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In physics, I have to solve some numerical problems based on the parallax method. I learnt the parallax method from Khan Academy and Wikipedia. (I found a really helpful Khan Academy video parallax in observing stars and stellar distance using parallax.) For stellar parallax, I know that the distance $d=\tan(90^{\circ}-\theta) \ \text{AU} = \cot \theta \ \text{AU} =1/\tan\theta \ \text{AU}$. But I am not sure how to solve problems based on diurnal parallax.

For the parallax method,

Knowing the distance between the observation points L ( basis ) and the displacement angle α, you can determine the distance to the object: $D = {\frac {L} {2 \ \sin \ \alpha / 2}}$.

For small angles (α - in radians): $D = {\frac {L} {\ \alpha}}$.

But how did we derived the formula $D = {\frac {L} {2 \ \sin \ \alpha / 2}}$?

Using parallax method, we can measure:

  1. the distance of sun or moon or any planet (Diurnal parallax method: solar parallax, lunar parallax)
  2. the diameter of sun or moon or any planet
  3. the distance of a nearby star (annual parallax or stellar parallax method)

For diurnal parallax (sun, moon, and planets within the solar system), What is the formula for measuring the distance and parallax angle (with an example)? In my textbook, it is derived from the arc length formula $s=r\theta$. Now once we know the distance we can solve for the diameter of the sun or moon or any planet using the same parallax formula. (This parallax angle or parallactic angle is usually given in the question/numerical problem and we have to solve for the distance or the diameter)


EDIT: Since I have started a bounty I want to make it clear what I want in the answer. I want the formulas, derivation of formulas, related examples and calculations. The following is an example numerical problem based on the parallax method (You don't necessarily need to solve them, just explain how to apply the formula):

(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon? (b) Moon is seen to be of (½)° diameter from the earth. What must be the relative size compared to the earth? (c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

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The derivation

Carefully observe the following triangle:

Parallax

We (the observers) are moving between the points $A$ and $B$, our object is in $C$.

By the definition of the sine: $$\sin\frac{\alpha}{2}=\frac{L}{2D}$$ and with expressing $D$: $$D=\frac{L}{2\sin(\alpha/2)}$$ but for small angles in radians: $$\sin x = x$$ so $$D=\frac{L}{2\alpha/2}=\frac{L}{\alpha}$$


Examples

Note that there are better methods to do these exercises, but we will use the ones we derived above.

(a) The Earth–Moon distance is about 60 Earth radii. What will be the diameter of the Earth (approximately in degrees) as seen from the Moon?

Exercise (a)

Here the total diameter of Earth is $L=2\rm\,R_E$. The distance is $D=60\rm\,R_E$. We express $\alpha$ in the formula above: $$\alpha=2\arcsin\frac{L}{2D}=2\arcsin\frac{2\rm\,R_E}{2\cdot 60\rm\,R_E}=1.91°$$

(b) Moon is seen to be of (½)° diameter from the Earth. What must be the relative size compared to the Earth?

Exercise (b)

Now, the situation is reverse. We have $\alpha=0.5°$ and again $D=60\rm\,R_E$. Now, we need to express $L$: $$L=2D\sin\frac{a}{2}=2\cdot 60\rm\,R_E\sin\frac{0.5°}{2}=0.52\rm\,R_E$$ We now divide both diameters: $$\frac{2R_M}{2R_E}=\frac{0.52\rm\,R_E}{2\rm\,R_E}=0.26$$

(c) From parallax measurement, the Sun is found to be at a distance of about 400 times the Earth–Moon distance. Estimate the ratio of Sun-Earth diameters.

Here the required data is the angular size of the Sun: $\alpha=0.5°$ (of course, you need to know that number). The distance is now $D=400\cdot 60\rm\,R_E=24000\rm\,R_E$. As before, we express $L$: $$L=2D\sin\frac{a}{2}=2\cdot 24000\rm\,R_E\sin\frac{0.5°}{2}=209.4\rm\,R_E$$ The ratio is now $$\frac{2R_S}{2R_E}=\frac{209.4\rm\,R_E}{2\rm\,R_E}=105\rm\,R_E$$

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  • $\begingroup$ It's not an isosceles triangle unless the observers happen to be in just the right positions. $\endgroup$
    – stretch
    Jul 13 at 11:44
  • $\begingroup$ @stretch It is. Observe the symmetry with rotating Earth. $\endgroup$
    – User123
    Jul 13 at 12:18
  • $\begingroup$ Does that mean you'd have a single observer who waits twelve hours between observations? The moon is moving: it's right ascension will have changed by about six degrees between those observations. Over the course of 24 hours today the Moon's right ascension will have changed by 11.7 degrees, and its declination by 4.6 degrees. The Earth - Moon distance in those same 24 hours will have decreased by 0.8 Earth radii. $\endgroup$
    – stretch
    Jul 13 at 14:09
  • $\begingroup$ @stretch The two observers observe at the same time on the antipodal points on the Earth's sphere. $\endgroup$
    – User123
    Jul 13 at 14:10
  • $\begingroup$ How do you find the antipodal points? The Moon's motion is irregular, north or south 25 degrees in declination. $\endgroup$
    – stretch
    Jul 13 at 16:38

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