In this article, we shall study the parallax and other methods to find the distance between two heavenly bodies.

Parallax:

Hold a pencil vertically in front of you at a certain distance against a point on a background like a wall. Now view the pencil through your left eye closing your right eye and then view the pencil through your right eye closing your left eye. You will notice that with respect to the fixed point on the wall the position of the pencil is changing. This phenomenon is known as parallax. The distance between the two points of observation is called the basis. In this example, the basis is the distance between the eyes.

Parallax is defined as the apparent shift of a body with respect to another, with the shift of eye. The distances between planets and stars from the Earth cannot be measured directly with a metre scale. Hence indirect method called the parallax method is used.

Example – 01:

Calculate the angle of a) 1^O (degree) b) 1’ (minute of arc or arcmin) and c) 1’’ (second of arc or arc second) in radians.

Solution:

Determination of the Distance Between a Faraway Planet or a Near Star and the Earth (Parallax Method – I):

To measure the distance d of a faraway planet or near star S₁ by the parallax method, we observe it from two different positions (observatories) A and B on the Earth, separated by distance b at the same time as shown in Fig. Some distant star S is taken as reference. We measure the angle between the two directions along which the planet is viewed at these two points. The∠ASB called parallax angle = θ = θ₁ + θ₂, Now the distance d << b. 

sinθ = θ = b/d
∴  d = b/θ

θ should be measured in radians.

Determination of the Distance Between a Faraway Planet or a Near Star and the Earth (Parallax Method – II):

To measure the distance d of a faraway planet, the moon or the near star (the sun) S₁ by the parallax method, we observe it from two diametrically opposite positions on the earth. A and B on the Earth, separated by distance b at the same time as shown in Fig. Some distant star S is taken as reference. We measure the angle between the two directions along which the planet is viewed at these two points. The∠ASB called parallax angle = θ = θ₁ + θ₂, Now the distance d << b. 

b = AB = 2R = 2 x radius of earth

sinθ = θ = b/d
∴  d = b/θ

θ should be measured in radians.

Example – 02:

The moon is observed from two diametrically opposite points A and B on Earth. The angle θ subtended at the moon by the two directions of observation is 1°54’. Given the diameter of the Earth to be about 1.276 x 10⁷ m, compute the distance of the moon from the Earth.

We have θ = 1^O54’ = 1 x 60’ + 54 = 114’
θ = 114’ x 2.91 x 10^-4 = 3.32 x 10^-2   rad

Also b = AB = diameter of earth = 1.276 x 10⁷ m

Ans: The distance of the moon from the earth is 3.84 x 10⁵ km

Example – 03:

Two parallax of a heavenly body measured from two points diametrically opposite on equator of Earth is 2.0′. Calculate the distance of the heavenly body if the radius of the earth is 6400 km.

We have θ = 2’ = 2 x 2.91 x 10^-4 = 5.82 x 10^-4   rad

Also b = AB = diameter of earth = 2 x 6400 km = 12800 km

Now, d = b/θ = 12800 / 5.82 x 10^-4

d = 2.2 x 10⁷ km = 2.2 x 10¹⁰ m

Ans: The distance of the heavenly body from the earth is 2.2 x 10¹⁰ m

Determination of The Distance of a Faraway Planet or a Near Star from the Earth (Parallax Method – III):

To measure the distance d of a faraway planet or the near star S₁ by the parallax method, we observe it from two diametrically opposite positions on the earth’s orbit around the sun at a time gap of six months. A and B on the Earth’s orbit, separated by distance b as shown in Fig. Some distant star S is taken as reference. We measure the angle between the two directions along which the planet is viewed at these two points. The∠ASB called parallax angle = θ = θ₁ + θ₂, Now the distance d << b.

b = AB = 2 AU = 2 x 1.496 x 10¹¹ m

sinθ = θ = b/d
∴  d = b/θ

θ should be measured in radians.

Example – 04:

When the observations are taken at an interval of 6 months, the angle of parallax for star is 0.4’’. Find the distance of star in light year and parsec.

θ = 0^.44’’ = 0.4 x 4.847 x 10^-6 = 1.939  x 10^-6  rad

Ans: The distance of the star from the earth is 16.3 lightyears or 5 parsec.

Determination of Diameter of the Moon or the Sun:

The angular diameter of the moon (or the sun) is the angle subtended by two diametrically opposite ends of the moon (or the sun) at a point on the Earth.

Let D be the diameter of the moon (or the sun) and d be the distance between the moon (or the sun) from the earth. D << d

arc length AB = D = d θ

θ should be measured in radians.

Example – 05:

The Sun’s angular diameter is measured to be 1920”. The distance D of the Sun from the Earth is 1.496 × 10¹¹ m. What is the diameter of the Sun?

Sun’s angular diameter θ = 1920″ = 1920 x 4.85 x 10^-6 = 9.31 x 10^-3 rad
Sun’s diameter = D = d θ

d = (1.496 × 10¹¹) x (9.31 x 10^-3 ) = 1.39 x 10⁹  m

Ans: The diameter of the sun is 1.39 x 10⁹ m

Example – 06:

The angle subtended by the Moon at a point on the Earth is 0^o31’. If the distance of the Moon from the Erath is 3.84 x 10⁵ km, find the diameter of the moon.

θ = 0^o31’ = 31 x 2.908 x 10^-4 = 9.01 x 10^-3 rad

Now D = d θ = 3.84 x 10⁵ x 9.01 x 10^-3 = 3.46 x 10³  km

Ans: Diameter of the moon is 3.46 x 10³ km

Determination of Angular Distance Between Two Stars:

Let θ₁  be the elevation of a distant star S₁ and θ₂  be the elevation of another distant star S₂ such that (θ₂ > θ₁). Then the quantity (θ₂ – θ₁) is called the angular distance between the stars.

It is to be noted that the smaller angular distance between two distant stars does not imply that the two stars are close to each other. There is a possibility that distance between them is very very large (hundreds of light-years).

Determination of the Distance of Inferior Planets from the Earth (Copernicus Method):

The planets which are closer to the Sun than the Earth are called inferior planets. The Mercury and Venus are two inferior planets. Copernicus assumed the orbits of inferior planets as perfectly circular.

The angle formed at the Earth between the Earth planet direction and the Earth-Sun Direction is called the planet elongation and is denoted by symbol ε. The planets elongation changes continuously.  When elongation acquires maximum value it appears to be farthest from the sun, At this time the Sun and the Earth subtends an angle of 90° at the planet.  This maximum elongation is noted.

SE = 1 AU = 1.496 x 10¹¹ m

Δ EPS is a right-angled triangle

sin ε = SP/ SE

∴  SP = SE sin ε

Example – 07:

Copernicus assumed the orbit of Mercury to be circular and estimated its orbital radius as 0.38 AU. Using this radius derive angle of maximum elongation for mercury and its distance from the Earth when the elongation is maximum.

We have SP = SE sin ε

Where SP = orbital radius of Planet = 0.38

SE = orbital radius of the Earth = 1 AU

0.38 = 1 x sin ε

ε = sin^-1(0.38) = 22°20′

Applying Pythagoras theorem to ΔEPS

SE² = EP² + SP²

∴ EP² = SE² – SP²

∴ EP² = (1)² – (0.38)² = (1+0.38)(1-0.38)

∴ EP² = 1.38 x 0.62 = 0.8556

∴ EP = 0.925 AU

∴ EP = 0.925 x 1.496 × 10¹¹  m = 1.384 ×  10¹¹  m

Ans: The angle of maximum elongation for Mercury is 22°20′ and the distance between the Earth and Mercury at maximum elongation is 1.384 × 10¹¹  m

Example – 08:

For Venus the angle of maximum elongation is found to be approximately equal to 47°. Determine the distance between the Sun and Venus and that between the Erath and the Venus. Also, find orbital period of Venus.

We have SP = SE sin ε

Where SP = orbital radius of Planet

SE = orbital radius of the Earth = 1 AU

SP = 1 x sin 47° = 1 x 0.7314 = 0.7314 AU

∴ SP = 0.7314 x 1.496 × 10¹¹  m = 1.094 ×  10¹¹  m

Applying Pythagoras theorem to ΔEPS

SE² = EP² + SP²

∴ EP² = SE² – SP²

∴ EP² = (1)² – (0.73)² = (1+0.73(1-0.73)

∴ EP² = 1.73 x 0.27 = 0.4671

∴ EP = 0.6834 AU

∴ EP = 0.6834 x 1.496 × 10¹¹  m = 1.022 ×  10¹¹  m

Ans: The distance between the sun and venus is 1.094 × 10¹¹  m and the distance between the Earth and Venus at maximum elongation is 1.022 × 10¹¹  m, Period of Venus is 226 days

Determination of the Distance of Superior Planets from the Sun:

The planets which are farther to the Sun than the Earth are called inferior planets. The Mars, The Jupiter, The Saturn, The Neptune and The Uranus are superior planets.

By Keppler’s period law of orbiting satellite “The square of the time period of the satellite is directly proportional to the cube of the semimajor axis of the orbit.

Example – 09:

A Period of a planet orbiting around the sun is half that of the earth. What is the orbital radius of the planet?

∴  r_P = 0.63 x 1.496 ×  10¹¹  m = 9.42 ×  10¹⁰  m

Thus r_P/ r_E = 0.63/1 = 0.63

Ans: The orbital size of the planet is smaller than that of the Earth by a factor of 0.63

Measurement of Distances of Star by Intensity Method (Optical or Spectroscopic Method):

This method uses the inverse square law of photometry, it states that “The intensity of illumination at a point varies inversely as the square of the distance from the source of light

Thus I ∝ 1/r²

This method is used to find the distance of a star from the earth. The assumption of this method is the intrinsic brightness of all the stars is the same. This method gives the approximate result.

In the above formula, I₁ is the apparent brightness of a star at a distance r₁ from the earth and I₂ is the apparent brightness of a star at a distance r₂ from the earth. If we know the intensity ratio and the distance of one star from the earth, the distance of another star from the earth can be calculated.

Measurement of Distances of Star by Doppler Shift Method (Optical or Spectroscopic Method):

When a body that is emitting radiation has a non-zero velocity relative to an observer, the wavelength of the emission will be shortened or lengthened, depending upon whether the body is moving towards or away from an observer. This change in observed wavelength or frequency is known as the Doppler shift.

If the object is moving towards an observer, then the emission will be blueshifted – i.e. the wavelength of the emission will be shortened, moving it towards the blue end of the spectrum. If the object is moving away from an observer, then the emission will be redshifted i.e. the wavelength of the emission will increase, moving it towards the red end of the spectrum.

Generally, by determining the redshift and using doppler’s formulae the distance of the star and its speed at which it is moving away from the earth can be found.

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Physics is a science of measurement. In science and engineering, we perform experiments. During experiments, we have to take readings. Thus all these experiments require some measurements to be made. During the production of mechanical products, we have to measure the parts so as to find whether the part is made as per the specifications. Thus measurements are necessary for production and quality control. A measurement is a quantitative description of one or more fundamental properties compared to a standard. To measure length is a very important step during the performance of experiments. Measurement can be done directly or indirectly. For direct methods metre scale, vernier callipers, micrometre screw gauges are used. In this article, we shall study the use of metre scale to measure length, diameter, etc.

When measurements are taken directly using tools, instruments, or other calibrated measuring devices, they are called direct measurements. e.g. Measurement of the length of a table by metre scale. When the measurement must be done through a formula or other calculations, the measurement is called indirect measurement. e.g. Measurement of the radius of the Earth.

Old Methods of Measurement of a Length

To measure lengths units used were a finger, palm, span, cubit, foot, yard, fathom, furlong etc.  The length of an inch was originally decided by Edward-II of England in the early 1300s as the length of three grains of barley laid end to end. Inch was divided into three parts called barley.

Measurement of Length:

The length is a fundamental quantity, it is used to measure a distance between two points in space. S.I. unit of length is a metre (m) and c.g.s. unit of length is centimetre (cm). Other practical units of measurement of length are micrometre (mm), millimetre (mm), kilometre (km), Angstrom (A°) etc.

Length of accessible objects can be measured directly using a metre scale, vernier callipers, micrometre screw gauge, measuring tapes, travelling microscopes etc. Length of non-accessible objects is measured indirectly.

Metre Scale:

one metre is divided into 100 equal parts, each part is called centimetre. Each centimetre is divided into 10 equal parts, each part is called millimetre. Engineering metre scale has centimetres marked on one edge and inches marked on another edge. Centimetres have decimal divisions while inches have fractional divisions. metre scale may have bevelled edges to avoid errors due to parallax.

Least count is the smallest measurement that can be taken accurately with an instrument or device. The least count of a metre scale is 1 mm or 0.1 cm.

Linear Measurements (Direct Method):

Place the metre scale along the object so that the ‘0’ mark on the metre scale coincides with one end of the object and reading at the other end of the scale indicates the length of the object. To avoid wear and tear off the end of the scale, sometimes the scale is placed along the object and readings at the ends of the object are taken. The length of the object is obtained by subtracting higher reading from lower reading.

The eye must be kept vertically above the end of the object so that the corresponding graduation can be read clearly. If the eye is not kept exactly vertically above the end of the object, it leads into error called parallax error.

In case, the end of the object lies between the two small divisions of the scale, the correct length is reported by noting the marking nearer to the end of the object. Limitation of metre scale is that it cannot measure the length of an object smaller than 1 mm or 0.1 cm.

Measurement of Length of a Curve

Thread Ruler Method (Direct Method):

In this case, a thread is laid along the curve. Then the length of the thread is measured using a metre scale.

Divider Ruler Method (Direct Method):

In this method, the curve is divided into small straight segments. Length of each such segment is measured using divider and scale. Then the total length of the curve can be obtained by adding lengths of all individual segments.

External diameter of cylinder or sphere

Blocks Ruler Method (Direct Method):

A cylinder or sphere whose external diameter is to be measured is placed between two blocks. The reading on the scale corresponding to the inner edge is x cm while that corresponding to the outer edge is y cm. Then the diameter is the absolute value of (x-y) cm.

Caliper Ruler Method (Direct Method):

To measure the external diameter of sphere or cylinder, an external calliper is used. In this method, the cylinder or sphere whose external diameter is to be measured is placed between two jaws of the calliper. The position of the calliper is fixed by tightening the screw. The distance between the jaws gives the external diameter of the cylinder or the sphere.

External Diameter of Cylinder: (Indirect Method):

A known number of turns (N) of a thin wire are wound on a cylinder whose diameter is to be measured. Then the wire is unwounded and straightened. Then its length is measured. The diameter of the rod can be obtained by the formula

To measure the internal diameter of the cylinder, an internal calliper is used.

External Diameter of Thin Wire: (Indirect Method):

A known number of turns of the wire whose diameter is to be measured are wound on a scale or on a rod of uniform diameter. The length of turns on the scale or the rod is measured. The diameter of the wire is calculated by dividing the length of turns by the number of turns.

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