D.1 · IB Physics
Gravitational field and potential
Around a spherical planet the field strength falls as 1/r² and the potential rises towards zero as −1/r. Field lines point radially inwards and the dashed equipotentials are drawn at equal steps of potential. Move a mass from point 1 to point 2 and read the work done.
Predict first
Start from the default settings (Earth, Point 1 at 1.00 R, Point 2 at 3.00 R). Move Point 1 out to 2.00 R. How does the field strength at point 1 change?
g = GM ÷ r² is an inverse-square law, so doubling r divides g by 2² = 4: from 9.81 to 2.45 N kg⁻¹. The potential V = −GM ÷ r would only halve.
Check it in the model: Move the Point 1 slider from 1.00 R to 2.00 R and compare Field strength at 1 (9.81 then 2.45 N kg⁻¹).
Arrow length at each point is proportional to g. Equipotentials are spaced at equal steps of V, so they spread out as the field weakens. The graph starts at the surface; the interior is not modelled.
V = −GM ÷ r
Ep = mV = −GMm ÷ r
W = mΔV = m(V₂ − V₁), any path
G = 6.67 × 10⁻¹¹ N m² kg⁻² ; r from the centre
Try this
- On Earth, set point 1 to 1.00 R and point 2 to 2.00 R. Compare g at the two points, then compare V₁ and V₂. Which quantity falls by a factor of four, and which by two?
- Use the preset "Earth: potential well" and look at the equipotentials. Why are they closer together near the surface?
- Compare the work to lift the same mass from R to 2R with the work from 2R to 3R. Why is the second one smaller?
Common error: using the height above the surface as r. A satellite 400 km up is at r = 6371 + 400 = 6771 km from Earth's centre; also, V is negative everywhere, so moving away makes V less negative and ΔV positive.