Parallax Method — Astronomical Distances

Try this: hold a finger up at arm’s length and close one eye, then the other, alternately. The finger appears to shift against the background. That apparent shift is parallax — the same trick astronomers use to measure distances to nearby stars and planets.

For Class 11 CBSE, parallax is the most-tested concept in the “Units and Measurements” chapter. JEE Main has occasional MCQs on it. Once you understand the geometry, the formula is one line — but the unit conversions trip up half the students.

Key Terms & Definitions

Parallax: The apparent shift in the position of an object when viewed from two different points.

Parallax angle (parallactic angle) $\theta$ : The angle subtended at the object by the line joining the two observation points (the baseline).

Baseline $b$ : The distance between the two observation points. For Earth-based stellar parallax, the baseline is typically Earth’s orbit diameter ( $2 \text{ AU}$ ).

Astronomical Unit (AU): Mean Earth-Sun distance, $1 \text{ AU} = 1.496 \times 10^{11}$ m.

Parsec (pc): The distance at which a baseline of $1 \text{ AU}$ subtends an angle of $1$ arcsecond. $1 \text{ pc} = 3.086 \times 10^{16}$ m $\approx 3.26$ light years.

Light year (ly): Distance light travels in one year, $\approx 9.46 \times 10^{15}$ m.

The Core Formula

If two observers separated by baseline $b$ see an object shifted by an angle $\theta$ (in radians), the distance to the object is:

D = \tfrac{b}{\theta}

(valid when $\theta$ is small; $\theta$ in radians)

The approximation $\tan\theta \approx \theta$ is excellent for small $\theta$ — and astronomical parallax angles are tiny (often a fraction of an arcsecond).

Why the Approximation Works

For $\theta < 0.1$ rad, $\tan\theta$ and $\theta$ agree to better than $0.3\%$ . Stellar parallax angles are typically $10^{-6}$ to $10^{-8}$ radians, so the approximation is essentially exact.

Unit Conversions You Must Memorise

$1° = \tfrac{\pi}{180} \text{ rad} \approx 1.745 \times 10^{-2}$ rad
$1' \text{ (arcminute)} = 1°/60$
$1'' \text{ (arcsecond)} = 1'/60 = 1°/3600 \approx 4.848 \times 10^{-6}$ rad

The arcsecond conversion is the one used in $9$ out of $10$ NCERT problems.

Worked Examples

Example 1 (NCERT/CBSE) — Distance to the Moon

The parallax of the Moon as seen from two points on Earth $1.276 \times 10^7$ m apart is $1°54'$ . Find the distance from Earth to the Moon.

$1°54' = 1° + 54/60 = 1.9°$ .

\theta = 1.9 \times \tfrac{\pi}{180} = 0.0332 \text{ rad}

D = \tfrac{b}{\theta} = \tfrac{1.276 \times 10^7}{0.0332} \approx 3.84 \times 10^8 \text{ m}

The actual mean distance to the Moon is about $3.84 \times 10^8$ m — a perfect match!

Example 2 (CBSE) — Distance to a Star

A star has a parallax angle of $0.5''$ as observed from two points $2$ AU apart. Find its distance in parsecs and metres.

By definition, $0.5''$ from $1 \text{ AU}$ baseline gives $1/0.5 = 2$ pc. With baseline doubled, the distance also doubles? No — the parsec definition uses half the baseline (parallax is conventionally defined as half the angular shift across $2$ AU).

For practical use: if the shift across $2 \text{ AU}$ is $\theta'' = 0.5''$ , the standard parallax (per $1$ AU) is $\theta'' / 2 = 0.25''$ and distance $= 1/0.25 = 4$ pc.

D = 4 \times 3.086 \times 10^{16} \approx 1.23 \times 10^{17} \text{ m}

Example 3 (JEE Main) — Diameter of an Object

A planet subtends an angle of $25''$ at the Earth. Its distance from Earth is $4.0 \times 10^{11}$ m. Find the diameter of the planet.

\alpha = 25 \times 4.848 \times 10^{-6} = 1.212 \times 10^{-4} \text{ rad}

d = D \alpha = 4.0 \times 10^{11} \times 1.212 \times 10^{-4} \approx 4.85 \times 10^7 \text{ m}

So the planet’s diameter is about $48{,}500$ km — roughly Neptune-sized.

Exam-Specific Tips

CBSE Class 11: This is a guaranteed $2$ – $3$ mark numerical. The favourite form: parallax angle in degrees-minutes-seconds, baseline in km, asking for distance. Conversion to radians is the only tricky step.

JEE Main: Less direct than CBSE. Watch for questions phrased as “angular diameter” or “angular size” — these use the same $D\theta$ relation but ask for object size, not distance.

NEET: Almost never appears, but understanding parallax helps you crack distance-related questions in optics.

Common Mistakes

Mistake 1: Forgetting to convert the angle to radians. The formula $D = b/\theta$ requires $\theta$ in radians. Plugging $\theta$ in degrees gives an answer off by a factor of $57$ .

Mistake 2: Using arcseconds without conversion. $1'' \approx 4.85 \times 10^{-6}$ rad. Many students leave $\theta$ as the literal number of arcseconds.

Mistake 3: Confusing parallax angle with angular diameter. Parallax angle is the shift seen from two viewpoints; angular diameter is the full angle subtended by the object’s width. They use the same formula form but answer different questions.

Mistake 4: Using $\tan\theta = \theta$ when $\theta$ is large. The approximation breaks down for $\theta > 0.1$ rad. For lab-scale parallax (e.g., a finger at arm’s length), use the exact $\tan$ .

Mistake 5: Confusing parsec with light year. $1$ pc $\approx 3.26$ ly, so a star at $5$ pc is $\approx 16.3$ ly away. Memorise this conversion.

Practice Questions

The parallax of a star is $0.1''$ when viewed from a baseline of $1$ AU. Find its distance in parsecs.
A tower subtends an angle of $30''$ at a point $1$ km away. Find the height of the tower.
A planet’s diameter is $1.4 \times 10^7$ m. From a distance of $5 \times 10^{11}$ m, what angle does it subtend?
The parallax of the Moon from two stations $1.276 \times 10^7$ m apart is $1°54'$ . Compare your computed distance with the standard value.
A star at $4$ parsecs has what parallax angle when measured against an Earth-orbit baseline?
Convert $0.05''$ to radians.
The Sun’s diameter is $1.39 \times 10^9$ m and Earth-Sun distance is $1$ AU. Find the angular diameter of the Sun.
Two stars are at the same distance but have parallax angles in the ratio $2:1$ when measured with different baselines. What is the ratio of the baselines?

Answers:

$D = 1/0.1 = 10$ pc.
$\alpha = 30 \times 4.848 \times 10^{-6}$ rad; $h = 1000\alpha \approx 0.145$ m.
$\theta = 1.4 \times 10^7 / 5 \times 10^{11} = 2.8 \times 10^{-5}$ rad $\approx 5.78''$ .
$D \approx 3.84 \times 10^8$ m, matches standard value.
$\theta = 1/4 = 0.25''$ .
$0.05 \times 4.848 \times 10^{-6} = 2.42 \times 10^{-7}$ rad.
$\theta = 1.39 \times 10^9 / 1.496 \times 10^{11} = 9.29 \times 10^{-3}$ rad $\approx 0.532° \approx 32'$ .
$2:1$ — same distance, $\theta \propto b$ .

Frequently Asked Questions

Q1: Why can’t we measure parallax of very distant stars?

The parallax angle scales as $1/D$ . For stars beyond $\sim 100$ pc, the angle becomes smaller than $0.01''$ , which is below the resolution of ground-based telescopes. Space-based missions (Hipparcos, Gaia) can measure smaller angles.

Q2: What’s the difference between heliocentric and geocentric parallax?

Heliocentric (annual) parallax uses Earth’s orbit diameter as baseline — for stars. Geocentric (diurnal) parallax uses Earth’s radius as baseline — for nearby objects like the Moon. Same formula, different baseline.

Q3: What is the parsec named after?

“Parsec” = “parallax of one arcsecond”. A star at $1$ pc shows a parallax of $1''$ when measured against a $1$ -AU baseline. It’s a unit defined by the geometry of the parallax method itself.

Q4: Why use parsecs instead of light years for stellar distances?

Astronomers prefer parsecs because they map directly to measurement. A parallax of $0.1''$ instantly means $10$ pc — no conversion. Light years are easier for popular communication but require multiplication by $3.26$ .

Q5: Can we use parallax for galaxies?

No. Galaxies are too far for any practical baseline. We use other distance ladder methods — Cepheid variables, Type Ia supernovae, redshift — for galactic and intergalactic distances. Parallax sets the foundation only out to about $1000$ pc.

Q6: What is the parallax of the nearest star, Proxima Centauri?

About $0.77''$ — corresponding to $1.30$ pc or $4.24$ light years. It’s also the largest stellar parallax we measure from Earth.