The Gravity Formula gives the gravitational force of attraction between two masses and is expressed as \( F = G rac{m_1 m_2}{r^2} \), where G is the universal gravitational constant. This fundamental formula is covered in NCERT Class 9 Chapter 10 (Gravitation) and revisited in Class 11 Chapter 8. It is equally critical for JEE Main, JEE Advanced, and NEET examinations, where gravitational force, acceleration due to gravity, and orbital mechanics appear every year. This article covers the complete Gravity Formula, its derivation, a full formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Gravity Formulas at a Glance
Quick reference for the most important gravity and gravitation formulas.
- Universal Gravitation: \( F = G \dfrac{m_1 m_2}{r^2} \)
- Acceleration due to gravity: \( g = \dfrac{GM}{R^2} \)
- Variation with height: \( g_h = g\left(1 – \dfrac{2h}{R}\right) \)
- Variation with depth: \( g_d = g\left(1 – \dfrac{d}{R}\right) \)
- Gravitational Potential Energy: \( U = -\dfrac{Gm_1 m_2}{r} \)
- Escape Velocity: \( v_e = \sqrt{\dfrac{2GM}{R}} \)
- Orbital Velocity: \( v_o = \sqrt{\dfrac{GM}{r}} \)
What is the Gravity Formula?
The Gravity Formula, also known as Newton’s Law of Universal Gravitation, states that every object in the universe attracts every other object 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 their centres.
Sir Isaac Newton formulated this law in 1687. The legend of an apple falling from a tree inspired him to think about the force pulling objects toward Earth. He realised that the same force governs the motion of the Moon around Earth.
In NCERT Class 9, Chapter 10 (Gravitation) introduces this concept for the first time. NCERT Class 11, Chapter 8 (Gravitation) extends it to orbital mechanics, escape velocity, and gravitational potential energy. The Gravity Formula is a cornerstone of classical mechanics. It explains tides, satellite motion, and planetary orbits.
The value of the universal gravitational constant G is \( 6.674 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \). The acceleration due to gravity on Earth’s surface is approximately \( g = 9.8 \, \text{m/s}^2 \).
Gravity Formula — Expression and Variables
The universal gravitational force between two point masses is given by:
\[ F = G \frac{m_1 m_2}{r^2} \]
The acceleration due to gravity at Earth’s surface is derived from the above as:
\[ g = \frac{GM}{R^2} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Gravitational Force | Newton (N) |
| G | Universal Gravitational Constant | N m² kg−² |
| m&sub1; | Mass of first object | kilogram (kg) |
| m&sub2; | Mass of second object | kilogram (kg) |
| r | Distance between the centres of the two masses | metre (m) |
| g | Acceleration due to gravity | m/s² |
| M | Mass of Earth (or central body) | kilogram (kg) |
| R | Radius of Earth (or central body) | metre (m) |
Derivation of the Gravity Formula
Newton observed that the gravitational force F between two masses satisfies two conditions. First, \( F \propto m_1 m_2 \). Second, \( F \propto \dfrac{1}{r^2} \). Combining both proportionalities gives \( F \propto \dfrac{m_1 m_2}{r^2} \). Introducing a proportionality constant G yields the full expression \( F = G \dfrac{m_1 m_2}{r^2} \). To find g, consider a mass m on Earth’s surface. The gravitational force on it equals \( F = mg \). Setting this equal to \( G \dfrac{Mm}{R^2} \) and cancelling m gives \( g = \dfrac{GM}{R^2} \). Substituting \( M = 5.972 \times 10^{24} \) kg and \( R = 6.371 \times 10^6 \) m produces \( g \approx 9.8 \, \text{m/s}^2 \).
Complete Gravitation Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Universal Gravitation | \( F = G \dfrac{m_1 m_2}{r^2} \) | G = gravitational constant, m&sub1;, m&sub2; = masses, r = distance | N | Class 9, Ch 10; Class 11, Ch 8 |
| Acceleration due to Gravity (surface) | \( g = \dfrac{GM}{R^2} \) | M = mass of Earth, R = radius of Earth | m/s² | Class 9, Ch 10 |
| Variation of g with Height | \( g_h = g\left(1 – \dfrac{2h}{R}\right) \) | h = height above surface (h << R) | m/s² | Class 11, Ch 8 |
| Variation of g with Depth | \( g_d = g\left(1 – \dfrac{d}{R}\right) \) | d = depth below surface | m/s² | Class 11, Ch 8 |
| Gravitational Potential Energy | \( U = -\dfrac{Gm_1 m_2}{r} \) | Negative sign indicates attractive force | Joule (J) | Class 11, Ch 8 |
| Gravitational Potential | \( V = -\dfrac{GM}{r} \) | V = potential at distance r from mass M | J/kg | Class 11, Ch 8 |
| Escape Velocity | \( v_e = \sqrt{\dfrac{2GM}{R}} \) | Minimum speed to escape gravitational pull | m/s | Class 11, Ch 8 |
| Orbital Velocity | \( v_o = \sqrt{\dfrac{GM}{r}} \) | r = orbital radius from centre of Earth | m/s | Class 11, Ch 8 |
| Time Period of Satellite | \( T = 2\pi\sqrt{\dfrac{r^3}{GM}} \) | T = orbital time period | second (s) | Class 11, Ch 8 |
| Kepler’s Third Law | \( T^2 \propto r^3 \) | T = period, r = semi-major axis | s² / m³ | Class 11, Ch 8 |
Gravity Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: Calculate the gravitational force between two objects of mass 50 kg and 80 kg separated by a distance of 2 m. (Take \( G = 6.674 \times 10^{-11} \) N m² kg−²)
Given: \( m_1 = 50 \) kg, \( m_2 = 80 \) kg, \( r = 2 \) m, \( G = 6.674 \times 10^{-11} \) N m² kg−²
Step 1: Write the Gravity Formula: \( F = G \dfrac{m_1 m_2}{r^2} \)
Step 2: Substitute the values: \( F = 6.674 \times 10^{-11} \times \dfrac{50 \times 80}{2^2} \)
Step 3: Simplify the numerator and denominator: \( F = 6.674 \times 10^{-11} \times \dfrac{4000}{4} \)
Step 4: Calculate: \( F = 6.674 \times 10^{-11} \times 1000 = 6.674 \times 10^{-8} \) N
Answer
Gravitational Force \( F = 6.674 \times 10^{-8} \) N
Example 2 (Class 11-12 Level)
Problem: A satellite orbits Earth at a height of 400 km above the surface. Calculate the orbital velocity of the satellite. (Take \( G = 6.674 \times 10^{-11} \) N m² kg−², mass of Earth \( M = 5.972 \times 10^{24} \) kg, radius of Earth \( R = 6.371 \times 10^6 \) m)
Given: \( h = 400 \) km \( = 4 \times 10^5 \) m, \( M = 5.972 \times 10^{24} \) kg, \( R = 6.371 \times 10^6 \) m
Step 1: Find the orbital radius: \( r = R + h = 6.371 \times 10^6 + 4 \times 10^5 = 6.771 \times 10^6 \) m
Step 2: Write the orbital velocity formula: \( v_o = \sqrt{\dfrac{GM}{r}} \)
Step 3: Substitute values: \( v_o = \sqrt{\dfrac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.771 \times 10^6}} \)
Step 4: Compute the numerator: \( GM = 3.986 \times 10^{14} \) m³ s−²
Step 5: Divide: \( \dfrac{3.986 \times 10^{14}}{6.771 \times 10^6} = 5.888 \times 10^7 \) m² s−²
Step 6: Take the square root: \( v_o = \sqrt{5.888 \times 10^7} \approx 7673 \) m/s \( \approx 7.67 \) km/s
Answer
Orbital Velocity \( v_o \approx 7.67 \) km/s
Example 3 (JEE/NEET Level)
Problem: A body is taken to a height equal to the radius of Earth above the surface. By what factor does the acceleration due to gravity change? Also find the percentage decrease in g.
Given: Height \( h = R \) (where R is the radius of Earth), surface gravity \( g = 9.8 \) m/s²
Step 1: Use the exact formula for g at height h: \( g_h = \dfrac{GM}{(R+h)^2} \)
Step 2: Substitute \( h = R \): \( g_h = \dfrac{GM}{(R+R)^2} = \dfrac{GM}{4R^2} \)
Step 3: Relate to surface g: Since \( g = \dfrac{GM}{R^2} \), we get \( g_h = \dfrac{g}{4} \)
Step 4: Calculate percentage decrease: \( \% \text{ decrease} = \dfrac{g – g_h}{g} \times 100 = \dfrac{g – g/4}{g} \times 100 = \dfrac{3}{4} \times 100 = 75\% \)
Note: The approximate formula \( g_h \approx g(1 – 2h/R) \) is valid only when \( h \ll R \). When h is comparable to R, always use the exact formula \( g_h = g\dfrac{R^2}{(R+h)^2} \).
Answer
At height h = R, the acceleration due to gravity becomes \( g/4 \), which is a 75% decrease from the surface value.
CBSE Exam Tips 2025-26
- Memorise the value of G: Always write \( G = 6.674 \times 10^{-11} \) N m² kg−² in your answer. CBSE deducts marks if you use an incorrect value.
- Use the correct formula for height: We recommend using the exact formula \( g_h = g \dfrac{R^2}{(R+h)^2} \) when h is large. Use the approximate formula only when the problem states \( h \ll R \).
- Show all substitution steps: CBSE Class 9 and Class 11 marking schemes award step marks. Write the formula, then substitute, then simplify — never skip steps.
- Units are compulsory: Write SI units at every step. A missing unit in the final answer can cost you half a mark in board exams 2025-26.
- Distinguish between g and G: G is the universal gravitational constant (same everywhere). g is the acceleration due to gravity (varies with location). Confusing the two is a common error examiners flag.
- Learn Kepler’s Laws: CBSE Class 11 board papers frequently ask one short-answer question on Kepler’s Third Law alongside the Gravity Formula. Our experts suggest practising derivation of \( T^2 \propto r^3 \) from the orbital velocity formula.
Common Mistakes to Avoid
- Using diameter instead of radius: The formula uses the distance between the centres of the two objects. Students often use the diameter of Earth instead of the radius when calculating g. Always use \( R = 6.371 \times 10^6 \) m.
- Forgetting the negative sign in potential energy: Gravitational potential energy is \( U = -\dfrac{Gm_1 m_2}{r} \). The negative sign is essential. It indicates a bound system. Dropping it leads to wrong answers in energy conservation problems.
- Applying the approximate g-height formula for large h: The formula \( g_h = g(1 – 2h/R) \) is a binomial approximation. It is valid only when \( h \ll R \). For h = R or larger, use the exact expression.
- Confusing orbital velocity with escape velocity: Orbital velocity is \( v_o = \sqrt{GM/r} \). Escape velocity is \( v_e = \sqrt{2GM/R} \). Note that \( v_e = \sqrt{2} \, v_o \) only when the orbit is at the surface.
- Ignoring that g = 0 at Earth’s centre: At depth d = R (the centre), \( g_d = g(1 – R/R) = 0 \). Many students assume g is maximum at the centre, which is incorrect.
JEE/NEET Application of the Gravity Formula
In our experience, JEE aspirants encounter the Gravity Formula in at least 2–3 questions per paper, spanning gravitation, circular motion, and energy conservation. NEET papers consistently include one question on variation of g and one on satellite motion.
Pattern 1: Variation of g with Height and Depth
JEE Main frequently asks students to compare g at a height h above Earth’s surface with g at a depth d below the surface. The key insight is that g decreases with both height and depth. However, the rate of decrease differs. At height h, \( g_h = g\dfrac{R^2}{(R+h)^2} \). At depth d, \( g_d = g\left(1 – \dfrac{d}{R}\right) \). For the same numerical value of h and d (both much smaller than R), g decreases faster with height than with depth.
Pattern 2: Satellite Energy and Binding Energy
JEE Advanced tests the total energy of a satellite in orbit. The kinetic energy of a satellite is \( KE = \dfrac{GMm}{2r} \). The potential energy is \( PE = -\dfrac{GMm}{r} \). Therefore, total energy \( E = -\dfrac{GMm}{2r} \). The negative total energy confirms the satellite is bound. The binding energy is \( \dfrac{GMm}{2r} \). Students must remember that as a satellite moves to a higher orbit, its speed decreases but its total energy increases (becomes less negative).
Pattern 3: Gravitational Field Intensity
NEET and JEE both test gravitational field intensity \( I = \dfrac{GM}{r^2} \), which equals g at the surface. A common question asks for the field at the midpoint between two masses. Students must apply the Gravity Formula separately for each mass and then use vector addition (directions may oppose each other). In our experience, drawing a clear diagram before solving such problems saves significant time in the exam.
FAQs on Gravity Formula
For more related Physics formulas, explore our complete guide on Terminal Velocity Formula, understand charge flow with the Electric Current Formula, and revise wave-particle duality using the de Broglie Wavelength Formula. You can also browse the full Physics Formulas hub for a complete list of NCERT-aligned formula articles. For official NCERT textbook content, visit ncert.nic.in.