In the previous section, we saw the details about parallax method. To find the distance between earth and another planet or star, we take observations from two places on the surface of the earth. In this section, we will see another procedure to find such distances. We will write it in steps:
1. In fig.2.7 below, the earth shown as a blue sphere is rotating around the sun (S).
• The green circle indicates the orbit of the earth around the sun.
• The white curved arrow indicates the direction of motion of the earth around the sun.
2. P is a planet or star.
• We want to find the distance of P from earth.
3. S1 is a 'very far away' star.
♦ Since it is very far away, it's position does not change when looked from the earth.
♦ So It acts as the reference.
4. Consider the instant at which the earth is at A.
• At that instant observe P.
(The line AP can be considered as the axis of the telescope).
• P will appear to be on the left side of S1
5. Now we wait for six months.
• Let the earth be at the position B at the instant when 6 months have passed.
• At that instant observe P.
(The line BP can be considered as the axis of the telescope)
• P will appear to be on the right side of S1
■ The earth makes a complete rotation around the sun in 12 months. So when 6 months pass after A, the earth would have completed half a rotation. That means, B is diametrically opposite to A.
6. A, B and P are three points in space.
• We get a triangle by joining the three points.
• This triangle can be represented on a sheet of paper. It is shown in fig.2.8 below:
• The sun S is in line with A and B. So S is on the base of the triangle ABP in fig.2.8.
7. When we take the observations mentioned in steps (4) and (5) above, we get angles θ1 and θ2
• So the parallax angle θ = (θ1+θ2).
8. Since P is far away, angle θ will be very small.
• Lines AP, SP and BP will be approximately of the same length.
• They can be considered as the radii of a circle with center at P.
9. 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:
$\mathbf\small{D=\frac{b}{\theta}}$
♦ Where θ is in radians.
10. The 'b' in this procedure has special significance:
• 'b' is two times the mean distance between earth and sun.
■ The 'mean distance between earth and sun' is a 'unit of length' in the SI system. It is called one astronomical unit.
• It's symbol is au.
• 1 au = 1.496 ×108 m.
• So b = 2 × 1.496 ×108 = 2.992 ×1011 m.
11. So the result in (9) becomes:
Eq.2.3: $\mathbf\small{D=\frac{2.992 \times 10^{11} \, \text{m}}{\theta}}$
Solved example 2.5
The nearest star to our solar system is Alpha Centauri. It is at a distance of 4.29 light years away from earth. How much parallax would this star show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?
Solution:
1. The D in fig.2.8 above is given to us:
• D = 4.29 ly.
• 1 ly = 9.461×1015 m.
• So D = 4.29 × 9.461 × 1015 = 4.05877 ×1015 m.
2. We have: $\mathbf\small{D=\frac{2.992 \times 10^{11} \, \text{m}}{\theta}}$
• Substituting for D, we get: $\mathbf\small{4.05877\times 10^{15}=\frac{2.992 \times 10^{8} \, \text{m}}{\theta}}$
$\mathbf\small{\Rightarrow \theta=\frac{2.992 \times 10^{8} \, \text{m}}{4.05877\times 10^{15}}=7.372\times 10^{-8}}$ rad
3. We have: $\mathbf\small{1''=4.847\times 10^{-6}\, \text{rad}}$
$\mathbf\small{\Rightarrow 1\, \text{rad}=\left( \frac{1}{4.847\times 10^{-6}} \right)''}$
• So we get: $\mathbf\small{7.372\times 10^{-8}\, \text{rad}=7.372\times 10^{-8} \times \left( \frac{1}{4.847\times 10^{-6}} \right)=1.521''}$
1. In fig.2.7 below, the earth shown as a blue sphere is rotating around the sun (S).
 |
| Fig.2.7 |
• The white curved arrow indicates the direction of motion of the earth around the sun.
2. P is a planet or star.
• We want to find the distance of P from earth.
3. S1 is a 'very far away' star.
♦ Since it is very far away, it's position does not change when looked from the earth.
♦ So It acts as the reference.
4. Consider the instant at which the earth is at A.
• At that instant observe P.
(The line AP can be considered as the axis of the telescope).
• P will appear to be on the left side of S1
5. Now we wait for six months.
• Let the earth be at the position B at the instant when 6 months have passed.
• At that instant observe P.
(The line BP can be considered as the axis of the telescope)
• P will appear to be on the right side of S1
■ The earth makes a complete rotation around the sun in 12 months. So when 6 months pass after A, the earth would have completed half a rotation. That means, B is diametrically opposite to A.
6. A, B and P are three points in space.
• We get a triangle by joining the three points.
• This triangle can be represented on a sheet of paper. It is shown in fig.2.8 below:
 |
| Fig.7.8 |
7. When we take the observations mentioned in steps (4) and (5) above, we get angles θ1 and θ2
• So the parallax angle θ = (θ1+θ2).
8. Since P is far away, angle θ will be very small.
• Lines AP, SP and BP will be approximately of the same length.
• They can be considered as the radii of a circle with center at P.
9. 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:
$\mathbf\small{D=\frac{b}{\theta}}$
♦ Where θ is in radians.
10. The 'b' in this procedure has special significance:
• 'b' is two times the mean distance between earth and sun.
■ The 'mean distance between earth and sun' is a 'unit of length' in the SI system. It is called one astronomical unit.
• It's symbol is au.
• 1 au = 1.496 ×108 m.
• So b = 2 × 1.496 ×108 = 2.992 ×1011 m.
11. So the result in (9) becomes:
Eq.2.3: $\mathbf\small{D=\frac{2.992 \times 10^{11} \, \text{m}}{\theta}}$
Now we will see a solved example.
Solved example 2.5
The nearest star to our solar system is Alpha Centauri. It is at a distance of 4.29 light years away from earth. How much parallax would this star show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?
Solution:
1. The D in fig.2.8 above is given to us:
• D = 4.29 ly.
• 1 ly = 9.461×1015 m.
• So D = 4.29 × 9.461 × 1015 = 4.05877 ×1015 m.
2. We have: $\mathbf\small{D=\frac{2.992 \times 10^{11} \, \text{m}}{\theta}}$
• Substituting for D, we get: $\mathbf\small{4.05877\times 10^{15}=\frac{2.992 \times 10^{8} \, \text{m}}{\theta}}$
$\mathbf\small{\Rightarrow \theta=\frac{2.992 \times 10^{8} \, \text{m}}{4.05877\times 10^{15}}=7.372\times 10^{-8}}$ rad
3. We have: $\mathbf\small{1''=4.847\times 10^{-6}\, \text{rad}}$
$\mathbf\small{\Rightarrow 1\, \text{rad}=\left( \frac{1}{4.847\times 10^{-6}} \right)''}$
• So we get: $\mathbf\small{7.372\times 10^{-8}\, \text{rad}=7.372\times 10^{-8} \times \left( \frac{1}{4.847\times 10^{-6}} \right)=1.521''}$
Next, we will see the definition of parsec. It is a unit of length in the SI system. It is used to measure distances between planets and stars. We will first write the basic steps:
1. From the discussions so far, we can write this:
• A triangle can be used to represent a large distance.
• That triangle will have 3 items:
(i) The angle at the apex.
• This is the parallax angle.
• It will be a very small quantity which should be specified in radians.
(ii) The two sides whose lengths are considered to be equal.
• This length of the sides is the 'large distance' which we are usually asked to find.
(iii) The base.
• This can be any one of the distances given below:
♦ (a) Distance between two points on the surface of the earth.
♦ (b) Distance between two diametrically opposite points on the earth's orbit around the sun.
‣ This is same as the diameter of the earths orbit.
‣ This is also same as '2 times au'.
♦ (c) Any other specified distance.
■ Knowing any two items, the third can be found.
2. The three items are shown in fig.7.8(b) above.
• In our present case we want to find 'how much a parsec is in m'.
• That means, we have to find item (ii).
3. Items (i) and (iii) will be given:
Item (i):
• The angle at apex = 1''
• We have to convert it into radians.
• We know that $\mathbf\small{1''=4.847\times 10^{-6}\, \text{rad}}$
Item (iii):
• It is the radius of the earths' orbit.
• It is same as 1 au.
• We have to convert it into m.
• 1 au = 1.496 ×1011 m.
4. Now we apply the arc formula to the triangle:
• We have: $\mathbf\small{\text{angle}=\frac{\text{arc}}{\text{radius}}}$
• Substituting the values, we get: $\mathbf\small{4.847\times 10^{-6}=\frac{1.496\times 10^{11}}{1\;\text{parsec}}}$
$\mathbf\small{\Rightarrow 1\;\text{parsec}=\frac{1.496\times 10^{11}}{4.847\times 10^{-6}}=3.086 \times 10^{16}\, \text{m}}$
5. Based on the above calculations, we can write the definition of parsec:
• One parsec corresponds to the distance at which the mean radius of the earth's orbit subtends an angle of one second of arc.
1. From solved example 4.5, we have:
• Distance of Alpha Centauri from earth = 4.29 ly.
2. one ly = 9.461 ×1015 m.
• So the distance = 4.29 × 9.461 ×1015 = 4.0588 ×1016 m.
3. We have 1 parsec = 3.086 ×1016 m.
• So the distance in parsec = $\mathbf\small{\frac{4.0588\times 10^{16}}{3.086\times 10^{16}}=1.32 \; \text{parsec}}$
Solved example 2.6
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.
Solution:
1. Distance from Earth to Jupiter (D) = 824.7 million km.
• We have to convert this into m.
• 1 million = 106
• So 824.7 million km = 824.7 × 106 km = 824.7 × 109 m.
2. Angle 𝛂 = 35.72"
• We have to convert this into radians.
• $\mathbf\small{1''=4.847\times 10^{-6}\, \text{rad}}$
• So 35.72" = 35.72 × 4.847 × 10-6 rad.
3. We have Eq.2.2: $\mathbf\small{d=\alpha \, D}$
(See fig.2.6(c) of the previous section)
• Substituting the values, we get: d = 824.7 × 109 × 35.72 × 4.847 × 10-6 = 1.429 × 108 m.
1. From the discussions so far, we can write this:
• A triangle can be used to represent a large distance.
• That triangle will have 3 items:
(i) The angle at the apex.
• This is the parallax angle.
• It will be a very small quantity which should be specified in radians.
(ii) The two sides whose lengths are considered to be equal.
• This length of the sides is the 'large distance' which we are usually asked to find.
(iii) The base.
• This can be any one of the distances given below:
♦ (a) Distance between two points on the surface of the earth.
♦ (b) Distance between two diametrically opposite points on the earth's orbit around the sun.
‣ This is same as the diameter of the earths orbit.
‣ This is also same as '2 times au'.
♦ (c) Any other specified distance.
■ Knowing any two items, the third can be found.
2. The three items are shown in fig.7.8(b) above.
• In our present case we want to find 'how much a parsec is in m'.
• That means, we have to find item (ii).
3. Items (i) and (iii) will be given:
Item (i):
• The angle at apex = 1''
• We have to convert it into radians.
• We know that $\mathbf\small{1''=4.847\times 10^{-6}\, \text{rad}}$
Item (iii):
• It is the radius of the earths' orbit.
• It is same as 1 au.
• We have to convert it into m.
• 1 au = 1.496 ×1011 m.
4. Now we apply the arc formula to the triangle:
• We have: $\mathbf\small{\text{angle}=\frac{\text{arc}}{\text{radius}}}$
• Substituting the values, we get: $\mathbf\small{4.847\times 10^{-6}=\frac{1.496\times 10^{11}}{1\;\text{parsec}}}$
$\mathbf\small{\Rightarrow 1\;\text{parsec}=\frac{1.496\times 10^{11}}{4.847\times 10^{-6}}=3.086 \times 10^{16}\, \text{m}}$
5. Based on the above calculations, we can write the definition of parsec:
• One parsec corresponds to the distance at which the mean radius of the earth's orbit subtends an angle of one second of arc.
Let us calculate the distance of the star Alpha Centauri in terms of parsec:
• Distance of Alpha Centauri from earth = 4.29 ly.
2. one ly = 9.461 ×1015 m.
• So the distance = 4.29 × 9.461 ×1015 = 4.0588 ×1016 m.
3. We have 1 parsec = 3.086 ×1016 m.
• So the distance in parsec = $\mathbf\small{\frac{4.0588\times 10^{16}}{3.086\times 10^{16}}=1.32 \; \text{parsec}}$
Solved example 2.6
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.
Solution:
1. Distance from Earth to Jupiter (D) = 824.7 million km.
• We have to convert this into m.
• 1 million = 106
• So 824.7 million km = 824.7 × 106 km = 824.7 × 109 m.
2. Angle 𝛂 = 35.72"
• We have to convert this into radians.
• $\mathbf\small{1''=4.847\times 10^{-6}\, \text{rad}}$
• So 35.72" = 35.72 × 4.847 × 10-6 rad.
3. We have Eq.2.2: $\mathbf\small{d=\alpha \, D}$
(See fig.2.6(c) of the previous section)
• Substituting the values, we get: d = 824.7 × 109 × 35.72 × 4.847 × 10-6 = 1.429 × 108 m.
In the next section, we will see estimation of very small distances.
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