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Gravitational Force
Calculated Mathematically

Master Newton's Law of Universal Gravitation and learn to calculate the force between any two masses in the universe.

Physics

What is Gravitational Force?

Gravitational force is the attractive force that exists between any two objects that have mass. It is one of the four fundamental forces of nature and governs everything from falling apples to orbiting planets.

The Key Variables

\( F \)
Gravitational force between the two objects, measured in Newtons (N)
\( m_1, m_2 \)
Masses of the two objects, measured in kilograms (kg)
\( r \)
Distance between the centers of the two objects, in meters (m)
\( G \)
Universal gravitational constant: \( 6.674 \times 10^{-11} \; \text{N·m}^2/\text{kg}^2 \)

Visualizing the Interaction

\( m_1 \)
\( m_2 \)
\( r \)
\( F \)
\( F \)
Equal and opposite forces (Newton's 3rd Law)

Newton's Law of Universal Gravitation

Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The Inverse Square Relationship

Distance \( r \)
Force \( F \)
\( r = 1 \): strongest
\( r = 2 \): \( F = \tfrac{1}{4} \)
\( r = 3 \): \( F = \tfrac{1}{9} \)
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Gravitational Force vs. Weight

Gravitational Force Weight
Formula \( F = G \frac{m_1 m_2}{r^2} \) \( W = mg \)
Applies to Any two masses, anywhere Object near a planet's surface
Uses Universal constant \( G \) Local \( g \approx 9.8 \; \text{m/s}^2 \)
Scope Universal Special case of universal law

How to Calculate Step by Step

1
Identify the two masses \( m_1 \) and \( m_2 \) in kilograms. Convert if needed.
2
Find the distance \( r \) between their centers, in meters.
3
Write the formula: \( F = G \frac{m_1 m_2}{r^2} \)
4
Substitute values including \( G = 6.674 \times 10^{-11} \).
5
Compute carefully using scientific notation. Report the answer in Newtons.

Example 1: Two Masses on a Table

\( 50\;\text{kg} \)
\( 30\;\text{kg} \)
\( r = 2\;\text{m} \)

Example 2: Earth and the Moon

Earth
\( 5.97 \times 10^{24} \; \text{kg} \)
Moon
\( 7.35 \times 10^{22} \; \text{kg} \)
\( 3.84 \times 10^8 \; \text{m} \)

Example 3: Solving for Distance

Two 100 kg satellites experience a gravitational force of \( 1.112 \times 10^{-7} \; \text{N} \). How far apart are they?

What Happens When Variables Change?

Change Effect on \( F \) Factor
Double \( m_1 \) Force doubles \( \times 2 \)
Triple both masses Force x 9 \( \times 9 \)
Double \( r \) Force quartered \( \times \frac{1}{4} \)
Halve \( r \) Force x 4 \( \times 4 \)
Double \( m_1 \), double \( r \) Force halved \( \times \frac{1}{2} \)

Deriving \( g \) from Universal Gravity

Gravity on Other Worlds

Using \( g = G \frac{M}{R^2} \), we can calculate surface gravity anywhere:

Body Mass (kg) Radius (m) \( g \) (m/s²)
Earth \( 5.97 \times 10^{24} \) \( 6.371 \times 10^6 \) 9.82
Moon \( 7.35 \times 10^{22} \) \( 1.74 \times 10^6 \) 1.62
Mars \( 6.42 \times 10^{23} \) \( 3.39 \times 10^6 \) 3.73
Jupiter \( 1.90 \times 10^{27} \) \( 6.99 \times 10^7 \) 25.95

Common Mistakes to Avoid

✕ Forgetting to Square \( r \)
The distance must be squared in the denominator. Using \( r \) instead of \( r^2 \) gives a force that is far too large.
✕ Wrong Units
Masses must be in \(\text{kg}\) and distance in \(\text{m}\). Using \(\text{km}\) for distance (without converting) produces answers off by factors of \( 10^6 \).
✕ Confusing \( G \) with \( g \)
\( G = 6.674 \times 10^{-11} \; \text{N·m}^2/\text{kg}^2 \) is the universal constant. \( g = 9.8 \; \text{m/s}^2 \) is Earth's surface gravity. They are not interchangeable.

Question 1 of 5

Two objects of mass \( 10 \; \text{kg} \) each are separated by \( 1 \; \text{m} \). What is the gravitational force between them?

Question 2 of 5

If the distance between two objects is tripled, the gravitational force between them:

Question 3 of 5

What is the value and unit of the universal gravitational constant \( G \)?

Question 4 of 5

A 500 kg satellite orbits Earth at a distance of \( 7.0 \times 10^{6} \; \text{m} \) from Earth's center. What is the gravitational force on it? (Use \( M_\text{Earth} = 5.97 \times 10^{24} \; \text{kg} \))

Question 5 of 5

If both masses are doubled and the distance is also doubled, the gravitational force: