The Hubble's Law Formula states that the recessional velocity of a galaxy is directly proportional to its distance from the observer, expressed as \ ( v = H_0 \times d \). This fundamental principle of physical cosmology appears in NCERT Class 11 Physics and forms a key concept in modern astrophysics. For JEE aspirants and NEET students, understanding this formula builds a strong foundation in cosmology and wave optics (redshift). This article covers the formula expression, derivation, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and FAQs.
Key Hubble's Law Formulas at a Glance
Quick reference for the most important Hubble's Law formulas and related cosmology expressions.
- Hubble's Law: \( v = H_0 \times d \)
- Hubble Constant: \( H_0 \approx 70 \text{ km/s/Mpc} \)
- Recessional velocity from redshift: \( v = z \times c \)
- Redshift parameter: \( z = \dfrac{\lambda_{\text{obs}} – \lambda_{\text{emit}}}{\lambda_{\text{emit}}} \)
- Age of the universe (approx): \( t \approx \dfrac{1}{H_0} \)
- Doppler shift: \( \dfrac{\Delta \lambda}{\lambda} = \dfrac{v}{c} \)
What is Hubble's Law Formula?
Hubble's Law Formula is the observational principle in physical cosmology that describes how distant galaxies move away from us. It was formulated by astronomer Edwin Hubble in 1929 after he studied the light spectra of numerous galaxies. The law states that the recessional velocity of a galaxy is directly proportional to its distance from Earth.
In simple terms, galaxies that are farther away move away faster. This observation became the first direct observational evidence for the expansion of the universe. It strongly supports the Big Bang theory, which proposes that the universe originated from a single point and has been expanding ever since.
The concept of redshift is central to this law. When a galaxy moves away, the light it emits gets stretched toward the red end of the electromagnetic spectrum. Scientists measure this redshift to determine the recessional velocity of a galaxy. NCERT Class 11 Physics introduces this concept in the context of waves and modern physics. Students preparing for CBSE board exams and competitive exams like JEE and NEET must understand this formula thoroughly.
Hubble's Law Formula — Expression and Variables
The mathematical expression for Hubble's Law Formula is:
\[ v = H_0 \times d \]
Here, \( v \) is the recessional velocity of the galaxy, \( H_0 \) is the Hubble constant, and \( d \) is the proper distance to the galaxy from the observer.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( v \) | Recessional velocity of the galaxy | km/s (kilometres per second) |
| \( H_0 \) | Hubble constant | km/s/Mpc (kilometres per second per megaparsec) |
| \( d \) | Proper distance to the galaxy | Mpc (megaparsec) |
| \( z \) | Redshift parameter (dimensionless) | Dimensionless |
| \( c \) | Speed of light | \( 3 \times 10^5 \) km/s |
| \( \lambda_{\text{obs}} \) | Observed wavelength of light | nm or m |
| \( \lambda_{\text{emit}} \) | Emitted wavelength of light | nm or m |
Derivation of Hubble's Law Formula
Hubble derived his law empirically by plotting recessional velocities against distances for a large sample of galaxies. The derivation follows these logical steps:
Step 1: Hubble measured the redshift \( z \) of each galaxy using its light spectrum. The redshift is defined as \( z = \dfrac{\lambda_{\text{obs}} – \lambda_{\text{emit}}}{\lambda_{\text{emit}}} \).
Step 2: For small redshifts (non-relativistic galaxies), the recessional velocity is related to redshift by \( v = z \times c \), where \( c \) is the speed of light.
Step 3: Hubble independently estimated the distances \( d \) to each galaxy using standard candles such as Cepheid variable stars.
Step 4: Plotting \( v \) on the y-axis and \( d \) on the x-axis, Hubble obtained a straight line passing through the origin. The slope of this line gave the Hubble constant \( H_0 \).
Step 5: This linear relationship is written as \( v = H_0 \times d \), which is Hubble's Law Formula.
The accepted value of the Hubble constant today is approximately \( H_0 \approx 70 \) km/s/Mpc, although different measurement methods give slightly different values (67 to 73 km/s/Mpc).
Complete Cosmology Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Hubble's Law | \( v = H_0 d \) | v = recessional velocity, H₀ = Hubble constant, d = distance | km/s | Class 11, Modern Physics |
| Redshift Parameter | \( z = \dfrac{\lambda_{\text{obs}} – \lambda_{\text{emit}}}{\lambda_{\text{emit}}} \) | z = redshift, λ₀ₛₛ = observed wavelength, λₛₘᴉᴕ = emitted wavelength | Dimensionless | Class 11, Ch 15 (Waves) |
| Velocity from Redshift (non-relativistic) | \( v = z \cdot c \) | v = velocity, z = redshift, c = speed of light | km/s | Class 11, Modern Physics |
| Doppler Shift | \( \dfrac{\Delta \lambda}{\lambda} = \dfrac{v}{c} \) | Δλ = shift in wavelength, λ = original wavelength, v = source velocity | Dimensionless ratio | Class 11, Ch 15 (Waves) |
| Age of Universe (Hubble Time) | \( t_H = \dfrac{1}{H_0} \) | t₀ = Hubble time, H₀ = Hubble constant | seconds (converted to years) | Class 11, Modern Physics |
| Relativistic Redshift | \( 1 + z = \sqrt{\dfrac{1 + v/c}{1 – v/c}} \) | z = redshift, v = velocity, c = speed of light | Dimensionless | Class 12, Special Relativity |
| Luminosity Distance | \( d_L = (1+z) \cdot d \) | dⁿ = luminosity distance, z = redshift, d = proper distance | Mpc | Advanced Astrophysics |
| Critical Density of Universe | \( \rho_c = \dfrac{3H_0^2}{8\pi G} \) | ρ₃ = critical density, H₀ = Hubble constant, G = gravitational constant | kg/m³ | Advanced Cosmology |
| Friedmann Equation (simplified) | \( H^2 = \dfrac{8\pi G \rho}{3} \) | H = Hubble parameter, G = gravitational constant, ρ = density | s⁻² | Advanced Cosmology |
Hubble's Law Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: A galaxy is located at a distance of 200 Mpc from Earth. Using Hubble's constant \( H_0 = 70 \) km/s/Mpc, find the recessional velocity of the galaxy.
Given: Distance \( d = 200 \) Mpc, Hubble constant \( H_0 = 70 \) km/s/Mpc
Step 1: Write Hubble's Law Formula: \( v = H_0 \times d \)
Step 2: Substitute the values: \( v = 70 \times 200 \)
Step 3: Calculate: \( v = 14{,}000 \) km/s
Answer
The recessional velocity of the galaxy is 14,000 km/s, which is approximately 4.7% of the speed of light.
Example 2 (Class 11-12 Level)
Problem: The observed wavelength of a hydrogen spectral line from a distant galaxy is 486.5 nm. The emitted (laboratory) wavelength of the same line is 486.1 nm. Calculate (a) the redshift \( z \), (b) the recessional velocity, and (c) the distance of the galaxy using \( H_0 = 70 \) km/s/Mpc.
Given: \( \lambda_{\text{obs}} = 486.5 \) nm, \( \lambda_{\text{emit}} = 486.1 \) nm, \( c = 3 \times 10^5 \) km/s, \( H_0 = 70 \) km/s/Mpc
Step 1: Calculate the redshift: \( z = \dfrac{486.5 – 486.1}{486.1} = \dfrac{0.4}{486.1} \approx 8.23 \times 10^{-4} \)
Step 2: Find recessional velocity using \( v = z \cdot c \): \( v = 8.23 \times 10^{-4} \times 3 \times 10^5 \approx 246.9 \) km/s
Step 3: Apply Hubble's Law to find distance: \( d = \dfrac{v}{H_0} = \dfrac{246.9}{70} \approx 3.53 \) Mpc
Answer
(a) Redshift \( z \approx 8.23 \times 10^{-4} \) | (b) Recessional velocity \( \approx 247 \) km/s | (c) Distance \( \approx 3.53 \) Mpc
Example 3 (JEE/NEET Level)
Problem: Two galaxies A and B are observed from Earth. Galaxy A is at a distance of 150 Mpc and Galaxy B is at 450 Mpc. Using \( H_0 = 70 \) km/s/Mpc, find (a) the recessional velocities of both galaxies and (b) the ratio of their recessional velocities. Also estimate the approximate age of the universe using the Hubble time formula \( t_H = 1/H_0 \). (Take 1 Mpc = \( 3.086 \times 10^{19} \) km.)
Given: \( d_A = 150 \) Mpc, \( d_B = 450 \) Mpc, \( H_0 = 70 \) km/s/Mpc, 1 Mpc = \( 3.086 \times 10^{19} \) km
Step 1: Recessional velocity of Galaxy A: \( v_A = 70 \times 150 = 10{,}500 \) km/s
Step 2: Recessional velocity of Galaxy B: \( v_B = 70 \times 450 = 31{,}500 \) km/s
Step 3: Ratio of velocities: \( \dfrac{v_A}{v_B} = \dfrac{10{,}500}{31{,}500} = \dfrac{1}{3} \). This confirms that velocity scales linearly with distance, consistent with Hubble's Law.
Step 4: Convert \( H_0 \) to SI units: \( H_0 = \dfrac{70 \text{ km/s}}{1 \text{ Mpc}} = \dfrac{70 \times 10^3 \text{ m/s}}{3.086 \times 10^{22} \text{ m}} \approx 2.27 \times 10^{-18} \) s\(^{-1}\)
Step 5: Hubble time: \( t_H = \dfrac{1}{H_0} = \dfrac{1}{2.27 \times 10^{-18}} \approx 4.41 \times 10^{17} \) s \( \approx 13.97 \) billion years
Answer
(a) \( v_A = 10{,}500 \) km/s; \( v_B = 31{,}500 \) km/s | (b) Ratio \( v_A : v_B = 1 : 3 \) | Estimated age of universe \( \approx 13.97 \) billion years (consistent with the accepted value of ~13.8 billion years).
CBSE Exam Tips 2025-26
- Memorise the formula and constant: Always write \( v = H_0 d \) with \( H_0 \approx 70 \) km/s/Mpc. CBSE often asks you to state the formula and define each term.
- Link redshift to velocity: Many CBSE questions combine the Doppler redshift formula \( z = \Delta\lambda / \lambda \) with Hubble's Law. Practise two-step problems that go from wavelength shift to distance.
- State the significance: Examiners frequently ask for the significance of Hubble's Law. We recommend writing that it provides evidence for the expanding universe and supports the Big Bang theory.
- Unit conversion is crucial: The Hubble constant is given in km/s/Mpc. For calculations involving seconds and metres, convert using 1 Mpc = \( 3.086 \times 10^{22} \) m. Errors here cost marks.
- Distinguish Hubble constant from Hubble parameter: \( H_0 \) is the present-day value (constant). \( H(t) \) varies with time. CBSE questions at Class 12 level expect this distinction.
- Use diagrams: A simple velocity-distance graph (straight line through origin) earns presentation marks. Label axes correctly as “Recessional Velocity (km/s)” and “Distance (Mpc)”.
Common Mistakes to Avoid
- Using wrong units for distance: Many students substitute distance in kilometres instead of megaparsecs (Mpc) when using \( H_0 = 70 \) km/s/Mpc. Always match the units of distance with the units embedded in \( H_0 \).
- Confusing redshift with blueshift: Redshift (\( z > 0 \)) means the galaxy is moving away. Blueshift (\( z < 0 \)) means it is approaching. Andromeda shows blueshift. Do not apply Hubble's Law to approaching objects.
- Applying Hubble's Law at very large redshifts: The simple formula \( v = H_0 d \) is valid only for relatively nearby galaxies (small \( z \)). At very high redshifts, relativistic corrections are needed. Mention this limitation in long-answer questions.
- Forgetting the direction of the relationship: Hubble's Law says velocity is proportional to distance, not the other way around. The independent variable is distance; velocity is the dependent variable.
- Mixing up Hubble time with the actual age of the universe: Hubble time \( t_H = 1/H_0 \) gives an estimate, not the exact age. The actual age (~13.8 billion years) depends on the cosmological model. State this clearly in answers.
JEE/NEET Application of Hubble's Law Formula
In our experience, JEE aspirants encounter Hubble's Law Formula most frequently in the context of modern physics and wave optics. NEET also tests this concept under physical world and units. Here are the key application patterns to master:
Application Pattern 1: Redshift-to-Distance Calculation
JEE problems often give the observed and emitted wavelengths of a spectral line. You must first compute the redshift \( z = \Delta\lambda / \lambda \), then find velocity using \( v = zc \), and finally find the distance using \( d = v / H_0 \). This three-step chain is the most common question format. Practise it until it is automatic.
Application Pattern 2: Ratio and Proportionality Problems
Because \( v \propto d \), JEE often asks: “If galaxy B is three times farther than galaxy A, how do their recessional velocities compare?” The answer follows directly from the linear relationship. No calculation is needed — just proportional reasoning. These are high-scoring, low-effort questions.
Application Pattern 3: Estimating the Age of the Universe
Both JEE and NEET have asked students to estimate the age of the universe using \( t_H = 1/H_0 \). The key skill is converting \( H_0 \) from km/s/Mpc to s\(^{-1}\) using the conversion 1 Mpc = \( 3.086 \times 10^{22} \) m. Our experts suggest practising this unit conversion separately until it takes less than 30 seconds.
In our experience, students who understand the physical meaning of Hubble's Law — that space itself is expanding, not that galaxies are flying through space — score significantly better on conceptual MCQs in JEE Advanced and NEET.
FAQs on Hubble's Law Formula
Explore More Physics Formulas
Strengthen your physics preparation by exploring these related formula articles on ncertbooks.net. For wave and optics concepts connected to Hubble's Law, read our detailed guide on the Electric Flux Formula, which covers field-based reasoning essential for modern physics. If you are working on mechanics and kinematics, our Average Acceleration Formula article provides step-by-step solved examples at CBSE and JEE levels. For rotational motion concepts that complement cosmological calculations, visit our Angular Speed Formula page. You can also browse the complete Physics Formulas Hub for a comprehensive list of all NCERT-aligned formula articles. For official NCERT textbook content, refer to the NCERT official website.