Work Done by Gravitational Force Formula: Complete Guide with Examples
The work done by gravitational force is a fundamental concept in physics that describes how gravity affects objects when they move under its influence. Whether lifting a book, dropping a ball, or launching a satellite, understanding the work done by gravity helps explain energy transfer in mechanical systems. This article explores the formula for calculating gravitational work, its derivation, practical applications, and frequently asked questions to provide a comprehensive understanding of this essential physics principle.
Formula for Work Done by Gravitational Force
The work done by gravitational force depends on the force applied and the displacement of the object. For near-Earth scenarios, where the gravitational acceleration g is approximately constant, the formula simplifies to:
W = mgh
Where:
- W = work done by gravity (in joules, J)
- m = mass of the object (in kilograms, kg)
- g = acceleration due to gravity (≈ 9.81 m/s² on Earth)
- h = vertical displacement (in meters, m)
For large-scale or orbital motion, where gravity varies with distance, the general formula uses integration:
W = ∫ F · dr = –GMm (1/r₂ – 1/r₁)
Where:
- G = gravitational constant (6.Here's the thing — 674 × 10⁻¹¹ N·m²/kg²)
- M = mass of the larger body (e. g.
Derivation of the Formula
Near-Earth Approximation (W = mgh)
When an object moves vertically near Earth’s surface, the gravitational force (F = mg) acts downward. If the object is displaced upward by height h, the angle between force and displacement is 180°, making the work negative:
W = F · d · cos(θ) = mg · h · cos(180°) = –mgh
This negative sign indicates gravity opposes upward motion. Conversely, when the object falls downward, h is negative, and work done by gravity is positive Most people skip this — try not to..
Universal Gravitation (W = –GMm (1/r₂ – 1/r₁))
For motion in a gravitational field (e.g., planetary orbits), the gravitational force varies with distance: F = GMm/r². The work done is calculated by integrating this force over the path:
W = ∫₍ᵣ₁₎₍ᵣ₂₎ (GMm/r²) dr = –GMm [1/r]₍ᵣ₁₎₍ᵣ₂₎ = –GMm (1/r₂ – 1/r₁)
This formula accounts for the inverse-square nature of gravity and applies to any radial motion in a gravitational field Easy to understand, harder to ignore..
When to Use the Formula
The mgh formula is suitable for:
- Objects moving vertically near Earth’s surface (e.Still, g. , elevators, falling apples).
- Situations where g is constant and distances are small compared to Earth’s radius.
The universal formula is used for:
- Satellites orbiting Earth or other celestial bodies.
- Objects moving through space where gravity changes significantly with distance.
- Calculating gravitational potential energy differences.
Step-by-Step Examples
Example 1: Lifting a Box Near Earth’s Surface
A 5 kg box is lifted vertically to a shelf 2 meters above the ground. Calculate the work done by gravity And that's really what it comes down to. Nothing fancy..
Solution:
- Mass (m) = 5 kg
- Height (h) = 2 m
- Gravity (g) = 9.81 m/s²
W = –mgh = –(5)(9.81)(2) = –98.1 J
The negative sign indicates gravity opposes the upward motion.
Example 2: Launching a Satellite
A 1000 kg satellite is moved from Earth’s surface (radius = 6.37 × 10⁶ m) to a circular orbit 400 km above the surface. Calculate the work done by gravity Took long enough..
Solution:
- G = 6.674 × 10⁻¹¹ N·m²/kg²
- M (Earth) = 5.97 × 10²⁴ kg
- r₁ = 6.37 × 10⁶ m
- r₂ = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m
W = –GMm (1/r₂ – 1/r₁)
Plugging in values:
W = –(6.674 × 10⁻¹¹)(5.97 × 10²⁴)(1000) [1/(6.77
