Physics Formulae Mechanics Attractive Force

Subset – Definition and Properties

Learn to use the attractive force formula to calculate the gravitational pull between two objects. This page explains th...
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Definition

Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. This fundamental force governs the motion of planets, satellites, and falling objects, unifying terrestrial and celestial mechanics under a single mathematical framework.

Historical Development:

  • Isaac Newton (1687): Formulated the law in his seminal work Principia Mathematica, with the key insight that the same force causing an apple to fall is responsible for keeping the Moon in orbit around the Earth.
  • Henry Cavendish (1798): Performed the first experimental measurement of the gravitational constant, G, using a highly sensitive torsion balance. This experiment was famously dubbed "weighing the Earth."
  • Albert Einstein (1915): Developed the theory of General Relativity, which refined the understanding of gravity as a curvature of spacetime caused by mass and energy. Newtonian gravity remains an excellent approximation for most non-relativistic scenarios.

Physical Properties

The gravitational attractive force is a fundamental, long-range interaction that possesses several key physical properties defining its behavior in classical mechanics.

PropertyDetails
NatureVector quantity, possessing both magnitude and direction.
SI UnitsNewton (N)
MagnitudeDirectly proportional to the product of the two interacting masses and inversely proportional to the square of the distance between their centers.
DirectionAlways attractive, directed along the straight line connecting the centers of the two masses.
Conservative ForceThe work done by gravity on an object moving between two points is independent of the path taken. This implies that mechanical energy (kinetic + potential) is conserved in a system under only gravitational influence.
Dimensional Formula[M][L][T]^-2
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Diagram & Visualization

r m₁ m₂ F F
Two masses (m₁ and m₂) exert an equal and opposite attractive gravitational force (F) on each other, separated by a distance (r).
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Key Formulas

\[ F = G \frac{m_1 m_2}{r^2} \]
Magnitude of Gravitational Force
\[ \vec{F}_{12} = -G \frac{m_1 m_2}{r^2} \hat{r}_{12} \]
Vector Form of Gravitational Force
\[ U = -G \frac{m_1 m_2}{r} \]
Gravitational Potential Energy
\[ g = \frac{GM}{r^2} \]
Gravitational Field Strength (Acceleration)
\[ G \approx 6.674 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2} \]
Gravitational Constant
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Variables

SymbolQuantitySI UnitDescription
\( F \)Gravitational ForceNewton (N)The mutual attractive force between the two masses.
\( G \)Gravitational ConstantN⋅m²/kg²An empirical physical constant determining the strength of gravity.
\( m_1, m_2 \)MassKilogram (kg)The masses of the two interacting objects.
\( r \)Separation DistanceMeter (m)The distance between the centers of mass of the two objects.
\( \vec{F}_{12} \)Force VectorNewton (N)The vector force exerted on object 1 by object 2.
\( \hat{r}_{12} \)Unit VectorDimensionlessThe unit vector pointing from object 1 towards object 2.
\( U \)Gravitational Potential EnergyJoule (J)The potential energy of the system of two masses due to their gravitational attraction.
\( g \)Gravitational Field Strengthm/s²The acceleration due to gravity experienced by an object in the gravitational field of a mass M.
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Derivation

We can derive the familiar value for acceleration due to gravity on Earth's surface, \( g \approx 9.8 \, \mathrm{m/s^2} \), from Newton's Law of Universal Gravitation and Newton's Second Law of Motion.

Step 1: Start with Newton's Second Law, where the force on an object is its mass times its acceleration. In this case, the force is the object's weight (W), and the acceleration is \( g \).

\[ F = W = m_{object}g \]

Step 2: State Newton's Law of Universal Gravitation for an object of mass \( m_{object} \) on the surface of the Earth (mass \( M_{Earth} \), radius \( R_{Earth} \)).

\[ F = G \frac{M_{Earth} m_{object}}{R_{Earth}^2} \]

Step 3: Equate the two expressions for the force \( F \).

\[ m_{object}g = G \frac{M_{Earth} m_{object}}{R_{Earth}^2} \]

Step 4: Cancel the mass of the object, \( m_{object} \), from both sides. This shows that the acceleration due to gravity is independent of the object's mass.

\[ g = G \frac{M_{Earth}}{R_{Earth}^2} \]
Derived formula for gravitational acceleration

Step 5: Substitute the known values for G, Earth's mass, and Earth's radius to calculate the numerical value of \( g \).

\[ g = (6.674 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2}) \frac{(5.972 \times 10^{24} \, \mathrm{kg})}{(6.371 \times 10^6 \, \mathrm{m})^2} \approx 9.81 \, \mathrm{m/s^2} \]
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Types & Special Cases

While the fundamental law of universal gravitation is always applicable, its calculation is often simplified for specific physical scenarios and mass distributions.

Type / CaseDescriptionWhen to Use
Uniform Gravitational FieldAn approximation where the gravitational force is assumed to be constant in magnitude and direction over a region of space. The force is calculated as F = mg.For objects near the surface of a large celestial body (like Earth), where changes in distance from the center are negligible compared to the body's radius.
Point Mass InteractionThe general case described by Newton's law, where objects are treated as if their entire mass is concentrated at a single point.When the distance between the objects is much larger than their individual sizes, or for objects that are true point particles.
Spherically Symmetric BodyA body whose mass density depends only on the distance from its center. It exerts a gravitational force on an external object as if its entire mass were concentrated at its center.For calculating the force exerted by planets, stars, and other nearly spherical objects on bodies outside of them.
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Worked Example

Given two point masses, \(m_1 = 500 \, \mathrm{kg}\) and \(m_2 = 1500 \, \mathrm{kg}\), separated by a distance of \(r = 2 \, \mathrm{m}\). Calculate the magnitude of the gravitational force between them.
  1. Start with Newton's Law of Universal Gravitation: \[ F = G \frac{m_1 m_2}{r^2} \]
  2. Substitute the known values: \( G = 6.674 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2} \), \( m_1 = 500 \, \mathrm{kg} \), \( m_2 = 1500 \, \mathrm{kg} \), and \( r = 2 \, \mathrm{m} \).
  3. Calculate the product of the masses: \( m_1 m_2 = 500 \times 1500 = 750,000 \, \mathrm{kg^2} \)
  4. Calculate the square of the distance: \( r^2 = 2^2 = 4 \, \mathrm{m^2} \)
  5. Substitute these values into the formula: \[ F = (6.674 \times 10^{-11}) \frac{750000}{4} \]
  6. Perform the final calculation: \[ F = (6.674 \times 10^{-11}) \times 187500 \approx 1.25 \times 10^{-5} \, \mathrm{N} \]
The magnitude of the gravitational force between the two masses is approximately \(1.25 \times 10^{-5} \, \mathrm{N}\).
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Try It

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Applications

Space Exploration: The formula is essential for calculating satellite orbits, planning interplanetary trajectories for probes like Voyager and the Mars rovers, executing gravity-assist maneuvers to save fuel, and maintaining the orbit of the International Space Station.

Astronomy and Astrophysics: Used to understand planetary motion within our solar system, model binary star systems, calculate the mass of galaxies (leading to the inference of dark matter), and describe the large-scale structure of the universe.

Geophysics and Earth Sciences: Explains the ocean tides caused by the gravitational pull of the Moon and Sun. It is also used in gravitational surveying to detect variations in Earth's density, which can indicate mineral deposits or geological structures.

Precision Technology: The principles of gravity are crucial for the operation of Global Positioning Systems (GPS), which must account for relativistic effects. It also underpins the technology behind gravitational wave detectors like LIGO, which measure tiny distortions in spacetime.

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Real-World Examples

Calculate the gravitational force between Earth and the Moon. Given: M_Earth = 5.97 × 10²⁴ kg, M_Moon = 7.35 × 10²² kg, and the average distance r_EM = 3.84 × 10⁸ m.
  1. Write the formula: \[ F_{EM} = G \frac{M_E M_M}{r_{EM}^2} \]
  2. Substitute the values: \[ F_{EM} = (6.67 \times 10^{-11}) \frac{(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2} \]
  3. Calculate the result: \[ F_{EM} = 1.99 \times 10^{20} \text{ N} \]
The gravitational force between the Earth and the Moon is approximately 1.99 × 10²⁰ N. This immense force is what keeps the Moon in orbit.
A 1000 kg spacecraft is in a circular orbit 400 km above Earth's surface. Calculate its orbital speed. Given: h = 4×10⁵ m, R_Earth = 6.37×10⁶ m, M_Earth = 5.97×10²⁴ kg.
  1. First, find the total orbital radius: \( r = R_{Earth} + h = 6.37\times10^6 + 4\times10^5 = 6.77\times10^6 \, \mathrm{m} \).
  2. For a circular orbit, the gravitational force provides the necessary centripetal force: \[ \frac{GM_E m}{r^2} = \frac{mv^2}{r} \]
  3. Solve for the orbital speed, \( v \): \[ v = \sqrt{\frac{GM_E}{r}} \]
  4. Substitute the values: \[ v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.77 \times 10^6}} \]
  5. Calculate the final speed: \[ v \approx 7670 \text{ m/s} \]
The spacecraft's orbital speed is approximately 7670 m/s, or about 27,600 km/h.
In a recreation of the Cavendish experiment, two 15 kg lead balls are placed with their centers 0.12 m apart. Calculate the gravitational force between them.
  1. Use the universal gravitation formula: \[ F = G \frac{m_1 m_2}{r^2} \]
  2. Substitute the given values: \[ F = (6.67 \times 10^{-11}) \frac{(15)(15)}{(0.12)^2} \]
  3. Calculate the force: \[ F = (6.67 \times 10^{-11}) \frac{225}{0.0144} \approx 1.04 \times 10^{-6} \text{ N} \]
The gravitational force between the two lead balls is approximately 1.04 × 10⁻⁶ N. This extremely small force highlights the weakness of gravity at human scales and the sensitivity required for Cavendish's experiment.
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Real-World Scenarios

Planetary Orbits
The Sun's gravitational force continuously pulls on planets, keeping them in stable orbits. This attractive force depends on their masses and the distance between them.
Ocean Tides
The Moon's gravitational pull creates bulges in Earth's oceans, causing high and low tides. This demonstrates how attractive force creates tangible effects across vast distances.
Feeling Weight
Your weight is the attractive gravitational force between your mass and the Earth's mass. This force constantly pulls you towards the planet's center.

Planetary Orbits: The Sun's immense mass exerts a continuous gravitational pull on Earth, forcing it into a stable elliptical orbit. This gravitational lock is what determines the length of our year and creates the seasons.

Ocean Tides: The gravitational forces from the Moon and, to a lesser extent, the Sun pull on Earth's oceans. This creates bulges of water on the sides of the Earth facing and opposite the Moon, resulting in the predictable rise and fall of tides.

Formation of Celestial Bodies: On a cosmic scale, gravity is the architect of the universe. It pulls vast clouds of gas and dust together to form stars and planets, and it holds trillions of stars together to form galaxies like our own Milky Way.

Feeling Weight: The sensation of weight is the direct result of the gravitational force between your body's mass and the mass of the entire Earth. This force pulls you towards the Earth's center, and the ground pushes back with an equal and opposite force that you feel.

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Limitations

⚠️ This formula is an approximation that works extremely well for weak gravitational fields and low velocities. For objects moving near the speed of light or in very strong gravitational fields (like near a black hole), Einstein's theory of General Relativity is required for accurate predictions.
⚠️ The formula assumes objects are point masses or spherically symmetric bodies. For irregularly shaped objects at close distances, the calculation of gravitational force requires complex integration over the volumes of both objects.
💡 The formula describes a two-body system. In reality, every object is influenced by the gravity of all other objects. For precise calculations, like spacecraft trajectories, the gravitational pulls of the Sun, Moon, and other planets must be accounted for (this is known as the n-body problem).

Common Mistakes

⚠️ Forgetting to square the distance (r). The force follows an inverse square law, so \( F \propto 1/r^2 \), not \( 1/r \). This is one of the most frequent algebraic errors.
⚠️ Using incorrect distance measurements. The distance \( r \) must be measured from the center of mass of one object to the center of mass of the other. For orbital problems, this means adding the planet's radius to the altitude above the surface (\( r = R_{planet} + h \)).
⚠️ Unit conversion errors. All masses must be in kilograms (kg) and all distances must be in meters (m) to be consistent with the SI units of the gravitational constant G (\( \mathrm{N \cdot m^2/kg^2} \)). Using kilometers for distance is a common error.
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Units and Dimensions

Dimensional analysis ensures the formula is consistent. The dimensions are expressed in terms of Mass (M), Length (L), and Time (T).

QuantitySymbolSI UnitDimensional Formula
ForceFNewton (kg·m/s²)[M][L][T]⁻²
MassmKilogram (kg)[M]
DistancerMeter (m)[L]
Gravitational ConstantGN·m²/kg²[M]⁻¹[L]³[T]⁻²

Checking the dimensions of the law of gravitation:

\[ [F] = [G] \frac{[m_1][m_2]}{[r^2]} = ([M]^{-1}[L]^3[T]^{-2}) \frac{[M][M]}{[L]^2} = [M][L][T]^{-2} \]

The resulting dimensions match the dimensions of force, confirming the consistency of the equation.

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Study Strategy

1 🧠 Grasp the Fundamentals
  • Thoroughly read the DEFINITION section to understand that this force is universal, attracting any two objects with mass.
  • Identify each variable in the formula: F (Force), G (Gravitational Constant), m1 and m2 (masses), and r (distance).
  • Focus on the relationships: force is directly proportional to the product of the masses and inversely proportional to the square of the distance.
  • Internalize that 'r' is the distance between the centers of mass, which is critical for objects like planets and satellites.
2 📝 Commit the Formula to Memory
  • Write the full formula, F = G * (m1 * m2) / r², repeatedly until you can do it from memory without error.
  • Create a mnemonic to remember the components, like 'Funny Giraffes Munch Mangoes over Red Squares' for F, G, m1, m2, and r².
  • Verbally explain the formula's meaning to a study partner or out loud to yourself, defining each term and its role.
  • Use flashcards for the formula itself and for the value and units of the gravitational constant, G (6.674 × 10⁻¹¹ N⋅m²/kg²).
3 ✍️ Practice with Problems
  • Start with basic calculations: find the force between Earth and the Moon using their known masses and distance.
  • Review the COMMON_MISTAKES section and actively check your work to ensure you are always squaring the distance 'r'.
  • For orbital problems, practice the specific skill mentioned in COMMON_MISTAKES: add the planet's radius to the satellite's altitude to get the correct 'r'.
  • Challenge yourself by solving for other variables, such as rearranging the formula to find the mass of a planet or the required orbital distance.
4 🌍 Connect to Real-World Physics
  • Read the APPLICATIONS section and visualize how this formula is used to calculate trajectories for space probes like Voyager.
  • Connect the formula to astronomy by considering how it governs the orbits of planets, moons, and even binary star systems.
  • Think about the International Space Station. How does this force keep it in orbit while its tangential velocity keeps it from falling?
  • Explore the concept of gravity assists mentioned in APPLICATIONS. Understand how a spacecraft can use a planet's gravity to change its speed and direction.
Master this universal law by understanding its concepts, practicing diligently, and connecting it to the cosmos it governs.

Frequently Asked Questions

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