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Work Done by Gravity

Work done by gravity depends only on vertical displacement, not the path taken.
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Derivation

The result

When a body of mass mm moves through a vertical displacement hh (downward), gravity does work:

Wgravity=mghW_{gravity} = mgh

If the body moves upward by hh, gravity does negative work: Wgravity=mghW_{gravity} = -mgh.

Crucially: the work done by gravity depends only on the vertical displacement, not on the path taken.

Derivation

Gravity is a constant force: Fg=mg\vec{F}_g = mg downward.

For a body moving at angle θ\theta to the vertical over displacement dd:

W=mgdcosθW = mgd\cos\theta

The vertical component of displacement is dcosθ=hd\cos\theta = h.

Wgravity=mghW_{gravity} = mgh

where hh is the vertical drop (positive downward).

Path independence — the key property

Consider three paths from point A (height HH) to point B (height 00):

  1. Straight vertical drop
  2. Inclined ramp
  3. Curved path

For all three:

Wgravity=mgHW_{gravity} = mgH

Only the vertical displacement matters. The horizontal component of displacement contributes nothing — gravity is vertical, so it has no component along any horizontal direction.

This path independence is the defining property of a conservative force. Gravity is conservative.

Sign convention

Taking upward as positive for displacement:

  • Body moves down by hh: displacement =h= -h, force =mg= -mg (downward in our convention means... )

Cleaner to state directly:

Wgravity=mg×(vertical drop)W_{gravity} = mg \times (\text{vertical drop})

  • Moving down: Wgravity>0W_{gravity} > 0 (gravity aids motion)
  • Moving up: Wgravity<0W_{gravity} < 0 (gravity opposes motion)
  • Moving horizontally: Wgravity=0W_{gravity} = 0 (gravity perpendicular to displacement)

Work done by gravity in a complete loop

If a body starts and ends at the same height (any closed path):

Wgravity=mg×0=0W_{gravity} = mg \times 0 = 0

Net work done by gravity over any closed path is zero. This is another way of stating that gravity is conservative.

Connection to gravitational potential energy

Work done by gravity = decrease in gravitational potential energy:

Wgravity=ΔU=(UfUi)=mghimghfW_{gravity} = -\Delta U = -(U_f - U_i) = mgh_i - mgh_f

When the body falls (decreases height), PE decreases and gravity does positive work — the PE converts to KE.

Note
This formula uses the approximation that $g$ is constant — valid near Earth's surface. For large distances from Earth (satellites, rockets), $g$ varies with height and $W = \int F \, dr$ must be used with $F = \frac{GMm}{r^2}$, giving $W = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$.