Gravitational potential energy
When mgh is not enough
Lift a school bag onto a shelf. You do work against gravity, and the bag stores it as potential energy. Near the ground we write this as m g h.
Gravity is a conservative force. The work done against it depends only on where you start and where you end, not on the path. That is why we can define a potential energy at all.
m g h assumes g is the same at every height. For a bag on a shelf, that is fine. Watch the bar of g beside the bag: it hardly changes.
But now lift a satellite thousands of kilometres up. g gets smaller as you go, so each extra kilometre costs less work than the one before. m g h will overestimate the work.
To get it right, we add up the work in tiny steps, using the true force G M m over r squared at each step. This needs integration, which you can see in the textbook.
The result: the work needed to lift m from a distance r₁ to a distance r₂ from the centre is G M m times one over r₁ minus one over r₂.
Zero at infinity
Potential energy needs a zero level. On a table, we might pick the floor. For gravity on a planetary scale, the natural choice is infinitely far away, where the pull is zero.
With U equal to zero at infinity, the potential energy at distance r from the centre of the Earth is U equals minus G M m over r.
Why negative? Start at infinity with zero. As the mass falls in, gravity does positive work and the potential energy goes down, below zero.
Watch the graph. U is lowest at the surface, minus G M m over R, and climbs towards zero far away, but never quite reaches it. It is like a deep well.
Climbing out of the well from the surface to infinity costs G M m over R. For each kilogram, that is g R, about 63 million joules.
The potential energy per unit mass is called the gravitational potential, V equals minus G M over r. Its unit is joule per kilogram.
And m g h? Zoom in near the surface. For a small climb h, the curve is almost a straight line, and the change in U is very nearly m g h. So m g h is a special case.
Try it: the gravity well
- The blue curve is U = −GMm/r for 1 kg. The dashed orange line is mgh.
- Pick the ISS: mgh and the true ΔU almost match.
- Go higher and watch mgh run away from the curve.
- mgh is too big by exactly h/R of the true value.
Quick check
A body of mass m is lifted from the Earth’s surface to a height equal to the Earth’s radius R. What is the gain in its potential energy?
- mgR
- mgR/2
- mgR/4
- 2mgR
Answer
B. ΔU = GMm(1/R − 1/2R) = GMm/2R. Since GM = gR², this is mgR/2. Using mgh would give mgR, twice too much, because g is much smaller at height R.
Negative energy
Why is the gravitational potential energy U = −GMm/r negative everywhere?
- Because the mass m is below the ground
- Because we chose U = 0 at infinity, and gravity pulls inwards, so U decreases as the body comes closer
- Because gravity is a non-conservative force
- Because G is a negative constant
Answer
B. U is zero at infinity by choice. Gravity attracts, so as a body comes in from infinity gravity does positive work and U drops below zero. G is positive, and gravity is a conservative force.
Summary
- Potential energy (U = 0 at infinity):
U = −GMm / r - Gravitational potential:
V = −GM / r - Work to lift from r₁ to r₂:
W = GMm (1/r₁ − 1/r₂) - Near the surface (h ≪ R):
ΔU ≈ mgh
- Gravity is conservative, so potential energy depends only on position.
- With the zero at infinity, U is negative everywhere and rises towards zero as r grows.
- mgh is an approximation that works only for heights much smaller than the Earth’s radius.
- To escape from the surface, each kilogram needs GM/R = gR ≈ 63 MJ.