In the previous section, we saw the units for measuring lengths (small and large ) in the SI system. We also saw the reason why the distant objects like moon and stars appear to be stationary even when we move. In this section, we will see the indirect methods used for measuring large distances.
An important method for measuring large distances is the parallax method.
First we will see some basics about this method. We will write them in steps.
1. The fig.2.4 below shows the view of a class room directly from above.
• In such a view, the black board will be seen as a thick black line.
• On the black board, a point P is marked in red color.
2. An observer stands in front of the black board and looks towards P.
• An object Q is placed in front of the observer.
3. The observer closes his right eye and views Q through his left eye.
• This is shown in fig.2.4(b).
• The observer is able to see Q through the left eye.
• He is able to see Q because, the ray of light from Q reaches the left eye.
• This ray is indicated by the blue dashed line in fig.b.
4. But this ray appears to come from the right side of P.
• So Q appears to be on the right side of P.
• Thus we get an interesting result:
♦ When Q is viewed with both eyes open, the observer sees it’s actual position.
♦ When Q is viewed with only the left eye, the observer sees that it is shifted to the right of P.
5. Next, the observer closes his left eye and views Q through his right eye.
• This is shown in fig.2.4(c).
• The observer is able to see Q through the right eye.
• He is able to see Q because, the ray of light from Q reaches the right eye.
• This ray is indicated by the blue dashed line in fig.c.
6. But this ray appears to come from the left side of P.
• So Q appears to be on the left side of P.
• Thus we get an interesting result:
♦ When Q is viewed with both eyes open, the observer sees it’s actual position.
♦ When Q is viewed with only the right eye, the observer sees that it is shifted to the left of P.
7. In both figs.(b) and (c), there is no change in the position of P.
• This is because, P is at a far more distance from the observer than Q.
• Recall that in the previous section, we saw this:
The near objects like trees move past the observer while he is moving in a car. But the distant object like the moon appears to be stationary.
• So in our present case, P is stationary.
■ The apparent shift in the position of an object (with respect to a stationary object) when viewed from different points is called parallax.
8. Since P is stationary, it serves as a ‘bench mark’ or 'reference' to measure the following two angles.
(i) Angle θ1 in fig.b (ii) Angle θ2 in fig.c
• Once those angles are measured, we get the total angle theta shown in fig.c
■ This theta is called the parallax angle.
■ The distance between the two points of observation is called the basis.
9. In our present case, the basis is the distance between the two eyes.
• If we know the basis and the parallax angle, we can calculate the distance between the observer and Q.
• For this, we use trigonometry as shown in fig.2.5(a) below:
• In fig.2.5(a), b is the basis
♦ The contribution from the left side triangle for making up b is D tan θ1
♦ The contribution from the right side triangle for making up b is D tan θ2
10. So we get b = D tan θ1 + D tan θ2
⇒ b = D(tan θ1 + tan θ2)
⇒ $\mathbf\small{D=\frac{b}{(\tan\, \theta_1+\tan\, \theta_2)}}$
■ This method of calculating distances is called Parallax method.
11. For small distances like 'distances within the class room', we do not need this method.
• This method is used for finding large distances like from earth to other planets or stars. It can be explained using fig.2.5(b) above.
12. A and B are two points on the surface of the earth.
• S is a planet or star in space. We want the distance D from the earth to S.
• For that, we make use of a 'very far away star' S1 as the reference.
• Recall that, a ‘very far away object’ will appear to be at the same spot.
♦ We can look at it from different places. It’s position will not change.
• But that is not the case with a near object like S.
♦ It will change positions.
13. First we observe S from A.
• S will appear to be on the right side of S1
• In this position, the angle θ1 is measured.
(The line AS can be considered as the axis of the telescope used for observation)
14. Then we observe S from B.
• S will appear to be on the left side of S1
• In this position, the angle θ2 is measured.
(The line BS can be considered as the axis of the telescope used for observation)
15. As before, we get: $\mathbf\small{D=\frac{b}{(\tan\, \theta_1+\tan\, \theta_2)}}$.
1. The parallax angle θ is given by θ = (θ1+θ2) .
• We know that, since S is very far away, θ will be very small.
2. Suppose that S is the center of a circle.
• Then SA and SB can be considered as radii.
3. The arc AB should normally be a curve, which is a part of the circle.
• But since θ is very small, AB can be considered to be a straight line.
4. That means, since θ is very small, the following two items are equal:
(i) arc AB (ii) straight length AB.
• From our math classes, we know that $\mathbf\small{\text{angle}=\frac{\text{arc}}{\text{radius}}}$.
• So we get $\mathbf\small{\theta=\frac{b}{D}}$
• From this we get:
Eq.2.1: $\mathbf\small{D=\frac{b}{\theta}}$
♦ Where θ is in radians.
1. PQ is the diameter of the planet or star.
• We can say: P and Q are two diametrically opposite points on the planet.
• Length of PQ = d.
• So d is the diameter of the planet. We want to find d.
2. From the same position A on the surface of the earth, we make two observations.
• First we observe end P.
(The line AP can be considered as the axis of the telescope used for observation)
3. Then we observe end Q.
(The line AQ can be considered as the axis of the telescope used for observation)
4. When the two observations are made, the instrument will give the angle 𝛂.
■ 𝛂 is called the angular diameter of the planet or star.
5. We have $\mathbf\small{\text{angle}=\frac{\text{arc}}{\text{radius}}}$
• So we get $\mathbf\small{\alpha=\frac{d}{D}}$
• From this we get:
Eq.2.2: $\mathbf\small{d=\alpha \, D}$
An important method for measuring large distances is the parallax method.
First we will see some basics about this method. We will write them in steps.
1. The fig.2.4 below shows the view of a class room directly from above.
 |
| Fig.2.4 |
• On the black board, a point P is marked in red color.
2. An observer stands in front of the black board and looks towards P.
• An object Q is placed in front of the observer.
3. The observer closes his right eye and views Q through his left eye.
• This is shown in fig.2.4(b).
• The observer is able to see Q through the left eye.
• He is able to see Q because, the ray of light from Q reaches the left eye.
• This ray is indicated by the blue dashed line in fig.b.
4. But this ray appears to come from the right side of P.
• So Q appears to be on the right side of P.
• Thus we get an interesting result:
♦ When Q is viewed with both eyes open, the observer sees it’s actual position.
♦ When Q is viewed with only the left eye, the observer sees that it is shifted to the right of P.
5. Next, the observer closes his left eye and views Q through his right eye.
• This is shown in fig.2.4(c).
• The observer is able to see Q through the right eye.
• He is able to see Q because, the ray of light from Q reaches the right eye.
• This ray is indicated by the blue dashed line in fig.c.
6. But this ray appears to come from the left side of P.
• So Q appears to be on the left side of P.
• Thus we get an interesting result:
♦ When Q is viewed with both eyes open, the observer sees it’s actual position.
♦ When Q is viewed with only the right eye, the observer sees that it is shifted to the left of P.
7. In both figs.(b) and (c), there is no change in the position of P.
• This is because, P is at a far more distance from the observer than Q.
• Recall that in the previous section, we saw this:
The near objects like trees move past the observer while he is moving in a car. But the distant object like the moon appears to be stationary.
• So in our present case, P is stationary.
■ The apparent shift in the position of an object (with respect to a stationary object) when viewed from different points is called parallax.
8. Since P is stationary, it serves as a ‘bench mark’ or 'reference' to measure the following two angles.
(i) Angle θ1 in fig.b (ii) Angle θ2 in fig.c
• Once those angles are measured, we get the total angle theta shown in fig.c
■ This theta is called the parallax angle.
■ The distance between the two points of observation is called the basis.
9. In our present case, the basis is the distance between the two eyes.
• If we know the basis and the parallax angle, we can calculate the distance between the observer and Q.
• For this, we use trigonometry as shown in fig.2.5(a) below:
 |
| Fig.2.5 |
♦ The contribution from the left side triangle for making up b is D tan θ1
♦ The contribution from the right side triangle for making up b is D tan θ2
10. So we get b = D tan θ1 + D tan θ2
⇒ b = D(tan θ1 + tan θ2)
⇒ $\mathbf\small{D=\frac{b}{(\tan\, \theta_1+\tan\, \theta_2)}}$
■ This method of calculating distances is called Parallax method.
11. For small distances like 'distances within the class room', we do not need this method.
• This method is used for finding large distances like from earth to other planets or stars. It can be explained using fig.2.5(b) above.
12. A and B are two points on the surface of the earth.
• S is a planet or star in space. We want the distance D from the earth to S.
• For that, we make use of a 'very far away star' S1 as the reference.
• Recall that, a ‘very far away object’ will appear to be at the same spot.
♦ We can look at it from different places. It’s position will not change.
• But that is not the case with a near object like S.
♦ It will change positions.
13. First we observe S from A.
• S will appear to be on the right side of S1
• In this position, the angle θ1 is measured.
(The line AS can be considered as the axis of the telescope used for observation)
14. Then we observe S from B.
• S will appear to be on the left side of S1
• In this position, the angle θ2 is measured.
(The line BS can be considered as the axis of the telescope used for observation)
15. As before, we get: $\mathbf\small{D=\frac{b}{(\tan\, \theta_1+\tan\, \theta_2)}}$.
Another method:
• We know that, since S is very far away, θ will be very small.
2. Suppose that S is the center of a circle.
• Then SA and SB can be considered as radii.
3. The arc AB should normally be a curve, which is a part of the circle.
• But since θ is very small, AB can be considered to be a straight line.
4. That means, since θ is very small, the following two items are equal:
(i) arc AB (ii) straight length AB.
• From our math classes, we know that $\mathbf\small{\text{angle}=\frac{\text{arc}}{\text{radius}}}$.
• So we get $\mathbf\small{\theta=\frac{b}{D}}$
• From this we get:
Eq.2.1: $\mathbf\small{D=\frac{b}{\theta}}$
♦ Where θ is in radians.
• Once we find the distance D to the planet, we can find the diameter of that planet. It can be explained using fig.2.5(c) above.
• We can say: P and Q are two diametrically opposite points on the planet.
• Length of PQ = d.
• So d is the diameter of the planet. We want to find d.
2. From the same position A on the surface of the earth, we make two observations.
• First we observe end P.
(The line AP can be considered as the axis of the telescope used for observation)
3. Then we observe end Q.
(The line AQ can be considered as the axis of the telescope used for observation)
4. When the two observations are made, the instrument will give the angle 𝛂.
■ 𝛂 is called the angular diameter of the planet or star.
5. We have $\mathbf\small{\text{angle}=\frac{\text{arc}}{\text{radius}}}$
• So we get $\mathbf\small{\alpha=\frac{d}{D}}$
• From this we get:
Eq.2.2: $\mathbf\small{d=\alpha \, D}$
Now we will see some solved examples.
Solved example 2.1
Convert the following angle measures into radians.
(a) 1o (b) 1' (c) 1''
[Hint: 360o = 2𝝅 rad, 1o = 60' and 1' = 60'']
Solution:
Part (a)
1. We have: 360o = 2𝝅 rad.
2. So $\mathbf\small{1^o=\frac{2 \pi}{360}=\frac{\pi}{180}\, \text{rad}}$
3. Substituting the value of 𝝅, we get: $\mathbf\small{1^o=\frac{3.14}{180}\, \text{rad}}$ = 1.745×10-2 rad.
Part (b)
1. We have: 1o = 60'
2. Applying the result obtained in part (a), we get: 1o = 1.745×10-2 rad = 60'.
3. So $\mathbf\small{1'=\frac{1.745 \times 10^{-2}}{60}=2.908\times 10^{-4}\, \text{rad}}$
Part (c)
1. We have: 1' = 60''
2. Applying the result obtained in part (b), we get: 1' = 2.908×10-4 rad = 60''
3. So $\mathbf\small{1''=\frac{2.908 \times 10^{-4}}{60}=4.847\times 10^{-6}\, \text{rad}}$
Solved example 2.2
A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m, and looks at O and C again. Since O is very distant, the direction 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 θ = 40o (θ is known as ‘parallax’) estimate the distance of the tower C from his original position A.
Solution:
1. In fig.2.6(a) below, point A is the initial position of the man
• He looks in the direction AO.
2. B is the final position of the man.
• When he is at B, he can see the distant object O. For that, he does not have to make any change in the direction of vision. This is because, distant objects do not change position. Since there is no change in direction for viewing O, the red dashed line BD is parallel to the initial direction AO.
3. But to view C from B, he has to make a shift in the direction of view.
• This shift is 40o. This is indicated by the ∠CBD.
4. AO and BD are parallel.
• So ∠ACB = ∠CBD = 40o
5. In the right triangle ABC, we have: tan 40 = 100⁄AC
⇒ AC = 100⁄tan 40 = 119 m.
Solved example 2.3
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 1o 54′. Given the diameter of the Earth to be about 1.276 × 107 m, compute the distance of the moon from the Earth.
Solution:
1. In the fig.2.6(b) above, A and B are the end points of a diameter of the earth.
• Given that AB = 1.276 × 107 m.
• We have: $\mathbf\small{D=\frac{b}{\theta}}$
2. In our present case:
• b = 1.276 × 107 m.
• θ = 1o 54′
♦ We have to conv.ert this 'angle in deg' to 'angle in rad'.
♦ 1o 54′ = 114'
♦ 1' = 2.908 ×10-4 rad.
♦ So 114' = 114 × 2.908 ×10-4 = 3.31 × 10-2 rad.
3. Substituting the values, we get:
$\mathbf\small{D=\frac{1.276 \times 10^{7}}{3.315 \times 10^{-2}}=3.85 \times 10^{8}}$ m.
Solved example 2.4
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 ?
Solution:
1. In the fig.2.6(c) above, P and Q are the end points of a diameter of the sun.
• We have: d = 𝛂D
2. In our present case:
• D = 1.496 × 1011 m.
• 𝛂 = 1920''
♦ We have to convert this 'angle in deg' to 'angle in rad'.
♦ 1'' = 4.847 ×10-6 rad.
♦ So 1920'' = 1920 × 4.847 ×10-6 = 9306.24 × 10-6 rad.
3. Substituting the values, we get:
d = 9306.24 × 10-6 × 1.496 × 1011 m = 1.392 × 109 m.
Solved example 2.1
Convert the following angle measures into radians.
(a) 1o (b) 1' (c) 1''
[Hint: 360o = 2𝝅 rad, 1o = 60' and 1' = 60'']
Solution:
Part (a)
1. We have: 360o = 2𝝅 rad.
2. So $\mathbf\small{1^o=\frac{2 \pi}{360}=\frac{\pi}{180}\, \text{rad}}$
3. Substituting the value of 𝝅, we get: $\mathbf\small{1^o=\frac{3.14}{180}\, \text{rad}}$ = 1.745×10-2 rad.
Part (b)
1. We have: 1o = 60'
2. Applying the result obtained in part (a), we get: 1o = 1.745×10-2 rad = 60'.
3. So $\mathbf\small{1'=\frac{1.745 \times 10^{-2}}{60}=2.908\times 10^{-4}\, \text{rad}}$
Part (c)
1. We have: 1' = 60''
2. Applying the result obtained in part (b), we get: 1' = 2.908×10-4 rad = 60''
3. So $\mathbf\small{1''=\frac{2.908 \times 10^{-4}}{60}=4.847\times 10^{-6}\, \text{rad}}$
Solved example 2.2
A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m, and looks at O and C again. Since O is very distant, the direction 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 θ = 40o (θ is known as ‘parallax’) estimate the distance of the tower C from his original position A.
Solution:
1. In fig.2.6(a) below, point A is the initial position of the man
 |
| Fig.2.6 |
2. B is the final position of the man.
• When he is at B, he can see the distant object O. For that, he does not have to make any change in the direction of vision. This is because, distant objects do not change position. Since there is no change in direction for viewing O, the red dashed line BD is parallel to the initial direction AO.
3. But to view C from B, he has to make a shift in the direction of view.
• This shift is 40o. This is indicated by the ∠CBD.
4. AO and BD are parallel.
• So ∠ACB = ∠CBD = 40o
5. In the right triangle ABC, we have: tan 40 = 100⁄AC
⇒ AC = 100⁄tan 40 = 119 m.
Solved example 2.3
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 1o 54′. Given the diameter of the Earth to be about 1.276 × 107 m, compute the distance of the moon from the Earth.
Solution:
1. In the fig.2.6(b) above, A and B are the end points of a diameter of the earth.
• Given that AB = 1.276 × 107 m.
• We have: $\mathbf\small{D=\frac{b}{\theta}}$
2. In our present case:
• b = 1.276 × 107 m.
• θ = 1o 54′
♦ We have to conv.ert this 'angle in deg' to 'angle in rad'.
♦ 1o 54′ = 114'
♦ 1' = 2.908 ×10-4 rad.
♦ So 114' = 114 × 2.908 ×10-4 = 3.31 × 10-2 rad.
3. Substituting the values, we get:
$\mathbf\small{D=\frac{1.276 \times 10^{7}}{3.315 \times 10^{-2}}=3.85 \times 10^{8}}$ m.
Solved example 2.4
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 ?
Solution:
1. In the fig.2.6(c) above, P and Q are the end points of a diameter of the sun.
• We have: d = 𝛂D
2. In our present case:
• D = 1.496 × 1011 m.
• 𝛂 = 1920''
♦ We have to convert this 'angle in deg' to 'angle in rad'.
♦ 1'' = 4.847 ×10-6 rad.
♦ So 1920'' = 1920 × 4.847 ×10-6 = 9306.24 × 10-6 rad.
3. Substituting the values, we get:
d = 9306.24 × 10-6 × 1.496 × 1011 m = 1.392 × 109 m.
In the above discussion, to find the distance to a planet or star, we observed them from two different spots on the earth. In the next section, we will see a method in which the observations are made from two different positions of the earth.
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