2.3 Measurement of length

Measurement of Length

You are already familiar with some direct methods for the measurement of length. For example, a metre scale is used for lengths from \(10^{-3} \mathrm{~m}\) to \(10^{2}\) m. A vernier callipers is used for lengths to an accuracy of \(10^{-4} \mathrm{~m}\). A screw gauge and a spherometer can be used to measure lengths as less as to \(10^{-5} \mathrm{~m}\). To measure lengths beyond these ranges, we make use of some special indirect methods.

Large distances such as the distance of a planet or a star from the earth cannot be measured directly with a metre scale. An important method in such cases is the parallax method.

To measure the distance \(D\) of a faraway planet \(S\) by the parallax method, we observe it from two different positions (observatories) A and \(\mathrm{B}\) on the Earth, separated by distance \(\mathrm{AB}=b\) at the same time as shown in Fig. 2.2. We measure the angle between the two directions along which the planet is viewed at these two points. The \(\angle\) ASB in Fig. 2.2 represented by the symbol \(\theta\) is called the parallax angle or parallactic angle.

As the planet is very far away, \(\frac{b}{D}<<1\), and therefore, \(\theta\) is very small. Then we approximately take \(\mathrm{AB}\) as an arc of length \(b\) of \(\mathrm{a}\) circle with centre at \(\mathrm{S}\) and the distance \(D\) as the radius \(\mathrm{AS}=\mathrm{BS}\) so that \(\mathrm{AB}=b=D \theta\) where \(\theta\) is in radians.
\(D=\frac{b}{\theta}\)

    Figure 2.2 Parallax method

Having determined \(D\), we can employ a similar method to determine the size or angular diameter of the planet. If \(d\) is the diameter of the planet and \(\alpha\) the angular size of the planet (the angle subtended by \(d\) at the earth),we have \(\alpha=d / D\)

The angle \(\alpha\) can be measured from the same location on the earth. It is the angle between the two directions when two diametrically opposite points of the planet are viewed through the telescope. Since \(D\) is known, the diameter \(d\) of the planet can be determined using the equation \(\alpha=d / D\).

Example 2.1: 

A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower \(\mathrm{C}\) and spots a very distant object \(\mathrm{O}\) in line with AC. He then walks perpendicular to \(\mathrm{AC}\) up to \(\mathrm{B}\), a distance of \(100 \mathrm{~m}\), and looks at \(\mathrm{O}\) and \(\mathrm{C}\) again. Since \(\mathrm{O}\) is very distant, the direction \(\mathrm{BO}\) is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle \(\theta\) \(=40^{\circ} \quad(\theta\) is known as ‘parallax’) estimate the distance of the tower \(\mathrm{C}\) from his original position A.

Answer: We have, parallax angle \(\theta=40^{\circ}\); From Fig. 2.3, AB \(=\mathrm{AC} \tan \theta\)
\(\mathrm{AC}=\mathrm{AB} / \tan \theta=100 \mathrm{~m} / \tan 40^{\circ}\)
\(=100 \mathrm{~m} / 0.8391=119 \mathrm{~m}\)

Example 2.2: 

The moon is observed from two diametrically opposite points A and B on Earth. The angle \(\theta\) subtended at the moon by the two directions of observation is \(1^{\circ} 54^{\prime}\). Given the diameter of the Earth to be about \(1.276 \times 10^{7} \mathrm{~m}\), compute the distance of the moon from the Earth.

Answer:  We have \(\theta=1^{\circ} 54^{\prime}=114^{\prime}\)
\(=(114 \times 60)^{\prime \prime} \times\left(4.85 \times 10^{-6}\right)\) rad
\(
=3.32 \times 10^{-2} \text { rad, }
\)
since \(1^{\prime \prime}=4.85 \times 10^{-6}\) rad.
Also \(b=A B=1.276 \times 10^{7} \mathrm{~m}\)
Hence from the following equation, we can calculate the earth-moon
distance, \(\quad D=b / \theta\)
\(
=\frac{1.276 \times 10^{7}}{3.32 \times 10^{-2}}=3.84 \times 10^{8} \mathrm{~m}
\)

Example:

The Sun’s angular diameter is measured to be \(1920^{\prime \prime}\). The distance \(D\) of the Sun from the Earth is \(1.496 \times 10^{11} \mathrm{~m}\). What is the diameter of the Sun?

Answer: Sun’s angular diameter \(\alpha\)
\(
\begin{aligned}
&=1920^{\prime \prime} \\
&=1920 \times 4.85 \times 10^{-6} \mathrm{rad} \\
&=9.31 \times 10^{-3} \mathrm{rad}
\end{aligned}
\)
Sun’s diameter
\(
\begin{aligned}
d &=\alpha D \\
&=\left(9.31 \times 10^{-3}\right) \times\left(1.496 \times 10^{11}\right) \mathrm{m} \\
&=1.39 \times 10^{9} \mathrm{~m}
\end{aligned}
\)

Range of Lengths:

The sizes of the objects we come across in the universe vary over a very wide range. These may vary from the size of the order of \(10^{-14} \mathrm{~m}\) of the tiny nucleus of an atom to the size of the order of \(10^{26} \mathrm{~m}\) of the extent of the observable universe. Table \(2.2\) gives the range and order of lengths and sizes of some of these objects.

Table2.3: Range and order of lengths

\(\begin{array}{|l|l|}
\hline \text { Size of object or distance } & \text { Length (m) } \\
\hline \text { Size of a proton } & 10^{-15} \\
\hline \text { Size of atomic nucleus } & 10^{-14} \\
\hline \text { Size of hydrogen atom } & 10^{-10} \\
\hline \text { Length of typical virus } & 10^{-8} \\
\text { Wavelength of light } & 10^{-7} \\
\text { Size of red blood corpuscle } & 10^{-5} \\
\hline \text { Thickness of a paper } & 10^{-4} \\
\hline \text { Height of the Mount Everest above sea level } & 10^{4} \\
\hline \text { Radius of the Earth } & 10^{7} \\
\hline \text { Distance of moon from the Earth } & 10^{8} \\
\hline \text { Distance of the Sun from the Earth } & 10^{11} \\
\hline \text { Distance of Pluto from the Sun } & 10^{13} \\
\hline \text { Size of our galaxy } & 10^{21} \\
\hline \text { Distance to Andromeda galaxy } & 10^{23} \\
\hline \text { Distance to the boundary of observable universe } & 10^{26} \\
\hline
\end{array}\)