IB Physics Topic 4.1: Gravitational fields | Notes & Flashcards | Aimnova

IB Physics — Gravitational fields

Topic 4.1 of IB Physics covers Gravitational fields, which is part of Unit 4: Fields. Students explore key concepts including Newton's law of gravitation and field strength, Kepler's laws and orbital motion, Circular orbits and satellites, Gravitational potential energy and escape speed. A strong understanding of gravitational fields is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Gravitational fields

Key Idea: Topic 4.1 is about gravity as a field — how a mass fills the space around it with a pull, and what that pull does to planets, satellites and rockets. It ties together four ideas: how strong the pull is (Newton's law F = GMm/r² and the field strength g = GM/r²), how things orbit (Kepler's laws, with T² ∝ r³), why a satellite stays up (gravity is the centripetal force, giving v = √(GM/r)), and the energy of being in a gravity well (the negative potential energy, and the escape speed). It is examined on Paper 1A (quick MCQs — inverse-square reasoning, recognising T² ∝ r³, the direction of a satellite's acceleration, why escape speed ignores the rocket's mass) and on Paper 2 (substitute into g = GM/r², compare two orbits with T²/r³, 'weigh' a central body with M = 4π²r³/(GT²), or use energy conservation for escape and impact speeds).

📐 Key formulas

Three of these are given in the data booklet (D.1) — you pick the right one rather than memorising it. The orbit and energy results below them are built from the given equations, so know how each one comes about.

F = G\frac{m_{1}m_{2}}{r^{2}}

Newton's law of gravitation (given). Attractive, and inverse-square in the distance r — double r and the force drops to a quarter.

$F$: gravitational force between the masses (N)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$m_{1}, m_{2}$: the two masses (kg)
$r$: distance between their centres (m)

g = \frac{F}{m} = G\frac{M}{r^{2}}

Gravitational field strength (given). The force per kilogram, and the same number as the free-fall acceleration; unit N kg⁻¹.

$g$: gravitational field strength (N kg⁻¹), also the free-fall acceleration
$F$: gravitational force on the small mass (N)
$m$: the small mass placed in the field (kg)
$M$: mass of the planet or star making the field (kg)
$r$: distance from the centre of M (m)

V = -\frac{GM}{r}

Gravitational potential (given). Energy per kilogram (J kg⁻¹); negative everywhere, zero at infinity.

$V$: gravitational potential — energy per kilogram (J kg⁻¹)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the planet or star (kg)
$r$: distance from the centre of M (m)

v = \frac{2\pi r}{T}

Speed around a circle (given, circular motion). Distance once around (2πr) over the time for one orbit (T) — used to turn an orbital speed into a period.

$v$: orbital speed (m s⁻¹)
$r$: orbit radius (m)
$T$: orbital period — time for one full orbit (s)

v = \sqrt{\frac{GM}{r}}

Orbital speed for a circular orbit. Built by setting gravity equal to the centripetal force (GMm/r² = mv²/r); the orbiting mass cancels. A bigger radius gives a smaller speed.

$v$: orbital speed (m s⁻¹)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the central body (kg)
$r$: orbit radius from the centre of the central body (m)

T^{2} = \frac{4\pi^{2}r^{3}}{GM}

Kepler's third law for a circular orbit. Built from g = GM/r² and the centripetal a = 4π²r/T². The key idea is just T² is proportional to r³.

$T$: orbital period — time for one full orbit (s)
$r$: orbital radius — distance from the central body's centre (m)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the central body being orbited (kg)

E_{p} = -\frac{GMm}{r}

Gravitational potential energy of a mass m (= mV). Always negative, zero at infinity, in joules.

$E_{p}$: gravitational potential energy of the object (J)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the planet or star (kg)
$m$: mass of the object placed in the field (kg)
$r$: distance from the centre of M (m)

v_{esc} = \sqrt{\frac{2GM}{r}}

Escape speed. From energy conservation (½mv² = GMm/r); the escaping object's mass cancels, so it depends only on the planet's M and r.

$v_{esc}$: escape speed — the launch speed needed to escape (m s⁻¹)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the planet or star (kg)
$r$: distance from the centre of M at launch — usually its radius (m)

🧭 Which equation, and when?

The most-tested decision in this topic: is the question about the field at a point, an orbit's timing or speed, or the energy of being in the well?

🛰️ The four big results — what each depends on

🪐 Kepler's three laws at a glance

🔄 The inverse-square shortcut

✍️ IB-style worked examples

IB-style question — field strength and the inverse-square law (g = GM/r²)

A planet has mass 5.4 × 10²⁴ kg and radius 6.0 × 10⁶ m. (a) Find the gravitational field strength at its surface. (b) State the field strength at a point three times as far from the centre. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².

Solution:

(a) Start with the given field-strength formula:
$g = G\frac{M}{r^{2}}$
Substitute M = 5.4 × 10²⁴, r = 6.0 × 10⁶:
$g = \frac{(6.67\times10^{-11})(5.4\times10^{24})}{(6.0\times10^{6})^{2}}$
Work it out — keep the unit:
$g = 10.0\ \text{N kg}^{-1}$
(b) At three times the distance, divide by 3² = 9 (inverse-square):
$g_{far} = \frac{10.0}{9} = 1.1\ \text{N kg}^{-1}$

Final answer:

g ≈ 10 N kg⁻¹ at the surface, and ≈ 1.1 N kg⁻¹ three times farther out — nine times smaller, because g ∝ 1/r².

IB-style question — compare two orbits (T² ∝ r³)

Two moons, P and Q, orbit the same planet. Moon Q's orbital radius is 4.0 times that of moon P, and moon P's period is 1.5 days. Find moon Q's orbital period.

Solution:

Same central body, so use Kepler's third law as a ratio (G and M cancel):
$\frac{T_{Q}^{2}}{r_{Q}^{3}} = \frac{T_{P}^{2}}{r_{P}^{3}}$
Rearrange for the period ratio squared:
$\left(\frac{T_{Q}}{T_{P}}\right)^{2} = \left(\frac{r_{Q}}{r_{P}}\right)^{3} = 4.0^{3} = 64$
Take the square root to get the period ratio:
$\frac{T_{Q}}{T_{P}} = \sqrt{64} = 8.0$
Multiply by moon P's period:
$T_{Q} = 8.0 \times 1.5 = 12\ \text{days}$

Final answer:

TQ = 12 days — a 4× bigger orbit takes 8× longer, since 4³ = 64 and √64 = 8. No need for the planet's mass.

IB-style question — weighing a central body (T² = 4π²r³/(GM))

A planet orbits a star in a circle of radius 1.8 × 10¹¹ m with a period of 1.6 × 10⁷ s. Determine the mass of the star. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².

Solution:

Gravity provides the centripetal force, giving Kepler's third law:
$T^{2} = \frac{4\pi^{2}r^{3}}{GM}$
Rearrange to make the star's mass M the subject:
$M = \frac{4\pi^{2}r^{3}}{G\,T^{2}}$
Substitute r = 1.8 × 10¹¹, T = 1.6 × 10⁷:
$M = \frac{4\pi^{2}(1.8\times10^{11})^{3}}{(6.67\times10^{-11})(1.6\times10^{7})^{2}}$
Work it out — keep the unit:
$M = 1.3\times10^{31}\ \text{kg}$

Final answer:

M ≈ 1.3 × 10³¹ kg. The orbiting planet's mass cancels out, so measuring just r and T 'weighs' the star.

IB-style question — escape speed and energy (v_esc = √(2GM/r))

A moon has mass 8.0 × 10²² kg and radius 1.8 × 10⁶ m. (a) Find the escape speed from its surface. (b) State whether a 2000 kg lander or a 500 kg probe needs the greater launch speed to escape. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².

Solution:

(a) Escape = supply enough kinetic energy to climb the whole well (½mv² = GMm/r); the mass cancels:
$v_{esc} = \sqrt{\frac{2GM}{r}}$
Substitute M = 8.0 × 10²², r = 1.8 × 10⁶:
$v_{esc} = \sqrt{\frac{2(6.67\times10^{-11})(8.0\times10^{22})}{1.8\times10^{6}}}$
Work out inside the root, then square-root — keep the unit:
$v_{esc} = \sqrt{5.9\times10^{6}} = 2.4\times10^{3}\ \text{m s}^{-1}$
(b) The launched mass does not appear in the formula — it cancelled out.
$v_{esc} = \sqrt{\frac{2GM}{r}}\quad(\text{no }m)$

Final answer:

vₑₛc ≈ 2.4 × 10³ m s⁻¹ (about 2.4 km s⁻¹). Both craft need the same escape speed — the launched mass cancels (though the heavier lander needs more energy, ½mv²).

✅ Quick self-check

Tap each card to reveal the answer.

🎯 Highest-yield exam reminders

Exam Tips

g = GM/r² and F = GMm/r² are both inverse-SQUARE: multiply the distance by n and you divide g (or F) by n². Three times farther → one ninth, not one third.
Field strength g is also the free-fall acceleration, and it does not depend on the falling mass — so all masses dropped at one place accelerate equally. N kg⁻¹ and m s⁻² are the same unit for g.
For two orbits round the SAME body, use TA²/rA³ = TB²/rB³ — the constant 4π²/(GM) cancels, so you never need G or M. Watch the powers: T is squared, r is cubed.
Gravity is the centripetal force for an orbit (GMm/r² = mv²/r). No period given for a speed? Use v = √(GM/r). Linking T and r, or finding the central mass? Use T² = 4π²r³/(GM), i.e. M = 4π²r³/(GT²).
Orbit radius r is measured from the CENTRE of the planet — a satellite's height above the surface is r minus the planet's radius. A geostationary satellite has a 24 h period over the equator, so it stays above one fixed point.
Eₚ = -GMm/r and V = -GM/r are always negative and zero at infinity — never drop the minus sign. Many 'launch or impact speed' questions are energy conservation: kinetic energy gained = depth of the well climbed.
Escape speed vₑₛc = √(2GM/r) depends only on the planet (M and r), not on the escaping object — and vₑₛc ∝ √M, so quadrupling the mass at fixed radius doubles it.

IB Physics — Gravitational fields

Key concepts in Gravitational fields

Key Idea: Topic 4.1 is about gravity as a field — how a mass fills the space around it with a pull, and what that pull does to planets, satellites and rockets. It ties together four ideas: how strong the pull is (Newton's law F = GMm/r² and the field strength g = GM/r²), how things orbit (Kepler's laws, with T² ∝ r³), why a satellite stays up (gravity is the centripetal force, giving v = √(GM/r)), and the energy of being in a gravity well (the negative potential energy, and the escape speed). It is examined on Paper 1A (quick MCQs — inverse-square reasoning, recognising T² ∝ r³, the direction of a satellite's acceleration, why escape speed ignores the rocket's mass) and on Paper 2 (substitute into g = GM/r², compare two orbits with T²/r³, 'weigh' a central body with M = 4π²r³/(GT²), or use energy conservation for escape and impact speeds).

📐 Key formulas

F = G\frac{m_{1}m_{2}}{r^{2}}

Newton's law of gravitation (given). Attractive, and inverse-square in the distance r — double r and the force drops to a quarter.

$F$: gravitational force between the masses (N)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$m_{1}, m_{2}$: the two masses (kg)
$r$: distance between their centres (m)

g = \frac{F}{m} = G\frac{M}{r^{2}}

Gravitational field strength (given). The force per kilogram, and the same number as the free-fall acceleration; unit N kg⁻¹.

$g$: gravitational field strength (N kg⁻¹), also the free-fall acceleration
$F$: gravitational force on the small mass (N)
$m$: the small mass placed in the field (kg)
$M$: mass of the planet or star making the field (kg)
$r$: distance from the centre of M (m)

V = -\frac{GM}{r}

Gravitational potential (given). Energy per kilogram (J kg⁻¹); negative everywhere, zero at infinity.

$V$: gravitational potential — energy per kilogram (J kg⁻¹)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the planet or star (kg)
$r$: distance from the centre of M (m)

v = \frac{2\pi r}{T}

Speed around a circle (given, circular motion). Distance once around (2πr) over the time for one orbit (T) — used to turn an orbital speed into a period.

$v$: orbital speed (m s⁻¹)
$r$: orbit radius (m)
$T$: orbital period — time for one full orbit (s)

v = \sqrt{\frac{GM}{r}}

Orbital speed for a circular orbit. Built by setting gravity equal to the centripetal force (GMm/r² = mv²/r); the orbiting mass cancels. A bigger radius gives a smaller speed.

$v$: orbital speed (m s⁻¹)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the central body (kg)
$r$: orbit radius from the centre of the central body (m)

T^{2} = \frac{4\pi^{2}r^{3}}{GM}

Kepler's third law for a circular orbit. Built from g = GM/r² and the centripetal a = 4π²r/T². The key idea is just T² is proportional to r³.

$T$: orbital period — time for one full orbit (s)
$r$: orbital radius — distance from the central body's centre (m)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the central body being orbited (kg)

E_{p} = -\frac{GMm}{r}

Gravitational potential energy of a mass m (= mV). Always negative, zero at infinity, in joules.

$E_{p}$: gravitational potential energy of the object (J)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the planet or star (kg)
$m$: mass of the object placed in the field (kg)
$r$: distance from the centre of M (m)

v_{esc} = \sqrt{\frac{2GM}{r}}

Escape speed. From energy conservation (½mv² = GMm/r); the escaping object's mass cancels, so it depends only on the planet's M and r.

$v_{esc}$: escape speed — the launch speed needed to escape (m s⁻¹)
$G$: gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
$M$: mass of the planet or star (kg)
$r$: distance from the centre of M at launch — usually its radius (m)

🧭 Which equation, and when?

The most-tested decision in this topic: is the question about the field at a point, an orbit's timing or speed, or the energy of being in the well?

🛰️ The four big results — what each depends on

🪐 Kepler's three laws at a glance

🔄 The inverse-square shortcut

✍️ IB-style worked examples

IB-style question — field strength and the inverse-square law (g = GM/r²)

Solution:

(a) Start with the given field-strength formula:
$g = G\frac{M}{r^{2}}$
Substitute M = 5.4 × 10²⁴, r = 6.0 × 10⁶:
$g = \frac{(6.67\times10^{-11})(5.4\times10^{24})}{(6.0\times10^{6})^{2}}$
Work it out — keep the unit:
$g = 10.0\ \text{N kg}^{-1}$
(b) At three times the distance, divide by 3² = 9 (inverse-square):
$g_{far} = \frac{10.0}{9} = 1.1\ \text{N kg}^{-1}$

Final answer:

g ≈ 10 N kg⁻¹ at the surface, and ≈ 1.1 N kg⁻¹ three times farther out — nine times smaller, because g ∝ 1/r².

IB-style question — compare two orbits (T² ∝ r³)

Two moons, P and Q, orbit the same planet. Moon Q's orbital radius is 4.0 times that of moon P, and moon P's period is 1.5 days. Find moon Q's orbital period.

Solution:

Same central body, so use Kepler's third law as a ratio (G and M cancel):
$\frac{T_{Q}^{2}}{r_{Q}^{3}} = \frac{T_{P}^{2}}{r_{P}^{3}}$
Rearrange for the period ratio squared:
$\left(\frac{T_{Q}}{T_{P}}\right)^{2} = \left(\frac{r_{Q}}{r_{P}}\right)^{3} = 4.0^{3} = 64$
Take the square root to get the period ratio:
$\frac{T_{Q}}{T_{P}} = \sqrt{64} = 8.0$
Multiply by moon P's period:
$T_{Q} = 8.0 \times 1.5 = 12\ \text{days}$

Final answer:

TQ = 12 days — a 4× bigger orbit takes 8× longer, since 4³ = 64 and √64 = 8. No need for the planet's mass.

IB-style question — weighing a central body (T² = 4π²r³/(GM))

A planet orbits a star in a circle of radius 1.8 × 10¹¹ m with a period of 1.6 × 10⁷ s. Determine the mass of the star. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².

Solution:

Gravity provides the centripetal force, giving Kepler's third law:
$T^{2} = \frac{4\pi^{2}r^{3}}{GM}$
Rearrange to make the star's mass M the subject:
$M = \frac{4\pi^{2}r^{3}}{G\,T^{2}}$
Substitute r = 1.8 × 10¹¹, T = 1.6 × 10⁷:
$M = \frac{4\pi^{2}(1.8\times10^{11})^{3}}{(6.67\times10^{-11})(1.6\times10^{7})^{2}}$
Work it out — keep the unit:
$M = 1.3\times10^{31}\ \text{kg}$

Final answer:

M ≈ 1.3 × 10³¹ kg. The orbiting planet's mass cancels out, so measuring just r and T 'weighs' the star.

IB-style question — escape speed and energy (v_esc = √(2GM/r))

Solution:

(a) Escape = supply enough kinetic energy to climb the whole well (½mv² = GMm/r); the mass cancels:
$v_{esc} = \sqrt{\frac{2GM}{r}}$
Substitute M = 8.0 × 10²², r = 1.8 × 10⁶:
$v_{esc} = \sqrt{\frac{2(6.67\times10^{-11})(8.0\times10^{22})}{1.8\times10^{6}}}$
Work out inside the root, then square-root — keep the unit:
$v_{esc} = \sqrt{5.9\times10^{6}} = 2.4\times10^{3}\ \text{m s}^{-1}$
(b) The launched mass does not appear in the formula — it cancelled out.
$v_{esc} = \sqrt{\frac{2GM}{r}}\quad(\text{no }m)$

Final answer:

vₑₛc ≈ 2.4 × 10³ m s⁻¹ (about 2.4 km s⁻¹). Both craft need the same escape speed — the launched mass cancels (though the heavier lander needs more energy, ½mv²).

✅ Quick self-check

Tap each card to reveal the answer.

🎯 Highest-yield exam reminders

Exam Tips

g = GM/r² and F = GMm/r² are both inverse-SQUARE: multiply the distance by n and you divide g (or F) by n². Three times farther → one ninth, not one third.
Field strength g is also the free-fall acceleration, and it does not depend on the falling mass — so all masses dropped at one place accelerate equally. N kg⁻¹ and m s⁻² are the same unit for g.
For two orbits round the SAME body, use TA²/rA³ = TB²/rB³ — the constant 4π²/(GM) cancels, so you never need G or M. Watch the powers: T is squared, r is cubed.
Gravity is the centripetal force for an orbit (GMm/r² = mv²/r). No period given for a speed? Use v = √(GM/r). Linking T and r, or finding the central mass? Use T² = 4π²r³/(GM), i.e. M = 4π²r³/(GT²).
Orbit radius r is measured from the CENTRE of the planet — a satellite's height above the surface is r minus the planet's radius. A geostationary satellite has a 24 h period over the equator, so it stays above one fixed point.
Eₚ = -GMm/r and V = -GM/r are always negative and zero at infinity — never drop the minus sign. Many 'launch or impact speed' questions are energy conservation: kinetic energy gained = depth of the well climbed.
Escape speed vₑₛc = √(2GM/r) depends only on the planet (M and r), not on the escaping object — and vₑₛc ∝ √M, so quadrupling the mass at fixed radius doubles it.

Key concepts in Gravitational fields

📐 Key formulas

🧭 Which equation, and when?

🛰️ The four big results — what each depends on

🪐 Kepler's three laws at a glance

🔄 The inverse-square shortcut

✍️ IB-style worked examples

IB-style question — field strength and the inverse-square law (g = GM/r²)

IB-style question — compare two orbits (T² ∝ r³)

IB-style question — weighing a central body (T² = 4π²r³/(GM))

IB-style question — escape speed and energy (v_esc = √(2GM/r))

✅ Quick self-check

🎯 Highest-yield exam reminders

Exam Tips

What you'll learn in Topic 4.1

Study resources — 4.1 Gravitational fields

Newton's law of gravitation and field strength

Kepler's laws and orbital motion

Circular orbits and satellites

Gravitational potential energy and escape speed

Ready to study Gravitational fields?

Ready to practice?

Key concepts in Gravitational fields

📐 Key formulas

🧭 Which equation, and when?

🛰️ The four big results — what each depends on

🪐 Kepler's three laws at a glance

🔄 The inverse-square shortcut

✍️ IB-style worked examples

IB-style question — field strength and the inverse-square law (g = GM/r²)

IB-style question — compare two orbits (T² ∝ r³)

IB-style question — weighing a central body (T² = 4π²r³/(GM))

IB-style question — escape speed and energy (v_esc = √(2GM/r))

✅ Quick self-check

🎯 Highest-yield exam reminders

Exam Tips

What you'll learn in Topic 4.1

Study resources — 4.1 Gravitational fields

Newton's law of gravitation and field strength

Kepler's laws and orbital motion

Circular orbits and satellites

Gravitational potential energy and escape speed

Ready to study Gravitational fields?

Ready to practice?