D.1.10 (HL)—Work in gravitational fields

Use the potential difference

For a mass $m$ moving from point 1 to point 2, external work in the stated convention equals the change in gravitational potential energy. Potential is scalar, so only the endpoints matter.

Worked example — changing orbit

For $m=850\,\mathrm{kg}$, $V_{g,1}=-5.07\times10^7\,\mathrm{J\,kg^{-1}}$ and $V_{g,2}=-5.40\times10^7\,\mathrm{J\,kg^{-1}}$, $W_{\mathrm{on}}=850[(-5.40)-(-5.07)]\times10^7=-2.8\times10^9\,\mathrm{J}$. The negative result means the satellite must lose energy to enter the lower orbit.

Distinguish the work agent

The work done by the gravitational field is the negative of the work done on the mass by an external agent when the motion is quasistatic:
$$W_{\text{field}}=-m\Delta V_g=m(V_{g,1}-V_{g,2})$$
Moving outward raises potential toward zero, so the field does negative work; moving inward lowers potential, so the field does positive work.

Read a graph

If a potential–distance graph gives $V_{g,1}$ and $V_{g,2}$, use their difference, not the area under the graph. An area under a force–distance graph can represent work, but a potential graph already gives work per unit mass through its vertical difference.

Common trap

Check whether the question asks for work by the field or work done on the mass. Reversing the order of the potential values changes the sign.