Use dimensional analysis (Section 1-8) to obtain the form for the centripetal acceleration,\({a_R} = {v^2}/r\).

Short Answer

Expert verified

The centripetal or radial acceleration is \(\left( {\frac{{{v^2}}}{r}} \right) = \;\frac{L}{{{T^2}}}\; = \left( {{a_R}} \right)\).

Step by step solution

01

Find the centripetal acceleration \({{\bf{a}}_{\bf{R}}} = \;{{\bf{v}}^{\bf{2}}}/{\bf{r}}\).

It is known that the dimension of the acceleration, which is the right-hand side of the given equation, is equal to:

\(\left( {{a_R}} \right) = \frac{L}{{{T^2}}}\)

The speed is given by:

\(v = \;\frac{{{\rm{Distance}}}}{{{\rm{Time}}}}\)

02

Find dimensions of the speed

Now, the dimension of the speed is\(\left( v \right) = \frac{L}{T}\), and the radius speed has a dimension of\(\left( r \right) = L\). So, the dimension of the left-hand side is equal to:

\(\begin{aligned}\left( {\frac{{{v^2}}}{r}} \right) &= \frac{{\frac{{{L^2}}}{{{T^2}}}}}{L}\\ &= \frac{L}{{{T^2}}}\end{aligned}\)

Therefore,

\(\left( {{a_R}} \right) = \;\left( {\frac{{{v^2}}}{r}} \right)\)

So, the equation is homogenous.

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Table 5-3 Principal Moons of Jupiter

Moon

Mass(kg)

Period
(Earth days)

Mean distance from Jupiter (km)

Io

\({\bf{8}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)

1.77

\({\bf{422 \times 1}}{{\bf{0}}^{\bf{3}}}\)

Europe

\({\bf{4}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)

3.55

\({\bf{671 \times 1}}{{\bf{0}}^{\bf{3}}}\)

Ganymede

\({\bf{15 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)

7.16

\({\bf{1070 \times 1}}{{\bf{0}}^{\bf{3}}}\)

Callisto

\({\bf{11 \times 1}}{{\bf{0}}^{{\bf{22}}}}\)

16.7

\({\bf{1883 \times 1}}{{\bf{0}}^{\bf{3}}}\)

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