Gravitational Fields and Space Problem Solving

This final lesson brings together everything from the topic — gravitational fields, orbits, energy, astrophysics, and cosmology — through extended worked examples and problem-solving strategies. These are the kinds of multi-step questions that appear on Papers 1 and 3, and strong performance requires fluent use of all the key equations.

Essential Equations Summary

Before tackling problems, here is every equation you need for this topic:

Equation	What it gives you
F = GMm/r²	Gravitational force between two masses
g = GM/r²	Gravitational field strength at distance r
g = F/m	Definition of field strength
V = −GM/r	Gravitational potential at distance r
E_p = −GMm/r	Gravitational potential energy
v = √(GM/r)	Orbital speed (circular orbit)
T² = 4π²r³/(GM)	Orbital period (Kepler's third law)
v_esc = √(2GM/r)	Escape velocity
E_k = GMm/(2r)	Kinetic energy in circular orbit
E_total = −GMm/(2r)	Total energy in circular orbit
g = −dV/dr	Field strength from potential gradient
λ_max = 2.898 × 10⁻³/T	Wien's displacement law
L = 4πr²σT⁴	Stefan-Boltzmann law
I = L/(4πd²)	Intensity at distance d
d = 1/p	Distance from parallax (parsecs)
v = H₀d	Hubble's law
z = Δλ/λ₀ ≈ v/c	Redshift

Problem-Solving Strategy

flowchart TD
    A["Read the question carefully"] --> B["Identify the physics:\nGravity? Orbits? Energy? Stars? Cosmology?"]
    B --> C["List known quantities\nConvert units (km→m, hours→s)"]
    C --> D["Select the right equation(s)"]
    D --> E["Check: is r the orbital radius\nor the surface radius?\nr = R + h for orbits!"]
    E --> F["Substitute and calculate"]
    F --> G["Check: Is the answer physically\nreasonable? Right order of magnitude?"]
    G --> H["Give correct units and\nappropriate significant figures"]

Problem Set 1: Gravitational Fields and Forces

Problem 1.1: Neutral Point Between Earth and Moon

Find the distance from Earth's centre where the gravitational field strength is zero, given:

M_E = 5.97 × 10²⁴ kg, M_M = 7.35 × 10²² kg, d = 3.84 × 10⁸ m

At the neutral point (distance x from Earth):

$\frac{GM_E}{x^2} = \frac{GM_M}{(d-x)^2}$

$\frac{d-x}{x} = \sqrt{\frac{M_M}{M_E}} = \sqrt{\frac{7.35 \times 10^{22}}{5.97 \times 10^{24}}} = \sqrt{0.01231} = 0.1109$

$d - x = 0.1109x \implies d = 1.1109x$

$x = \frac{3.84 \times 10^8}{1.1109} = 3.457 \times 10^8 \text{ m}$

This is about 346,000 km from Earth, or about 38,000 km from the Moon. The neutral point is much closer to the less massive body.

Problem 1.2: g Inside the Earth

Inside a uniform sphere of density ρ at distance r from the centre (r < R), only the mass within radius r contributes:

$M_{\text{enclosed}} = \frac{4}{3}\pi r^3 \rho$

$g = \frac{GM_{\text{enclosed}}}{r^2} = \frac{G \times \frac{4}{3}\pi r^3 \rho}{r^2} = \frac{4}{3}\pi G \rho r$

Inside a uniform sphere, g is proportional to r — it increases linearly from zero at the centre to GM/R² at the surface. This is very different from the 1/r² behaviour outside.

Location	r (×10⁶ m)	g (N kg⁻¹)
Centre	0	0
Quarter radius	1.59	2.45
Half radius	3.19	4.91
Three-quarter radius	4.78	7.36
Surface	6.37	9.81

(Assuming uniform density — the real Earth is denser at the core, so g actually increases slightly before decreasing inside the mantle.)

Problem Set 2: Orbits and Satellite Calculations

Problem 2.1: Complete Satellite Calculation

A 1200 kg communications satellite is placed in geostationary orbit (T = 86,164 s) around Earth (M = 5.97 × 10²⁴ kg).

Step 1: Orbital radius

$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} = \left(\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times (86164)^2}{4\pi^2}\right)^{1/3}$

$= \left(\frac{3.984 \times 10^{14} \times 7.424 \times 10^9}{39.48}\right)^{1/3} = (7.494 \times 10^{22})^{1/3} = 4.216 \times 10^7 \text{ m}$

Step 2: Orbital speed

$v = \frac{2\pi r}{T} = \frac{2\pi \times 4.216 \times 10^7}{86164} = 3073 \text{ m s}^{-1} \approx 3.07 \text{ km s}^{-1}$

Step 3: Gravitational field strength at orbit

$g = \frac{GM}{r^2} = \frac{3.984 \times 10^{14}}{(4.216 \times 10^7)^2} = \frac{3.984 \times 10^{14}}{1.777 \times 10^{15}} = 0.224 \text{ N kg}^{-1}$

Step 4: All three energies

$E_k = \frac{GMm}{2r} = \frac{3.984 \times 10^{14} \times 1200}{2 \times 4.216 \times 10^7} = 5.67 \times 10^9 \text{ J} = 5.67 \text{ GJ}$

$E_p = -\frac{GMm}{r} = -11.34 \text{ GJ}$

$E_{\text{total}} = -5.67 \text{ GJ}$

Step 5: Escape velocity from this orbit

$v_{\text{esc}} = \sqrt{2} \times v = 1.414 \times 3073 = 4345 \text{ m s}^{-1}$

Problem 2.2: Comparing Orbits Using Kepler's Third Law

A planet has two moons. Moon A orbits at r_A = 2.0 × 10⁸ m with period T_A = 10 days. Moon B orbits at r_B = 5.0 × 10⁸ m. Find T_B.

$\frac{T_B^2}{T_A^2} = \frac{r_B^3}{r_A^3}$

$T_B = T_A \times \left(\frac{r_B}{r_A}\right)^{3/2} = 10 \times \left(\frac{5.0}{2.0}\right)^{3/2} = 10 \times 2.5^{1.5} = 10 \times 3.953 = 39.5 \text{ days}$

Problem 2.3: Mass of a Planet from Satellite Data

A satellite orbits an unknown planet at altitude h = 500 km above the surface. The planet has radius R = 4,000 km. The satellite's period is 8,200 s. Find the planet's mass.

r = R + h = 4,000 + 500 = 4,500 km = 4.5 × 10⁶ m

$M = \frac{4\pi^2 r^3}{GT^2} = \frac{4\pi^2 \times (4.5 \times 10^6)^3}{6.674 \times 10^{-11} \times 8200^2}$

$= \frac{39.48 \times 9.112 \times 10^{19}}{6.674 \times 10^{-11} \times 6.724 \times 10^7} = \frac{3.597 \times 10^{21}}{4.487 \times 10^{-3}} = 8.02 \times 10^{23} \text{ kg}$

This is similar to the mass of Mars (6.42 × 10²³ kg) — a plausible rocky planet.

Problem Set 3: Energy and Escape

Problem 3.1: Can a Molecule Escape?

On a planet with escape velocity 2,500 m s⁻¹ and surface temperature 400 K, can a nitrogen molecule (mass = 4.65 × 10⁻²⁶ kg) escape thermally?

The average KE of a gas molecule at temperature T is:

$E_k = \frac{3}{2}k_BT = \frac{3}{2} \times 1.38 \times 10^{-23} \times 400 = 8.28 \times 10^{-21} \text{ J}$

KE needed to escape:

$E_{\text{esc}} = \frac{1}{2}mv_{\text{esc}}^2 = \frac{1}{2} \times 4.65 \times 10^{-26} \times 2500^2 = 1.45 \times 10^{-19} \text{ J}$

Ratio: E_esc/E_k = 1.45 × 10⁻¹⁹ / 8.28 × 10⁻²¹ = 17.5

The average molecule has only about 1/18 of the energy needed to escape. However, the Maxwell-Boltzmann distribution means a tiny fraction of molecules in the high-speed tail can exceed escape velocity. Over geological time, this slow leakage can strip a planet of its atmosphere — this is why Mars (low escape velocity, moderate temperature) has lost most of its atmosphere.

Problem Set 4: Astrophysics Calculations

Problem 4.1: Full HR Diagram Analysis

Three stars are observed:

Star	Luminosity (L☉)	Temperature (K)
P	10,000	25,000
Q	500	4,000
R	0.01	15,000

Classify each star and find its radius:

Star P (hot, luminous): Upper main sequence or supergiant.

$\frac{r_P}{r_\odot} = \sqrt{\frac{L_P/L_\odot}{(T_P/T_\odot)^4}} = \sqrt{\frac{10000}{(25000/5800)^4}} = \sqrt{\frac{10000}{345.6}} = \sqrt{28.94} = 5.4$

Radius ≈ 5.4 R☉. This is a hot main-sequence star (B-type, upper main sequence).

Star Q (cool, very luminous): Red giant region.

$\frac{r_Q}{r_\odot} = \sqrt{\frac{500}{(4000/5800)^4}} = \sqrt{\frac{500}{0.227}} = \sqrt{2203} = 46.9$

Radius ≈ 47 R☉. This is a red giant.

Star R (hot, very dim): White dwarf region.

$\frac{r_R}{r_\odot} = \sqrt{\frac{0.01}{(15000/5800)^4}} = \sqrt{\frac{0.01}{44.85}} = \sqrt{2.23 \times 10^{-4}} = 0.0149$

Radius ≈ 0.015 R☉ (about 10,400 km — slightly larger than Earth). This is a white dwarf.

Problem 4.2: Complete Cosmology Calculation

A galaxy shows the Hβ spectral line at 498.2 nm (rest wavelength 486.1 nm). Using H₀ = 72 km s⁻¹ Mpc⁻¹:

Step 1: Redshift: $z = \frac{498.2 - 486.1}{486.1} = \frac{12.1}{486.1} = 0.02489$

Step 2: Recession velocity: $v = zc = 0.02489 \times 3.0 \times 10^5 = 7467 \text{ km s}^{-1}$

Step 3: Distance: $d = \frac{v}{H_0} = \frac{7467}{72} = 103.7 \text{ Mpc}$

Step 4: Convert to metres: $d = 103.7 \times 3.09 \times 10^{22} = 3.20 \times 10^{24} \text{ m}$

Step 5: Light travel time (approximate lookback time): $t = \frac{d}{c} = \frac{3.20 \times 10^{24}}{3.0 \times 10^8} = 1.07 \times 10^{16} \text{ s} = 338 \text{ million years}$

We are seeing this galaxy as it was about 340 million years ago.

Problem Set 5: Multi-Concept Questions

Problem 5.1: Satellite to Escape — Full Energy Analysis

A 800 kg satellite is in circular orbit at r = 8.0 × 10⁶ m around Earth. Calculate:

(a) Orbital speed: $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.984 \times 10^{14}}{8.0 \times 10^6}} = \sqrt{4.98 \times 10^7} = 7057 \text{ m s}^{-1}$

(b) Total orbital energy: $E_{\text{total}} = -\frac{GMm}{2r} = -\frac{3.984 \times 10^{14} \times 800}{2 \times 8.0 \times 10^6} = -1.99 \times 10^{10} \text{ J}$

(c) Energy needed to escape: The satellite must go from E_total = −1.99 × 10¹⁰ J to E = 0 (just reaches infinity): $\Delta E = 0 - (-1.99 \times 10^{10}) = 1.99 \times 10^{10} \text{ J} \approx 19.9 \text{ GJ}$

(d) Speed increase needed: $v_{\text{esc}} = \sqrt{2} \times v_{\text{orbital}} = 1.414 \times 7057 = 9979 \text{ m s}^{-1}$

$\Delta v = 9979 - 7057 = 2922 \text{ m s}^{-1}$

Verification: Extra KE = ½m(v_esc² − v²) = ½ × 800 × (9979² − 7057²) = ½ × 800 × (9.958 × 10⁷ − 4.980 × 10⁷) = ½ × 800 × 4.978 × 10⁷ = 1.99 × 10¹⁰ J. This matches part (c).

Problem 5.2: Binary Star System

Two stars of equal mass M = 3.0 × 10³⁰ kg orbit their common centre of mass at separation d = 4.0 × 10¹¹ m. Find:

(a) The orbital radius of each star: Since the masses are equal, each orbits at r = d/2 = 2.0 × 10¹¹ m from the centre of mass.

(b) The orbital period: The gravitational force provides the centripetal force:

$\frac{GM^2}{d^2} = \frac{Mv^2}{r} = \frac{M \times (2\pi r/T)^2}{r} = \frac{4\pi^2 Mr}{T^2}$

$T^2 = \frac{4\pi^2 r d^2}{GM} = \frac{4\pi^2 \times 2.0 \times 10^{11} \times (4.0 \times 10^{11})^2}{6.674 \times 10^{-11} \times 3.0 \times 10^{30}}$

$= \frac{4\pi^2 \times 2.0 \times 10^{11} \times 1.6 \times 10^{23}}{2.002 \times 10^{20}} = \frac{1.264 \times 10^{36}}{2.002 \times 10^{20}} = 6.313 \times 10^{15}$

$T = \sqrt{6.313 \times 10^{15}} = 7.946 \times 10^{7} \text{ s} \approx 2.52 \text{ years}$

Common Exam Mistakes to Avoid

Mistake	Correction
Using altitude h instead of orbital radius r	r = R_planet + h
Forgetting that V and E_p are negative	Always include the minus sign
Saying astronauts are weightless because there is no gravity	They are in free fall; g is still ~8.7 N kg⁻¹ at ISS altitude
Confusing G (constant) and g (varies)	G is universal; g depends on location
Plotting HR diagram with T increasing right	Convention: T decreases to the right
Saying the Big Bang was an explosion in space	The Big Bang was an expansion of space itself
Using v = H₀d with inconsistent units	Match km s⁻¹ with Mpc, or m s⁻¹ with m
Forgetting to square r in F = GMm/r²	Most common arithmetic error

Summary

Multi-step problems require combining gravity, orbital mechanics, energy, and astrophysics equations
Always define r clearly: orbital radius (from centre), not altitude
Energy analysis: E_total = E_k + E_p = −GMm/(2r); always negative for bound orbits
For stellar problems, use Stefan's law as a ratio to eliminate constants
Cosmology problems: spectral line → redshift → velocity → distance (chain of equations)
Check answers are physically reasonable: right sign, right order of magnitude, right units

A-Level Deep Dive: Gravitational Fields and Space Problem Solving

Spec mapping

Edexcel 9PH0 Topic 12 — Gravitational fields and space is the synoptic capstone of the Edexcel A-Level Physics specification. It draws together Newton's law of gravitation, gravitational field strength and potential, orbital motion, escape velocity, stellar classification by luminosity and temperature, the Hertzsprung–Russell diagram, stellar evolution and cosmological recession (refer to the official specification document for exact wording). Topic 12 is examined predominantly on Paper 3 (General and Practical Principles in Physics), where extended-response and multi-step questions blend gravitational mechanics with thermal physics, electromagnetic radiation and data analysis. The Edexcel Physics formula booklet provides $F = Gm_1m_2/r^2$ , $g = -\Delta V_{grav}/\Delta r$ , $V_{grav} = -Gm/r$ , $L = \sigma T^4 4\pi r^2$ , Wien's law, and $v = H_0 d$ , but candidates must memorise vectorial reasoning, the sign convention for potential, and the link between redshift and velocity for non-relativistic recession.

Worked example with full mark scheme

Question (12 marks): A satellite of mass 1200 kg is to be placed in a circular orbit at altitude 600 km above the Earth's surface. Take $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ , $M_E = 5.97 \times 10^{24}\,\text{kg}$ , $R_E = 6.37 \times 10^6\,\text{m}$ .

(a) Calculate the orbital speed at this altitude. (3) (b) Calculate the additional kinetic energy required to lift the satellite from rest on Earth's surface into this orbit, ignoring atmospheric drag and Earth's rotation. (5) (c) Determine the minimum speed at which the satellite would need to be launched directly from the surface to escape Earth's gravity entirely. (2) (d) Explain qualitatively why the speed in (c) is independent of launch direction (provided no surface obstruction). (2)

Solution with mark scheme:

(a) Step 1 — equate gravitational force and centripetal force. Orbital radius $r = R_E + h = 6.37 \times 10^6 + 6.00 \times 10^5 = 6.97 \times 10^6\,\text{m}$ .

$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \implies v = \sqrt{\dfrac{GM}{r}}$

M1 — correct equation linking gravitational and centripetal force.

$v = \sqrt{\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.97 \times 10^6}} = \sqrt{5.71 \times 10^7} = 7560\,\text{m s}^{-1}$

M1 — correct substitution. A1 — answer $\approx 7.56 \times 10^3\,\text{m s}^{-1}$ to 3 sf.

(b) Step 1 — total mechanical energy in orbit. $E_{orbit} = -\dfrac{GMm}{2r}$ .

$E_{orbit} = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1200)}{2(6.97 \times 10^6)} = -3.43 \times 10^{10}\,\text{J}$

M1 — correct expression for total energy of a circular orbit (sum of KE and gravitational PE).

Step 2 — energy at rest on surface. $E_{surface} = -\dfrac{GMm}{R_E} = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1200)}{6.37 \times 10^6} = -7.50 \times 10^{10}\,\text{J}$ .

M1 — correct gravitational PE at surface (KE = 0).

Step 3 — energy difference.

$\Delta E = E_{orbit} - E_{surface} = -3.43 \times 10^{10} - (-7.50 \times 10^{10}) = 4.07 \times 10^{10}\,\text{J}$

M1 A1 — correct subtraction with consistent signs; answer $\approx 4.07 \times 10^{10}\,\text{J}$ (or $4.1 \times 10^{10}\,\text{J}$ to 2 sf).

$v_{esc} = \sqrt{\dfrac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}} = 1.12 \times 10^4\,\text{m s}^{-1}$

M1 A1 — correct formula and answer $\approx 11.2\,\text{km s}^{-1}$ .

(d) Escape speed depends only on the scalar quantity kinetic energy ( $\tfrac{1}{2}mv^2$ ) and gravitational potential, both scalar. The condition $\tfrac{1}{2}mv^2 \geq \dfrac{GMm}{R_E}$ contains no directional information, so any launch direction (provided unobstructed) gives the same minimum speed. B1 for scalar argument; B1 for explicit reference to KE and PE both being scalar.

Total: 12 marks (M5 A4 B3 split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 3 format

Question (10 marks): Astronomers observe a hydrogen absorption line of rest wavelength 656.3 nm appearing in a distant galaxy's spectrum at 689.1 nm.

(a) Calculate the recessional speed of the galaxy. (3) (b) Using $H_0 = 2.3 \times 10^{-18}\,\text{s}^{-1}$ , estimate the distance to the galaxy in megaparsecs (1 Mpc $= 3.09 \times 10^{22}\,\text{m}$ ). (3) (c) Estimate how long the light has been travelling. (2) (d) Discuss two assumptions in this calculation that limit its accuracy at large redshifts. (2)

Mark scheme decomposition by AO:

(a) M1 (AO1.1a) — using $\Delta\lambda/\lambda \approx v/c$ for non-relativistic redshift. M1 (AO2.1) — substituting $\Delta\lambda = 32.8\,\text{nm}$ , $\lambda = 656.3\,\text{nm}$ , giving $v = c \cdot 0.0500 = 1.50 \times 10^7\,\text{m s}^{-1}$ . A1 (AO2.5) — answer to 3 sf with units.

(b) M1 (AO1.1a) — using Hubble's law $v = H_0 d$ rearranged to $d = v/H_0$ . M1 (AO2.1) — $d = (1.50 \times 10^7)/(2.3 \times 10^{-18}) = 6.52 \times 10^{24}\,\text{m}$ . A1 (AO2.5) — converting: $d \approx 211\,\text{Mpc}$ (or $\approx 210\,\text{Mpc}$ ).

(c) M1 (AO2.1) — light-travel time $t = d/c$ . A1 (AO2.5) — $t = (6.52 \times 10^{24})/(3.00 \times 10^8) = 2.17 \times 10^{16}\,\text{s} \approx 6.9 \times 10^8\,\text{years}$ .