Work done moving a mass between two points
Work done by an external force in moving a mass m between two points is W = ΔU = m(φ_final − φ_initial). Since φ ∝ −1/r, doubling r halves the potential, making it less negative, so moving outwards gives positive work.
W = m(φ_f − φ_i) = (−GMm/r_f) − (−GMm/r_i)What this actually means
Subtract in the right order: FINAL minus INITIAL, always. Reverse it and you get a correct magnitude with the wrong sign, which in a sign-sensitive question is a wrong answer.
Work out the sign check first and the arithmetic second. Moving away from the planet means work done against gravity, so W is positive. Moving inwards means gravity does the work for you, so W is negative. If your sign disagrees with the direction, you have flipped the subtraction.
The scaling trick saves whole minutes. Since φ ∝ −1/r, if X and Y sit at R and 2R then φ_Y = φ_X/2. Given φ_X = −800 kJ kg⁻¹ you get φ_Y = −400 kJ kg⁻¹ immediately, and for 1 kg the work is (−400) − (−800) = +400 kJ.
That halving catches people out because dividing a negative number by two makes it BIGGER. −400 is greater than −800. The mass has gained energy by moving out, which is exactly what you expect.
Multiply by the actual mass at the end, not the middle. Potentials are per kilogram. A 3 kg mass over the same move does three times the work, so a question quoting φ in kJ kg⁻¹ and a mass in kg is testing whether you noticed.
This method sidesteps integration entirely, which is the whole reason potential was invented. You never need to know the path taken, only the two endpoints, because gravity is a conservative field.
Computing φ_initial − φ_final, or forgetting that a smaller-magnitude negative potential is a LARGER potential.
Prove it — watch it be true
- Place marker A at radius R and marker B at radius 2R with the work-done two-marker tool.
- Check the tool reports φ at B as exactly half the value at A, and the work as positive.
- Swap the two markers and confirm the work flips sign with the same magnitude.