The Universal Gravitation Formula, given by F = Gm₁m₂/r², describes the attractive force between any two masses in the universe. Introduced by Sir Isaac Newton, this law forms the backbone of Class 11 Physics (NCERT Chapter 8 — Gravitation). It is equally critical for JEE Main, JEE Advanced, and NEET, where gravitational force, orbital motion, and escape velocity questions appear every year. This article covers the formula expression, derivation, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Universal Gravitation Formulas at a Glance
Quick reference for the most important gravitation formulas used in CBSE and competitive exams.
- Newton’s Law of Gravitation: \( F = \dfrac{Gm_1 m_2}{r^2} \)
- Acceleration due to gravity: \( g = \dfrac{GM}{R^2} \)
- Variation of g with height: \( g_h = g\left(1 – \dfrac{2h}{R}\right) \) (for h << R)
- Variation of g with depth: \( g_d = g\left(1 – \dfrac{d}{R}\right) \)
- Escape velocity: \( v_e = \sqrt{\dfrac{2GM}{R}} \)
- Orbital velocity: \( v_o = \sqrt{\dfrac{GM}{r}} \)
- Time period of satellite: \( T = 2\pi\sqrt{\dfrac{r^3}{GM}} \)
What is the Universal Gravitation Formula?
The Universal Gravitation Formula states that every particle in the universe attracts every other particle with a force. This force is directly proportional to the product of their masses. It is also inversely proportional to the square of the distance between them. Newton published this law in his landmark work Principia Mathematica in 1687.
In the NCERT Class 11 Physics textbook, the Universal Gravitation Formula appears in Chapter 8 — Gravitation. It is one of the most fundamental laws in classical mechanics. The law applies universally — from falling apples to orbiting planets and distant galaxies. Understanding this formula helps students solve problems on satellite motion, tidal forces, and planetary orbits.
The formula introduces the universal gravitational constant G, whose value is \( 6.674 \times 10^{-11} \) N m² kg⁻². This constant is the same everywhere in the universe. That universality is precisely why Newton’s law carries the word “universal” in its name.
Universal Gravitation Formula — Expression and Variables
The mathematical expression for Newton’s Universal Gravitation Formula is:
\[ F = \frac{G m_1 m_2}{r^2} \]
Here, F is the magnitude of the gravitational force between two point masses \( m_1 \) and \( m_2 \) separated by a distance r. The force acts along the line joining the two masses and is always attractive in nature.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Gravitational Force | Newton (N) |
| G | Universal Gravitational Constant | N m² kg⁻² |
| m₁ | Mass of first object | kilogram (kg) |
| m₂ | Mass of second object | kilogram (kg) |
| r | Distance between the centres of the two masses | metre (m) |
Derivation of the Universal Gravitation Formula
Newton derived this law using two key observations. First, the gravitational force is proportional to each mass: \( F \propto m_1 \) and \( F \propto m_2 \). Combining these gives \( F \propto m_1 m_2 \). Second, using Kepler’s Third Law of planetary motion, Newton showed that the force falls off with the square of distance: \( F \propto \dfrac{1}{r^2} \). Combining both proportionalities gives \( F \propto \dfrac{m_1 m_2}{r^2} \). Introducing the proportionality constant G yields the complete formula: \( F = \dfrac{G m_1 m_2}{r^2} \). The value of G was first measured experimentally by Henry Cavendish in 1798 using a torsion balance.
Acceleration Due to Gravity from the Universal Gravitation Formula
When an object of mass m is placed on the surface of the Earth (mass M, radius R), the gravitational force equals the weight of the object: \( F = mg \). Equating with Newton’s formula:
\[ mg = \frac{GMm}{R^2} \implies g = \frac{GM}{R^2} \]
Substituting G = \( 6.674 \times 10^{-11} \) N m² kg⁻², M = \( 6 \times 10^{24} \) kg, and R = \( 6.4 \times 10^6 \) m gives g ≈ 9.8 m s⁻². This derivation is a favourite in CBSE board exams.
Complete Gravitation Formula Sheet
The table below covers all major gravitation formulas from NCERT Class 11 Chapter 8. Use this as a quick reference during revision.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Newton’s Law of Gravitation | \( F = \dfrac{Gm_1 m_2}{r^2} \) | G = gravitational constant, m₁, m₂ = masses, r = distance | N | Class 11, Ch 8 |
| Acceleration due to Gravity (surface) | \( g = \dfrac{GM}{R^2} \) | M = mass of Earth, R = radius of Earth | m s⁻² | Class 11, Ch 8 |
| g at height h (exact) | \( g_h = \dfrac{GM}{(R+h)^2} \) | h = height above surface | m s⁻² | Class 11, Ch 8 |
| g at height h (approximate, h << R) | \( g_h \approx g\left(1 – \dfrac{2h}{R}\right) \) | h = height, R = Earth’s radius | m s⁻² | Class 11, Ch 8 |
| g at depth d | \( g_d = g\left(1 – \dfrac{d}{R}\right) \) | d = depth below surface | m s⁻² | Class 11, Ch 8 |
| Gravitational Potential Energy | \( U = -\dfrac{GMm}{r} \) | M = source mass, m = object mass, r = distance from centre | J | Class 11, Ch 8 |
| Escape Velocity | \( v_e = \sqrt{\dfrac{2GM}{R}} \) | M = planet mass, R = planet radius | m s⁻¹ | Class 11, Ch 8 |
| Orbital Velocity | \( v_o = \sqrt{\dfrac{GM}{r}} \) | r = orbital radius from Earth’s centre | m s⁻¹ | Class 11, Ch 8 |
| Time Period of Satellite | \( T = 2\pi\sqrt{\dfrac{r^3}{GM}} \) | r = orbital radius | s | Class 11, Ch 8 |
| Kepler’s Third Law | \( T^2 \propto r^3 \) | T = time period, r = semi-major axis | — | Class 11, Ch 8 |
Universal Gravitation Formula — Solved Examples
Example 1 (Class 9–10 Level): Gravitational Force Between Two Objects
Problem: Calculate the gravitational force between two spheres of masses 20 kg and 30 kg placed 0.5 m apart. Take G = \( 6.674 \times 10^{-11} \) N m² kg⁻².
Given: m₁ = 20 kg, m₂ = 30 kg, r = 0.5 m, G = \( 6.674 \times 10^{-11} \) N m² kg⁻²
Step 1: Write the Universal Gravitation Formula: \( F = \dfrac{G m_1 m_2}{r^2} \)
Step 2: Substitute values: \( F = \dfrac{6.674 \times 10^{-11} \times 20 \times 30}{(0.5)^2} \)
Step 3: Calculate numerator: \( 6.674 \times 10^{-11} \times 600 = 4.004 \times 10^{-8} \) N m²
Step 4: Divide by r² = 0.25: \( F = \dfrac{4.004 \times 10^{-8}}{0.25} = 1.602 \times 10^{-7} \) N
Answer
Gravitational Force F = \( 1.602 \times 10^{-7} \) N
Example 2 (Class 11–12 Level): Acceleration Due to Gravity at Height
Problem: Find the acceleration due to gravity at a height of 3200 km above the Earth’s surface. Take R = 6400 km, g (surface) = 9.8 m s⁻².
Given: h = 3200 km = \( 3.2 \times 10^6 \) m, R = \( 6.4 \times 10^6 \) m, g = 9.8 m s⁻²
Step 1: Use the exact formula: \( g_h = \dfrac{g R^2}{(R + h)^2} \)
Step 2: Calculate (R + h): \( 6400 + 3200 = 9600 \) km = \( 9.6 \times 10^6 \) m
Step 3: Substitute: \( g_h = 9.8 \times \dfrac{(6.4 \times 10^6)^2}{(9.6 \times 10^6)^2} \)
Step 4: Simplify the ratio: \( \dfrac{6.4^2}{9.6^2} = \dfrac{40.96}{92.16} = \dfrac{4}{9} \)
Step 5: Calculate: \( g_h = 9.8 \times \dfrac{4}{9} = \dfrac{39.2}{9} \approx 4.36 \) m s⁻²
Answer
Acceleration due to gravity at 3200 km height = 4.36 m s⁻²
Example 3 (JEE/NEET Level): Orbital Velocity and Time Period of a Satellite
Problem: A satellite orbits the Earth at a height of 600 km above the surface. Find its orbital velocity and time period. Take R = 6400 km, g = 9.8 m s⁻², GM = gR².
Given: h = 600 km = \( 6 \times 10^5 \) m, R = \( 6.4 \times 10^6 \) m, so r = R + h = \( 7.0 \times 10^6 \) m
Step 1: Calculate GM using \( GM = gR^2 \):
\( GM = 9.8 \times (6.4 \times 10^6)^2 = 9.8 \times 4.096 \times 10^{13} = 4.014 \times 10^{14} \) m³ s⁻²
Step 2: Orbital velocity: \( v_o = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{4.014 \times 10^{14}}{7.0 \times 10^6}} \)
\( v_o = \sqrt{5.734 \times 10^7} \approx 7572 \) m s⁻¹ ≈ 7.57 km s⁻¹
Step 3: Time period: \( T = \dfrac{2\pi r}{v_o} = \dfrac{2\pi \times 7.0 \times 10^6}{7572} \)
\( T = \dfrac{4.398 \times 10^7}{7572} \approx 5808 \) s ≈ 96.8 minutes
Answer
Orbital velocity ≈ 7.57 km s⁻¹; Time period ≈ 96.8 minutes
CBSE Exam Tips 2025-26
- Memorise G precisely: Always write G = \( 6.674 \times 10^{-11} \) N m² kg⁻². Losing even one mark for a wrong constant is avoidable.
- Derive g from the Universal Gravitation Formula: CBSE frequently asks for a 2–3 mark derivation of g = GM/R². Practise this derivation until it is automatic.
- Know both forms of g at height: The approximate form \( g_h \approx g(1 – 2h/R) \) is only valid when h << R. Use the exact form otherwise. Examiners award marks for choosing the correct form.
- Unit consistency is critical: Always convert km to m before substituting. A common error is substituting R in km, which gives a wrong answer.
- Kepler’s Third Law numericals: We recommend practising at least five Kepler T² ∝ r³ problems. These appear almost every year in Section C or D of the CBSE board paper.
- Negative sign in potential energy: Gravitational potential energy is always negative: \( U = -GMm/r \). CBSE deducts marks if you write it as positive. Remember: bound systems have negative energy.
Common Mistakes to Avoid
- Confusing r with height h: In the Universal Gravitation Formula, r is the distance between the centres of the two masses. For a satellite at height h above Earth, r = R + h, not just h. Many students substitute h directly and get a wrong answer.
- Using the approximate formula when h is large: The formula \( g_h \approx g(1 – 2h/R) \) is a binomial approximation valid only for h << R. When h is comparable to R, always use \( g_h = gR^2/(R+h)^2 \).
- Forgetting the negative sign in gravitational potential energy: Gravitational PE is \( U = -GMm/r \). Students often write it as positive. This leads to errors in total energy and escape velocity calculations.
- Wrong units for G: G has units of N m² kg⁻². Students sometimes write N m kg⁻² (missing the square on metres). Always double-check dimensional analysis.
- Applying inverse square law incorrectly: If the distance doubles, the force becomes one-fourth (not one-half). Always square the ratio of distances when comparing gravitational forces.
JEE/NEET Application of the Universal Gravitation Formula
In our experience, JEE aspirants encounter the Universal Gravitation Formula in at least 2–3 questions per paper, either directly or embedded in multi-concept problems. Here are the most common application patterns.
Pattern 1: Ratio-Based Force Problems (JEE Main)
JEE Main frequently asks: “If the distance between two masses is halved, by what factor does the gravitational force change?” The answer follows directly from the inverse square law. If r becomes r/2, then \( F \propto 1/(r/2)^2 = 4/r^2 \), so force becomes 4 times. Our experts suggest always expressing the new force as a multiple of the original using ratios rather than substituting absolute values.
Pattern 2: Satellite and Orbital Energy (JEE Advanced)
JEE Advanced tests the relationship between orbital velocity, escape velocity, and total mechanical energy. The key results are: orbital velocity \( v_o = \sqrt{GM/r} \), escape velocity \( v_e = \sqrt{2GM/R} \), and total energy of a satellite \( E = -GMm/(2r) \). Note that \( v_e = \sqrt{2} \cdot v_o \) when r = R. JEE Advanced problems often ask for the minimum energy needed to move a satellite from one orbit to another. This requires calculating the difference in total energies.
Pattern 3: Variation of g for NEET
NEET consistently tests variation of g with altitude and depth. A typical NEET question gives a specific height or depth and asks for the new value of g. Students must remember that g decreases with both altitude and depth. At the centre of the Earth, g = 0. At the surface, g is maximum. NEET also tests the effect of Earth’s rotation on g, where g is minimum at the equator and maximum at the poles. Practise at least ten such problems from previous NEET papers (2018–2024) to recognise all variants quickly.
FAQs on Universal Gravitation Formula
For more Physics formulas, explore our complete Physics Formulas hub. You may also find our articles on the Photon Energy Formula, Heat Transfer Formula, and Shear Modulus Formula useful for your CBSE and JEE/NEET preparation. For the official NCERT syllabus, refer to the NCERT official website.