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.

Apply · Data & Reasoning

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 M_E = 5.97 × 10²⁴ kg throughout. 9 marks

Orbital radius r (m)	Altitude above surface (km)	U (J), calculate	Magnitude \|U\| (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 Calculate ΔU from the 400 km radial position to GEO. Hence state W_g and the external work for a quasistatic move. 2 marks

Stuck? Use U = −GMm/r. For 1.3, ΔU = U_GEO − U_LEO. Remember: r = R_Earth + h, and R_Earth = 6.37 × 10⁶ m.

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, M_E = 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 ΔU, W_g and quasistatic W_ext. 3 marks

2.3 Explain why the curve approaches but never crosses U = 0 at a finite radius. Why must trajectory classification use total mechanical energy rather than the sign of U alone? 2 marks

Stuck? For 2.1: U ∝ 1/r gives a hyperbolic curve. For 2.3: revisit the Key Insight callout and the Common Misconceptions box in the lesson.

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

Stuck? Revisit Card 3 “Comparison with mgh” and the Important callout in the lesson.

4. Predict and justify, a Martian surface scenario

Mars has mass M_Mars = 6.39 × 10²³ kg and mean surface radius R_Mars = 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 ΔU from Mars’ surface to each radial position. Hence state the quasistatic external work. 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

Stuck? For 4.1: ΔU = GMm(1/r₁ − 1/r₂) where r₁ = R_Mars for the surface, r₂ = orbit radius. For 4.2: compare h with R_Mars.

Answers, Do not peek before attempting

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 = U_GEO − U_LEO = (−9.43 × 10⁹) − (−5.88 × 10¹⁰) = +4.94 × 10¹⁰ J. Therefore W_g = −4.94 × 10¹⁰ J and, for a quasistatic move, W_ext = +4.94 × 10¹⁰ J [1 method + 1 signed result].

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: U_LEO ≈ −5.9 × 10¹⁰ J [1]; U_GEO ≈ −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, Reference and total energy (2 marks)

As r increases, U = −GMm/r approaches zero from below and reaches it only in the limit r → ∞ [1]. Because U is negative at every finite radius under this convention, bound/unbound motion must be classified using E = K + U, not the sign of U alone [1].

Q3, Compare-and-contrast table

When valid? Exact: point masses or non-overlapping spherically symmetric bodies with centre-to-centre r. 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: a radial energy change over a satellite-scale height. Approx: an everyday near-surface change, such as lifting a box 2 m.

Q4.1, Mars orbital energy (4 marks)

GM_Marsm = (6.67 × 10⁻¹¹)(6.39 × 10²³)(500) = 2.131 × 10¹⁶ J m.

Surface: r_surface = 3.39 × 10⁶ m.

Option A: r_LMO = 3.39 × 10⁶ + 4.00 × 10⁵ = 3.79 × 10⁶ m.

ΔU = GMm(1/r_surface − 1/r_LMO) = 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: r_B = 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; R_Mars = 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].