Free Cheatsheet · Topic D.1 · SL + AHL

IB Physics HL Gravitational Fields — Complete Cheatsheet

Every formula, derivation, trick and trap for IB Physics HL Topic D.1 — Kepler's laws, Newton's law of gravitation, orbital motion, gravitational potential, escape speed and satellite energetics. Hand-built by an IBO-certified Singapore tutor with 15+ years of IB experience.

Topic: D.1 Gravitational Fields Syllabus: SL + AHL Topic D.1 Read time: ~12 minutes Last updated: Apr 2026

Gravitational fields is the gateway topic of IB Physics HL Theme D — and one of the highest-yielding topics on the exam. Almost every IB session features a long Paper 2 question on satellite motion, orbital energetics or escape speed, and the AHL extensions (potential, equipotentials, total energy) are graded ferociously. Get this topic right and you bank an easy 15–20 marks; get it wrong and you'll keep tripping on the same algebraic traps for the rest of Theme D.

This cheatsheet condenses every formula, derivation and exam-mark trap from D.1 (SL core and AHL extension) into a single page you can revise from. If you want the printable PDF, the full set of notes, the worked tutorials and the marked-up solutions, scroll to the bottom for the download links and the gated full library.

§1 — Kepler's Laws & Newton's Law of Gravitation D.1

Kepler's three laws

  1. 1st Law (Ellipses): Every planet moves in an elliptical orbit with the Sun at one focus.
  2. 2nd Law (Equal Areas): A line from planet to Sun sweeps equal areas in equal times. Consequence: the planet moves faster at perihelion, slower at aphelion. This follows from conservation of angular momentum.
  3. 3rd Law (Harmonic): $T^2 \propto r^3$ for all bodies orbiting the same central mass: $$\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} = \frac{4\pi^2}{GM}$$

Newton's law of gravitation

Force:$\displaystyle F = \frac{GMm}{r^2}$,   $G = 6.67 \times 10^{-11}\;\mathrm{N\,m^2\,kg^{-2}}$
Trap$r$ is always the centre-to-centre distance. For a satellite at height $h$ above a planet of radius $R$: $r = R + h$, NOT $h$. This is the single most common D.1 numerical error.
TrickInverse-square law quick check: if $r \to 2r$, then $F \to F/4$. If $r \to 3r$, then $F \to F/9$. Use this to verify any answer.

§2 — Gravitational Field Strength D.1

Definition:$\displaystyle g = \frac{F}{m}$  [$\mathrm{N\,kg^{-1}} = \mathrm{m\,s^{-2}}$]
From Newton's law:$\displaystyle g = \frac{GM}{r^2}$   (at distance $r$ from centre of mass $M$)
At surface:$\displaystyle g_{\text{surface}} = \frac{GM}{R^2}$
Note$g$ is a vector — it points toward the mass. Field lines for a sphere point radially inward. Between two masses, the two fields oppose each other and you must add as vectors.

Neutral point between two masses

Between masses $M_1$ and $M_2$ separated by distance $D$, the neutral point ($g_{\text{net}} = 0$) is at distance $d$ from $M_1$:

$$\frac{d}{D-d} = \sqrt{\frac{M_1}{M_2}} \quad \Longrightarrow \quad d = \frac{D}{1 + \sqrt{M_2/M_1}}$$

The neutral point is always closer to the smaller mass.

TrapThe neutral point is where $\vec{g} = 0$, NOT where $V_g = 0$. Gravitational potential is never zero between two masses (it's negative everywhere except at infinity).

§3 — Orbital Motion & Kepler's Third Law Derivation D.1

Equate the gravitational force with the centripetal force on an orbiting satellite of mass $m$:

$$\frac{GMm}{r^2} = \frac{mv^2}{r} \quad \Longrightarrow \quad v^2 = \frac{GM}{r} \quad \Longrightarrow \quad \boxed{\,v = \sqrt{\frac{GM}{r}}\,}$$

Substitute $v = \dfrac{2\pi r}{T}$ to derive Kepler's 3rd law:

$$\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \quad \Longrightarrow \quad \boxed{\,T^2 = \frac{4\pi^2}{GM}\,r^3\,}$$

TrickOrbital speed $v \propto 1/\sqrt{r}$ — closer satellites orbit faster. Period $T \propto r^{3/2}$ — closer satellites have shorter periods. The satellite's mass $m$ cancels out of everything — orbits are independent of the orbiting mass.

§4 — Weightlessness D.1

Apparent weightlessness occurs when the astronaut and spacecraft are both in free fall with the same acceleration, so no contact (normal) force exists between them. Gravity is still acting — it's the contact force that has vanished.

TrapNEVER write "there is no gravity in space" in your answer. At the ISS altitude ($\sim 400$ km), $g \approx 8.7\;\mathrm{m\,s^{-2}}$ — only 11% less than at Earth's surface. The astronaut feels weightless because the floor is falling with them.

§5 — Gravitational Potential Energy & Potential D.1 AHL

Gravitational PE:$\displaystyle E_p = -\frac{GMm}{r}$   (always negative; zero at $r = \infty$)
Gravitational potential:$\displaystyle V_g = \frac{E_p}{m} = -\frac{GM}{r}$   [$\mathrm{J\,kg^{-1}}$]
Work done:$\displaystyle W = m\,\Delta V_g = m(V_f - V_i)$
TrickPotential is a scalar — when several masses contribute, add the potentials algebraically, not as vectors. This is much easier than adding $\vec{g}$ contributions, so if a question asks for either, answer the potential one first.
NoteThe gravitational field is conservative: the work done depends only on the start and end positions, not on the path. Work done around any closed loop is zero.

§6 — Potential Gradient, Equipotentials & Escape Speed D.1 AHL

Potential gradient:$\displaystyle g = -\frac{\Delta V_g}{\Delta r}$   ($g$ is the negative slope of the $V_g$-vs-$r$ graph)

Equipotential surfaces

  • Surfaces where $V_g = $ constant.
  • Always perpendicular ($\perp$) to the field lines.
  • No work done moving along an equipotential ($\Delta V_g = 0$).
  • For a point mass: concentric spheres.
  • Closer spacing $\Rightarrow$ stronger field.

Escape speed

Escape speed:$\displaystyle v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = \sqrt{2}\;v_{\text{orbital at surface}}$
TrapEscape speed does NOT depend on the mass of the escaping object. It also does not depend on the launch direction (ignoring atmospheric drag). A pebble and a rocket have the same $v_{\text{esc}}$ from Earth — about $11.2\;\mathrm{km\,s^{-1}}$.

§7 — Energy of an Orbiting Satellite D.1 AHL

Satellite energy at orbital radius $r$

QuantityExpressionSign
Kinetic energy$E_K = \dfrac{GMm}{2r}$positive
Potential energy$E_p = -\dfrac{GMm}{r}$negative
Total energy$E_T = -\dfrac{GMm}{2r}$negative (bound)

Key relations: $E_K = -\tfrac{1}{2}E_p$,   $E_T = -E_K = \tfrac{1}{2}E_p$.

Work to change orbit:$\displaystyle W = \Delta E_T = \frac{GMm}{2}\!\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$
TrickThe "drag paradox": atmospheric drag causes a satellite to spiral inward and speed up. As $r$ decreases, $v = \sqrt{GM/r}$ increases. $E_K$ rises, $E_p$ falls (more negative), $E_T$ falls (more negative — energy lost to heat).
TrapDo not write "drag slows the satellite". The instantaneous effect is deceleration, but the satellite drops to a lower, faster orbit. Net effect: speed increases.

§8 — Exam Attack Plan All sections

When you see this in the question — reach for that:

Question triggerReach for
"Find the orbital speed / period"$v = \sqrt{GM/r}$ or $T^2 = 4\pi^2 r^3 / (GM)$
"Compare two orbits"Use ratios: $v \propto r^{-1/2}$, $T \propto r^{3/2}$, $E_T \propto -1/r$
"Energy needed to change orbit"$W = \Delta E_T = \tfrac{GMm}{2}(1/r_1 - 1/r_2)$
"Escape from surface"$v_{\text{esc}} = \sqrt{2GM/R}$
"Field at point P from two masses"Resolve $\vec{g}_1$ and $\vec{g}_2$ as vectors, add components
"Potential at point P from several masses"$V_g = -G\sum M_i/r_i$ — scalar sum, watch the minus signs
"Work to move mass from A to B"$W = m(V_B - V_A)$
"Why does drag speed up a satellite?"$E_T$ falls $\Rightarrow$ $r$ falls $\Rightarrow$ $v = \sqrt{GM/r}$ rises
"Explain weightlessness in ISS"Both astronaut and station in free fall — no contact force

Worked Example — IB-Style Orbital Energetics

Question (HL Paper 2 style — 7 marks)

A satellite of mass $m = 1200\;\mathrm{kg}$ is in a circular orbit at altitude $h_1 = 400\;\mathrm{km}$ above the Earth's surface. It is to be raised to a geostationary orbit at altitude $h_2 = 36\,000\;\mathrm{km}$. Take $G = 6.67 \times 10^{-11}\;\mathrm{N\,m^2\,kg^{-2}}$, $M_E = 5.97 \times 10^{24}\;\mathrm{kg}$, $R_E = 6.37 \times 10^{6}\;\mathrm{m}$.

(a) State the orbital radius $r_1$ and $r_2$ in metres. (b) Calculate the work that must be done on the satellite to move it between the two orbits.

Solution

  1. State centre-to-centre radii (apply $r = R + h$):
    $r_1 = 6.37\times 10^6 + 4.00\times 10^5 = 6.77\times 10^6\;\mathrm{m}$  (A1)
    $r_2 = 6.37\times 10^6 + 3.60\times 10^7 = 4.24\times 10^7\;\mathrm{m}$  (A1)
  2. State the correct energy formula: $E_T = -\dfrac{GM_E m}{2r}$  (M1)
  3. Energy in low orbit:
    $E_{T,1} = -\dfrac{(6.67\times 10^{-11})(5.97\times 10^{24})(1200)}{2(6.77\times 10^6)} = -3.53 \times 10^{10}\;\mathrm{J}$  (A1)
  4. Energy in geostationary orbit:
    $E_{T,2} = -\dfrac{(6.67\times 10^{-11})(5.97\times 10^{24})(1200)}{2(4.24\times 10^7)} = -5.63 \times 10^{9}\;\mathrm{J}$  (A1)
  5. Work done = change in total energy:
    $W = \Delta E_T = E_{T,2} - E_{T,1} = -5.63\times 10^9 - (-3.53\times 10^{10}) = +2.97\times 10^{10}\;\mathrm{J}$  (M1)(A1)

Examiner's note: Two errors are punished here. (1) Forgetting that $r = R + h$, not $h$ — substituting $h$ alone gives wildly wrong values. (2) Forgetting the negative sign in $E_T$ and ending up with a negative $W$. The work done on the satellite to raise it to a higher orbit must be positive: a higher orbit means $E_T$ is closer to zero (less negative).

Common Student Questions

Is there gravity in space?
Yes. At the ISS altitude of about 400 km, $g \approx 8.7\;\mathrm{m\,s^{-2}}$ — only 11% less than at Earth's surface. Astronauts feel weightless because they and the spacecraft are both in free fall together, so there is no contact (normal) force between them. Saying "there is no gravity in space" is one of the fastest ways to lose marks in IB Physics.
What is $r$ in Newton's law of gravitation?
$r$ is the centre-to-centre distance between the two masses, not the surface-to-surface distance. For a satellite at altitude $h$ above a planet of radius $R$, you must use $r = R + h$. Substituting $h$ alone is the single most common D.1 numerical error and will multiply your final answer by huge factors when raised to the second or third power.
Why does atmospheric drag make a satellite speed up?
This is the classic "drag paradox". Drag dissipates energy, so total energy $E_T = -\dfrac{GMm}{2r}$ becomes more negative. That requires $r$ to decrease. Since $v = \sqrt{GM/r}$, a smaller $r$ means a larger $v$. The instantaneous deceleration from drag is more than offset by the drop in altitude. Net effect: the satellite drops to a lower, faster orbit and burns up.
Does escape speed depend on the mass of the object escaping?
No. $v_{\text{esc}} = \sqrt{2GM/R}$ depends only on the mass $M$ and radius $R$ of the planet. It also does not depend on the launch direction (ignoring atmosphere). A pebble and a rocket have the same escape speed from Earth — about $11.2\;\mathrm{km\,s^{-1}}$. What differs is the energy required: $E = \tfrac{1}{2} m v_{\text{esc}}^2$ scales with the escaping mass.
Where is the neutral point between two masses?
The neutral point is where the net gravitational field $\vec{g} = 0$, NOT where the gravitational potential $V_g = 0$. It lies on the line joining the two masses and is always closer to the smaller mass. Use $\dfrac{d}{D-d} = \sqrt{\dfrac{M_1}{M_2}}$. Note that $V_g$ is never zero at any finite distance between two masses — the two negative potentials add to give an even more negative total.

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