Physics › Gravitational fields › Gravitational potential
Gravitational potential
Potential prices every location in a gravitational field: the work per kilogram to arrive there from infinity. The values are negative, the zero lives infinitely far away, and one graph converts between potential and field strength in both directions.
Builds on Newton's law of gravitation and Conservation of energy.
IN THIS TOPIC
- Define gravitational potential with its zero at infinity, and use ΔW = mΔV.
- Use V = −GM/r and explain the significance of the negative sign.
- Connect the g and V graphs: g = −ΔV/Δr, and ΔV as the area under g against r.
WHAT YOU PROBABLY THINK
It takes energy to move anywhere in a gravitational field.
Pricing a location
The gravitational potential V at a point is the work done per unit mass to bring a small mass from infinity to that point. Infinity is where the zero lives: far from everything, a mass owes nothing. In the radial field of a mass M,
The negative sign carries the physics. Gravity attracts, so the field does work on an arriving mass; arriving costs less than nothing, and every location in the universe sits below zero. Moving away from a mass means climbing, V rising toward zero, and the picture to hold is a well: every planet sits at the bottom of one, and leaving means paying the full depth.
The potential difference between two points prices a move between them, and the energy bill for a mass m is
The free directions
Surfaces of constant potential, equipotentials, ring a mass like contour lines ring a hill, always crossing the field lines at right angles. Moving along an equipotential involves no change in V, so no work is done: a satellite in a circular orbit rides a single equipotential the whole way round, which is exactly why it needs no engine. The lie above dies here: only motion between equipotentials costs anything.
Two graphs, one dictionary
The spec wants fluency with the graphs of g and V against r, and with the translation rules between them. Field strength is the negative gradient of the potential:
steep potential, strong field, and the minus sign points the force downhill into the well. The dictionary reads the other way too:
the potential difference between two distances is the area under the g against r graph between them. Gradient in one direction, area in the other, the same pairing motion graphs taught in Year 12, now working on fields.
THE EXAM BIT
- Define V with both ingredients: work done per unit mass, brought from infinity. Omitting either half drops the mark.
- The negative sign has a stock explanation: gravity is attractive, the zero is at infinity, so every real location lies below zero. Learn it as a sentence.
- No work is done moving along an equipotential; a circular orbit is the standard example, and “why does the satellite need no fuel to maintain speed” is this fact in costume.
- ΔW = mΔV works with signs: moving outward, ΔV is positive and work must be supplied; falling inward, the field pays.
- The graph dictionary earns structured marks: g is minus the gradient of V against r; ΔV is the area under g against r. Quote whichever direction the data supports.
CHECK YOURSELF
At the Earth's surface V = −6.26 × 107 J kg−1, and at a satellite's orbit V = −0.94 × 107 J kg−1. Find the energy needed to lift a 1200 kg satellite between the two, ignoring its kinetic energy.
Show a hint
The price is m times the potential difference.
Show the answer
ΔV = (−0.94 × 107) − (−6.26 × 107) = +5.32 × 107 J kg−1: the climb is out of the well, so ΔV is positive.
ΔW = mΔV = 1200 × 5.32 × 107 = 6.4 × 1010 J.
Sixty-four gigajoules just to be there, before a single joule of orbital kinetic energy: launch costs are mostly the well's depth.
Potential prices arrival from infinity, and it's always negative.
Gradient gives g from V; area gives ΔV from g.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.