Physics › Gravitational fields › Gravitational potential
Gravitational potential
Gravitational potential attaches a number to every location in a field, the work per kilogram needed to arrive there from infinity. Every value comes out negative, the zero sits at infinity, and one pair of graphs converts between potential and field strength in both directions.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
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.
- Explain why no work is done moving along an equipotential surface.
- Connect the g and V graphs, with g the negative gradient of V and the size of ΔV the area under g against r.
COMMON MISCONCEPTION
It takes energy to move anywhere in a gravitational field.
Potential, measured from infinity
The gravitational potential V at a point is the work done per unit mass to bring a small mass from infinity to that point. The zero is at infinity, far from everything. In the radial field of a mass M,
The negative sign matters. Gravity attracts, so the field itself does the work as a mass arrives and nobody has to supply any energy; every location in the universe therefore sits below zero. Moving away from a mass means climbing, with V rising towards zero, and the picture to hold on to is a well. Every planet sits at the bottom of one, and leaving means climbing the full depth.
The potential difference between two points sets the energy transferred in moving between them, and for a mass m,
WORKED EXAMPLE
The potential at the Earth's surface
Find the gravitational potential at the Earth's surface (M = 5.97 × 1024 kg, R = 6.37 × 106 m).
V = −GM/r = −(6.67 × 10−11 × 5.97 × 1024)/(6.37 × 106).
V = −6.3 × 107 J kg−1. Every kilogram standing here sits some sixty-three megajoules below free space.
The sign is the physics. You are in a well, and the number is the depth of your shelf; lifting anything anywhere begins by paying against it.
The free directions
Surfaces of constant potential, equipotentials, ring a mass the way contour lines ring a hill, and they always cross the field lines at right angles. Moving along an equipotential changes V by nothing, so no work is done. A satellite in a circular orbit rides one equipotential the whole way round, takes no work from gravity, and needs no engine to keep going. Only motion between equipotentials costs anything.
The graph dictionary
You need fluency with the graphs of g and V against r, and with the rules that translate between them. Field strength is the negative gradient of the potential,
steep potential, strong field, and the minus sign aims the force downhill into the well. The dictionary reads the other way too.
the size of the potential difference between two distances is the area under the g against r graph between them, the graph plotting g as a positive magnitude. Gradient in one direction, area in the other, the same pairing motion graphs taught in Year 12, now working on fields.
GUIDED PRACTICE
g, recovered from the dictionary
Near the surface, the potential changes from −6.253 × 107 to −6.185 × 107 J kg−1 over a climb of 70 km. Use the graph dictionary to recover the field strength.
Show the working
g = |ΔV/Δr| = (6.8 × 105)/(7.0 × 104) = 9.7 N kg−1, the magnitude of the potential gradient.
Just below 9.8, as it must be for an average taken over a climb above the surface where the field is weakening. The gradient of the potential curve is the field, and the dictionary between the two graphs runs both ways.
ASSESSMENT FOCUS
- Define V with both ingredients, work done per unit mass and brought from infinity. Omit either half and the mark goes.
- The negative sign has a stock explanation. Gravity attracts and the zero sits at infinity, so every real location lies below it. Learn that as one 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 its speed” is the same fact asked backwards.
- ΔW = mΔV carries signs. Move outward and ΔV is positive, so work must be supplied; fall inward and the field does the work for you.
- The graph dictionary earns structured marks. g is minus the gradient of V against r, and the size of ΔV is the area under g against r. Quote whichever direction the data supports.
- Potential and potential energy are not the same quantity. V is per kilogram, in J kg−1; the potential energy of a mass m sitting at that point is mV, in joules. An answer carrying the wrong unit has usually confused the two ideas, and the arithmetic was never the problem.
CHECK YOURSELF
At the Earth's surface V = −6.25 × 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 energy needed is m times the potential difference.
Show the answer
ΔV = (−0.94 × 107) − (−6.25 × 107) = +5.31 × 107 J kg−1. The climb runs out of the well, so ΔV comes out positive.
= 1200 × 5.31 × 107 = 6.4 × 1010 J.
Sixty-four gigajoules simply to be there, before a single joule of orbital kinetic energy. Launch costs are mostly the depth of the well.
Potential is the work per kilogram to arrive from infinity, and it is always negative.
Gradient gives g from V; area gives ΔV from g.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the gravitational potential questions page.
WHERE TO GO NEXT
- Gradients and areas under graphs is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
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- Define gravitational potential with its zero at infinity, and use ΔW = mΔV.
- Use V = −GM/r and explain the significance of the negative sign.
- Explain why no work is done moving along an equipotential surface.
- Connect the g and V graphs, with g the negative gradient of V and the size of ΔV the area under g against r.
Open the full revision checklist to track your progress across the whole unit.