array}$

Compare the exponential coefficient of $M$ in Equation (V).

$\begin{array}{l}{b}_1+c_1-1=0\\ b_1-1-1=0\\ b_1=2\end{array}$

Compare the exponential coefficient of $L$ in Equation (V).

$\begin{array}{l}{a}_1+c_1+3=0\\ a_1-1+3=0\\ a_1=-2\end{array}$

Substitute $-1$ for $c_1$, $0$ for $b_1$ and $-2$ for $a_1$ in Equation (IV) to obtain first $\Pi$ term.

$\begin{array}{c}{\Pi }_1=r^{-2}{m}^2{F}^{-1}G\\ =\frac{m^{2}G}{r^{2}F}\end{array}$

Substitute $[M^0{L}^0{T}^0]$ for $\Pi$, $[M^1{L}^1{T}^{-2}]$ for dimensional formula of $F$. $[M^0{L}^1{T}^0]$ for dimensional formula of $r$, $[M^1{L}^0{T}^0]$, $[M^1{L}^0{T}^0]$ for the dimensional formula of $m_1$ and $[M^1{L}^0{T}^0]$ for the dimensional formula of $m_2$ in Equation (V) to obtain the second $\Pi$ term.

$\begin{array}{l}[M^0{L}^0{T}^0]=\left[{[M^0{L}^1{T}^0]}^{a_1}{[M^1{L}^0{T}^0]}^{b_1}{[M^1{L}^1{T}^{-2}]}^{c_1}[M^{-1}]\right]\\ [M^0{L}^0{T}^0]=\left[{[M]}^{b_2+c_2+1}{[L]}^{a_2+c_2}{[T]}^{-2c_2}\right]\end{array}$...... (VII)

Compare the exponential coefficient of $T$ in Equation (VII).

$\begin{array}{l}-2c_2=0\\ c_2=0\end{array}$

Compare the exponential coefficient of $M$ in Equation (VII).

$\begin{array}{l}{b}_2+c_2+1=0\\ b_2+0+1=0\\ b_2=-1\end{array}$

Compare the exponential coefficient of $L$ in Equation (VII).

$\begin{array}{l}{a}_2+c_2=0\\ a_2+0=0\\ a_2=0\end{array}$

Substitute $-1$ for $c_1$, $0$ for $b_1$ and $-2$ for $a_1$ in Equation (IV) to obtain first $\Pi$ term.

$\begin{array}{c}{\Pi }_2=r^0{m}_1^{-1}{F}^0{m}_2\\ =\frac{m_2}{m_1}\end{array}$...... (IX)
From Equation (VII) and Equation (IX).

${\Pi }_1=\varphi {\Pi }_2$...... (X)
Here, the constant is $\varphi$.

Substitute $\frac{m^{2}G}{r^{2}F}$ for ${\Pi }_1$ and $\frac{m_2}{m_1}$ for ${\Pi }_2$ in Equation (X).

$\begin{array}{l}\frac{m^{2}G}{r^{2}F}=\varphi \left(\frac{m_2}{m_1}\right)\\ F\varphi \left(\frac{m_2}{m_1}\right)=\frac{m^{2}G}{r^2}\\ F=\frac{\frac{m^{2}G}{r^{2}F}}{\varphi \left(\frac{m_2}{m_1}\right)}\end{array}$

Conclusion:

The required expression is $F=\frac{\frac{m^{2}G}{r^{2}F}}{\varphi \left(\frac{m_2}{m_1}\right)}$.

To determine

(c)

The functional relation between ${\Pi }_1$ and ${\Pi }_2$

Answer to Problem 67P

Any functional relation is possible between the two pi terms.

Explanation of Solution

Write the expression for the first pi term.

${\Pi }_1=\frac{m^{2}G}{r^{2}F}$..... (XI)
Write the expression for the second pi term.

${\Pi }_2=\frac{m_2}{m_1}$...... (XII)
Calculation:

Substitute $[M^1{L}^1{T}^{-2}]$ for dimensional formula of $F$. $[M^0{L}^1{T}^0]$ for dimensional formula of $r$, $[M^1{L}^0{T}^0]$, $[M^1{L}^0{T}^0]$ for the dimensional formula of $m_1$ in Equation (XI).

$\begin{array}{c}{\Pi }_1=\frac{\left[M^2\right]\left[M^{-1}{L}^3{T}^{-2}\right]}{\left[L^2\right]\left[M^1{L}^1{T}^{-2}\right]}\\ =\frac{\left[M^1{L}^3{T}^{-2}\right]}{\left[M^1{L}^3{T}^{-2}\right]}\\ =\left[M^0{L}^0{T}^0\right]\end{array}$

Substitute $\left[M^1\right]$ for $m_1$ and $\left[M^1\right]$ for $m_2$ in Equation (XII).

$\begin{array}{c}{\Pi }_2=\frac{\left[M^1\right]}{\left[M^1\right]}\\ =\left[M^0{L}^0{T}^0\right]\end{array}$

Conclusion:

Any functional relation is possible between the two pi terms.