Physics • Year 12 • Module 5 • Lesson 14
Gravitational Potential Energy
Apply U = −GMm/r and ΔU = GMm(1/r1 − 1/r2) to real-data scenarios, graph data, and the limits of the mgh approximation.
1. Interpret orbital data — GPE at different altitudes
The table below gives data for a 1000 kg satellite at several orbital radii above Earth. Use G = 6.67 × 10−11 N m2 kg−2 and ME = 5.97 × 1024 kg throughout. 9 marks
| Orbital radius r (m) | Altitude above surface (km) | U (J) — calculate | |U| = binding energy (J) |
|---|---|---|---|
| 6.77 × 106 | 400 | ||
| 7.37 × 106 | 1 000 | ||
| 2.00 × 107 | 13 630 | ||
| 4.22 × 107 | 35 830 (GEO) |
1.1 Calculate U and |U| for all four rows. Show working for one row as an example. 5 marks
1.2 Describe the trend in U as the orbital radius increases. Explain why U approaches zero rather than becoming a large positive number. 2 marks
1.3 How much energy must be supplied to move the satellite from the 400 km orbit to GEO? Write the formula and substitute values. 2 marks
2. Interpret graph — GPE versus orbital radius
The graph below shows GPE (in units of 1010 J) for a 1000 kg satellite as a function of orbital radius r. 7 marks
Figure 2. GPE of a 1000 kg satellite vs orbital radius. G = 6.67×10−11 N m2 kg−2, ME = 5.97×1024 kg. Illustrative data.
2.1 Describe the shape of the GPE–r curve and explain why it has this particular shape (not linear). 2 marks
2.2 Read the graph to estimate the GPE at GEO (r ≈ 4.22 × 107 m) and at LEO (r ≈ 6.77 × 106 m). Hence estimate the energy needed to move the satellite from LEO to GEO. 3 marks
2.3 Explain why the curve approaches but never crosses U = 0. What would it mean physically if U became positive? 2 marks
3. Compare U = −GMm/r with ΔU ≈ mgh
Complete the two-column comparison table. For each feature, write a concise contrasting description. 10 marks (1 per cell)
| Feature | Exact formula: U = −GMm/r | Approximation: ΔU ≈ mgh |
|---|---|---|
| When valid? | ||
| Reference level (zero GPE) | ||
| Sign of ΔU when object rises | ||
| How g (field strength) is treated | ||
| Example where you MUST use this formula |
4. Predict and justify — a Martian surface scenario
Mars has mass MMars = 6.39 × 1023 kg and mean surface radius RMars = 3.39 × 106 m. A 500 kg lander probe is on the surface. A mission planner considers two options to park it in orbit:
- Option A: Low Mars orbit (LMO) at altitude h = 400 km above the surface.
- Option B: Higher orbit at r = 1.00 × 107 m from Mars’ centre.
6 marks
4.1 Calculate the energy required to move the probe from Mars’ surface to each orbital option. Show full working including r values. 4 marks
4.2 Predict whether the mgh approximation (using Mars surface g = 3.72 m s−2) would give a reasonable estimate for either option. Justify your prediction without calculating. 2 marks
Q1.1 — GPE table (5 marks)
GM = (6.67 × 10−11)(5.97 × 1024) = 3.982 × 1014 N m2 kg−1; m = 1000 kg; GMm = 3.982 × 1017 J m.
r = 6.77 × 106 m: U = −3.982 × 1017 / 6.77 × 106 = −5.88 × 1010 J; |U| = 5.88 × 1010 J.
r = 7.37 × 106 m: U = −3.982 × 1017 / 7.37 × 106 = −5.40 × 1010 J; |U| = 5.40 × 1010 J.
r = 2.00 × 107 m: U = −3.982 × 1017 / 2.00 × 107 = −1.99 × 1010 J; |U| = 1.99 × 1010 J.
r = 4.22 × 107 m: U = −3.982 × 1017 / 4.22 × 107 = −9.43 × 109 J; |U| = 9.43 × 109 J.
Marking note: 1 mark for correct GMm; 1 mark per correct U value (any two rows) + 1 mark for correct |U| pattern.
Q1.2 — Trend in U (2 marks)
As r increases, U increases (becomes less negative) and approaches zero [1]. It cannot exceed zero because the zero-at-infinity convention sets U = 0 only at infinite separation; at any finite r there is still a gravitational attraction pulling the masses together, which means work has been done against the field [1].
Q1.3 — Energy to move from 400 km to GEO (2 marks)
ΔU = UGEO − ULEO = (−9.43 × 109) − (−5.88 × 1010) = 4.94 × 1010 J [1 formula + 1 correct answer]. This is approximately 49 GJ of work against gravity.
Q2.1 — Shape of GPE–r curve (2 marks)
The curve is a rectangular hyperbola (U ∝ −1/r): it is steep and negative at small r, and flattens asymptotically toward zero as r → ∞ [1]. The non-linear shape reflects the inverse-square nature of gravitational force; doubling r does not halve U, it makes it half as negative [1].
Q2.2 — Reading GPE from graph (3 marks)
From the graph: ULEO ≈ −5.9 × 1010 J [1]; UGEO ≈ −0.94 × 1010 J [1]. ΔU = (−0.94) − (−5.9) ≈ 5.0 × 1010 J [1]. (Accept values within reading tolerance of the graph.)
Q2.3 — Why U never reaches zero at finite r (2 marks)
As r increases, U = −GMm/r → 0 only as r → ∞; gravity always remains attractive at finite separation, so there is always some negative binding. A positive U would mean the object has more energy than at infinite separation, which would require a repulsive gravitational interaction — physically impossible for ordinary matter [1 for each point].
Q3 — Compare-and-contrast table
When valid? Exact: always (any separation, any altitude). Approx: only when h ≪ R (near-surface).
Reference level. Exact: zero at infinity (r → ∞). Approx: surface of the planet (h = 0).
Sign of ΔU when rising. Exact: ΔU is positive (becomes less negative). Approx: ΔU = mgh is positive.
How g is treated. Exact: g decreases with altitude (encoded in 1/r); field strength varies. Approx: g treated as constant = GM/R2 at the surface.
Example where MUST use this formula. Exact: any satellite or spacecraft problem (e.g. orbit change, escape energy). Approx: any everyday near-surface problem (e.g. lifting a box 2 m).
Q4.1 — Mars orbital energy (4 marks)
GMMarsm = (6.67 × 10−11)(6.39 × 1023)(500) = 2.131 × 1016 J m.
Surface: rsurface = 3.39 × 106 m.
Option A: rLMO = 3.39 × 106 + 4.00 × 105 = 3.79 × 106 m.
ΔU = GMm(1/rsurface − 1/rLMO) = 2.131 × 1016 × (2.950 × 10−7 − 2.639 × 10−7) = 2.131 × 1016 × 3.11 × 10−8 = 6.63 × 108 J [1 + 1 marks].
Option B: rB = 1.00 × 107 m.
ΔU = 2.131 × 1016 × (2.950 × 10−7 − 1.000 × 10−7) = 2.131 × 1016 × 1.950 × 10−7 = 4.16 × 109 J [1 + 1 marks].
Q4.2 — Validity of mgh for Mars (2 marks)
For Option A: h = 400 km; RMars = 3390 km; h/R ≈ 0.12 — about 12%, which is borderline (error ~12%) so mgh is not reliable [1]. For Option B: h ≈ 6600 km; h/R ≈ 1.9 — far larger than R, so mgh is completely invalid and the exact formula is essential [1].