Gravity

Gravity

The force which pulls masses towards each other. In high school math, we see this most often as the force which pulls objects downwards. Note: The force of gravity between two objects is jointly proportional to the mass of each object and inversely proportional to the square of the distance between between their centers of mass.

See also

Projectile motion

Key Formula

F = G\frac{m_1 \, m_2}{r^2}

Where:

$F$ = Gravitational force between the two objects (in newtons, N)
$G$ = Universal gravitational constant, approximately 6.674 × 10⁻¹¹ N·m²/kg²
$m_1$ = Mass of the first object (in kilograms)
$m_2$ = Mass of the second object (in kilograms)
$r$ = Distance between the centers of mass of the two objects (in meters)

Worked Example

Problem: Find the gravitational force between the Earth (mass 6.0 × 10²⁴ kg) and a 70 kg person standing on the surface. The radius of the Earth is 6.4 × 10⁶ m. Use G = 6.674 × 10⁻¹¹ N·m²/kg².

Step 1: Identify the known values and write the formula.

F = G\frac{m_1 \, m_2}{r^2}, \quad m_1 = 6.0 \times 10^{24}\text{ kg},\; m_2 = 70\text{ kg},\; r = 6.4 \times 10^{6}\text{ m}

Step 2: Calculate the numerator: G × m₁ × m₂.

G \cdot m_1 \cdot m_2 = 6.674 \times 10^{-11} \times 6.0 \times 10^{24} \times 70 = 2.803 \times 10^{16}\text{ N·m}^2

Step 3: Calculate the denominator: r².

r^2 = (6.4 \times 10^{6})^2 = 4.096 \times 10^{13}\text{ m}^2

Step 4: Divide to find the force.

F = \frac{2.803 \times 10^{16}}{4.096 \times 10^{13}} \approx 684\text{ N}

Step 5: Check: this is close to mg = 70 × 9.8 = 686 N, which confirms our answer is reasonable.

F \approx 684\text{ N}

Answer: The gravitational force on the person is approximately 684 N (about 154 pounds of force), consistent with what we expect from the simpler formula F = mg.

Another Example

This example uses the constant-acceleration form of gravity (g = 10 m/s²) in a projectile/free-fall problem, rather than Newton's universal gravitation formula between two masses.

Problem: A ball is dropped from a height of 80 meters near Earth's surface. Using the acceleration due to gravity g = 10 m/s², find how long it takes to hit the ground and its speed at impact. Ignore air resistance.

Step 1: Use the free-fall distance formula, where the initial velocity is 0.

h = \frac{1}{2}g\,t^2

Step 2: Substitute h = 80 m and g = 10 m/s², then solve for t.

80 = \frac{1}{2}(10)\,t^2 \implies 80 = 5\,t^2 \implies t^2 = 16 \implies t = 4\text{ s}

Step 3: Find the speed at impact using v = g·t.

v = 10 \times 4 = 40\text{ m/s}

Answer: The ball takes 4 seconds to hit the ground and reaches a speed of 40 m/s at impact.

Frequently Asked Questions

What is the difference between g and G in gravity formulas?

G (capital) is the universal gravitational constant, approximately 6.674 × 10⁻¹¹ N·m²/kg². It appears in Newton's law of universal gravitation and never changes. Lower-case g is the acceleration due to gravity at a specific location; near Earth's surface, g ≈ 9.8 m/s². The value of g actually comes from applying the universal law: g = GM/r², where M is Earth's mass and r is Earth's radius.

Why is gravity inversely proportional to the square of the distance?

Gravitational force spreads out over the surface area of a sphere as you move away from a mass. The surface area of a sphere is 4πr², so the force per unit area decreases with r². This is why doubling the distance between two objects reduces the gravitational force to one-quarter of its original value, not one-half.

When do you use F = mg instead of F = Gm₁m₂/r²?

Use F = mg when an object is near Earth's surface and you only need the gravitational force on it (its weight). This simplified formula assumes g is constant at about 9.8 m/s². Use the full formula F = Gm₁m₂/r² when you are dealing with two specific masses, objects far from Earth's surface, or problems involving other planets and celestial bodies.

Universal Gravitation (F = Gm₁m₂/r²) vs. Near-Surface Weight (F = mg)

	Universal Gravitation (F = Gm₁m₂/r²)	Near-Surface Weight (F = mg)
Definition	Force between any two masses anywhere in the universe	Force on an object due to a nearby planet's gravity
Formula	F = Gm₁m₂ / r²	F = mg
When to use	Two distinct masses, varying distances, orbital problems	Objects near Earth's surface, free-fall, projectile motion
Key constant	G = 6.674 × 10⁻¹¹ N·m²/kg²	g ≈ 9.8 m/s² (Earth) or 32 ft/s²
Assumptions	None — works universally	Assumes constant g (valid near planet's surface)

Why It Matters

Gravity appears throughout high school math and physics in projectile motion problems, free-fall calculations, and variation/proportion problems. Understanding the universal gravitation formula directly reinforces concepts of joint and inverse variation, since the force depends on the product of two masses (joint) and the reciprocal of distance squared (inverse). These same patterns show up in other real-world formulas, from electrical forces to light intensity.

Common Mistakes

Mistake: Forgetting to square the distance r in the denominator.

Correction: The law says inversely proportional to the square of the distance, so you must compute r² in the denominator. If the distance doubles, the force drops to 1/4, not 1/2.

Mistake: Confusing mass (kg) with weight (N) in gravity calculations.

Correction: Mass is a measure of how much matter an object contains (kilograms). Weight is the gravitational force on that mass (newtons). Weight = mass × g. A 70 kg person has a weight of about 686 N on Earth but only about 114 N on the Moon.

Related Terms

Joint Variation — Force varies jointly with both masses
Inverse Variation — Force varies inversely with distance squared
Center of Mass Formula — Distance r is measured between centers of mass
Projectile Motion Formula — Uses constant g for motion near Earth's surface
Force — Gravity is a specific type of force
Acceleration — g is the acceleration due to gravity
Inverse Square Law — Gravity follows an inverse-square relationship