Physics Grade 12 · Unit 15 · Gravitation

Physics Grade 12 · Unit 15 · Gravitation · Athens Academy Islamabad

Physics · Grade 12 · FBISE · Exam March 2027

This unit is a scoring unit. Almost every mark comes from four formulae and knowing which one the question is asking for. This guide walks you through exactly what the examiner wants you to write and draw — the definitions to reproduce word-for-word, the derivations that carry the long questions, and the six diagrams you should be able to sketch from memory. Learn these and Gravitation becomes one of the easiest units in the paper.

1. Newton's Law of Universal Gravitation

Everything in the universe pulls on everything else. You are being pulled towards the person sitting next to you right now — the pull is just far too weak to feel. Newton put this everyday-invisible force into one exact law, and that law is your first guaranteed mark.

Definition — learn word for word

Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Building the formula is a standard 2-mark request. Start from the two proportionalities and combine them:

$F \propto m_1 m_2 \qquad \text{and} \qquad F \propto \frac{1}{r^{2}}$

$F \propto \frac{m_1 m_2}{r^{2}} \qquad \Longrightarrow \qquad F = G\,\frac{m_1 m_2}{r^{2}}$

where $F$ is the gravitational force (N), $m_1$ and $m_2$ are the two masses (kg), $r$ is the distance between their centres (m), and $G$ is the universal gravitational constant, $G = 6.67 \times 10^{-11}~\text{N m}^{2}\,\text{kg}^{-2}$ .

Fig. 1 — The gravitational force is always attractive and forms an action–reaction pair. Reproduce this in the exam.

Two facts about this force are examinable in short questions. It is always attractive, and it does not depend on the medium between the masses — air, water or vacuum makes no difference. And because the two forces are equal in size but opposite in direction, the law is consistent with Newton's third law of motion. That last sentence is the exact wording examiners look for.

Why the apple falls but Earth doesn't visibly move

The force on the apple and on the Earth is equal. But $a = F/m$ , and Earth's mass is enormous, so Earth's acceleration is unmeasurably tiny. Same force, wildly different accelerations. This is a favourite short question.

Where students lose marks

The distance $r$ is measured centre to centre, not surface to surface — and it is squared. If masses double and the distance halves, the force does not double: it becomes $\dfrac{(2)(2)}{(1/2)^{2}} = 16$ times larger. Always deal with the square before you write the final number.

Worked Example 1 — Force between two bodies MEDIUM

Given	$m_1 = 45~\text{kg}$ · $m_2 = 50~\text{kg}$ · $r = 1~\text{m}$ · $G = 6.67 \times 10^{-11}~\text{N m}^{2}\,\text{kg}^{-2}$
Required	Gravitational force $F$ (N)
Formula	$F = G\,\dfrac{m_1 m_2}{r^{2}}$
Substitution	$F = 6.67 \times 10^{-11} \times \dfrac{45 \times 50}{(1)^{2}}$
Calculation	$F = 6.67 \times 10^{-11} \times 2250 = 1.5 \times 10^{-7}~\text{N}$

∴

F \approx 1.5 \times 10^{-7}~\text{N}

Examiner tip: such a tiny force is why we never feel the pull between everyday objects — quote this to explain "why don't two books on a desk attract each other?"

2. Gravitational Field Strength

A mass changes the space around it: any other mass placed nearby feels a pull. That region of influence is the gravitational field, and its strength at a point is what we now measure.

Definition

The gravitational field strength $g$ at a point is the gravitational force acting on a unit mass placed at that point:

$g = \frac{F}{m}$

where $g$ is in $\text{N kg}^{-1}$ , $F$ is the force (N) and $m$ is the mass (kg). It is a vector, pointing in the direction of the force. Numerically $g$ equals the acceleration due to gravity.

The deriving of $g$ at Earth's surface is a classic long-question part. Put a small mass $m_0$ on Earth's surface. Two expressions describe the same force. From the law of gravitation, with $M_E$ the mass of Earth and $R_E$ its radius:

$F = \frac{G M_E m_0}{R_E^{2}}$

From $g = F/m_0$ , the weight is $F = m_0\,g$ . Setting the two equal:

$m_0\,g = \frac{G M_E m_0}{R_E^{2}} \qquad \Longrightarrow \qquad g = \frac{G M_E}{R_E^{2}}$

The mass $m_0$ cancels. This is the single most important consequence in the whole unit: because $m_0$ cancels, light and heavy bodies fall at the same rate. The general form for any planet of mass $m$ and radius $r$ is:

$g = \frac{G m}{r^{2}}$

Closer to home

Put in Earth's numbers — $M_E = 6 \times 10^{24}~\text{kg}$ , $R_E = 6.4 \times 10^{6}~\text{m}$ — and you get $g = 9.8~\text{m s}^{-2}$ , the same value you already use in every mechanics question. It was never a magic number; it falls straight out of this formula.

2.1 Variation of g with altitude

Climb higher and $r$ grows, so $g$ falls. At height $h$ above the surface the distance from Earth's centre is $(R_E + h)$ , giving:

$g_h = \frac{G M_E}{(R_E + h)^{2}} = \frac{g R_E^{2}}{(R_E + h)^{2}}$

where $g_h$ is the field strength at height $h$ . The change is small near the surface — even at the top of Mount Everest $g$ is still about $9.8~\text{m s}^{-2}$ . It only drops noticeably at satellite heights, which is why $g$ is treated as constant for small changes in height — a standard one-line answer.

Table 2.1 — How g weakens with altitude (learn the pattern, not the digits)
Altitude (km)	$g_h$ (m s $^{-2}$ )	Where
0	9.8	Earth's surface
8.8	9.8	Mount Everest — barely changed
400	8.7	Space shuttle / ISS orbit
35,700	0.2	Communication satellites

2.2 Gravitational field lines

Field lines are the examiner's favourite quick diagram. For a point mass (or a uniform sphere seen from outside) they point radially inward, evenly in all directions. Where lines are close together the field is strong; where they spread apart it is weak.

Fig. 2 — Radial field lines of a point mass. Reproduce this in the exam — arrows must point towards the mass.

Uniform sphere shortcut

For any point outside a uniform sphere, treat the whole mass as if concentrated at its centre. That single sentence is why the same $g = Gm/r^{2}$ formula works for planets, moons and stars alike.

Worked Example 2 — Free-fall acceleration on Jupiter MEDIUM

Given	$m_J = 1.898 \times 10^{27}~\text{kg}$ · $r_J = 7.15 \times 10^{7}~\text{m}$ · astronaut mass $= 60~\text{kg}$
Required	Surface gravity $g_J$ (m s $^{-2}$ ) and astronaut's weight $W$ (N)
Formula	$g_J = \dfrac{G m_J}{r_J^{2}}$ and $W = m\,g_J$
Substitution	$g_J = 6.67 \times 10^{-11} \times \dfrac{1.898 \times 10^{27}}{(7.15 \times 10^{7})^{2}}$
Calculation	$g_J = 24.77~\text{m s}^{-2}$ ; $W = 60 \times 24.77 = 1486~\text{N}$

∴

g_J \approx 24.8~\text{m s}^{-2}

(about 2.53 × Earth's) and

W \approx 1486~\text{N}

Examiner tip: the same 60 kg astronaut weighs only

60 \times 9.8 = 588~\text{N}

on Earth. Mass is unchanged; weight depends on the planet's

g

3. Satellites and Orbital Velocity

A satellite is held in its path by nothing but gravity — no engine. The trick is that Earth's gravity supplies exactly the centripetal force needed to bend its straight-line motion into a circle.

Definition

Orbital velocity is the tangential speed a satellite needs to stay in a circular orbit at a given radius. Too slow and it falls back to Earth; too fast and it flies off into an elliptical orbit or escapes entirely.

The derivation is short and heavily examined. For a satellite of mass $m$ orbiting at radius $r$ , gravity provides the centripetal force:

$\frac{G M_E m}{r^{2}} = \frac{m v_o^{2}}{r}$

The satellite mass $m$ cancels and one $r$ cancels, leaving:

$v_o = \sqrt{\frac{G M_E}{r}}$

where $v_o$ is the orbital velocity (m s $^{-1}$ ), $M_E$ is the mass of the central body (kg) and $r = R_E + h$ is the orbital radius from Earth's centre (m). The result depends only on the central mass and the orbital radius — not on the satellite's own mass.

Fig. 3 — Gravity supplies the centripetal force; velocity is tangent to the orbit. Reproduce this in the exam.

Why higher satellites move slower

Because $v_o = \sqrt{GM_E/r}$ , a larger $r$ means a smaller $v_o$ . Satellites close to Earth race around; distant ones drift slowly. This is a guaranteed short question — the answer is simply " $v_o$ is inversely related to $\sqrt{r}$ ."

Worked Example 3 — Orbital velocity of the ISS MEDIUM

Given	$h = 400~\text{km} = 4.0 \times 10^{5}~\text{m}$ · $R_E = 6.4 \times 10^{6}~\text{m}$ · $M_E = 6 \times 10^{24}~\text{kg}$
Required	Orbital velocity $v_o$ (m s $^{-1}$ )
Formula	$v_o = \sqrt{\dfrac{G M_E}{R_E + h}}$
Substitution	$v_o = \sqrt{\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^{6} + 4.0 \times 10^{5}}}$
Calculation	$v_o = \sqrt{\dfrac{4.0 \times 10^{14}}{6.8 \times 10^{6}}} = 7.67 \times 10^{3}~\text{m s}^{-1}$

∴

v_o \approx 7.67~\text{km s}^{-1}

— the ISS circles Earth about 16 times a day

Common error: forgetting to add

R_E

h

. The orbital radius is measured from Earth's centre, so

r = R_E + h

, never just

h

4. Geostationary Satellites

Set the orbit at just the right radius and the satellite takes exactly 24 hours to go round — the same time Earth takes to spin. To someone on the ground it then hangs motionless in the sky. That is a geostationary satellite, and it is why your dish antenna can be bolted in one fixed direction.

Definition

A geostationary satellite orbits directly above the equator with an orbital period of 24 hours, moving west to east, so it stays fixed above one point on Earth's surface.

To find its orbital radius, combine the orbital-velocity formula with the fact that speed = distance ÷ time over one full circle. The circumference is $2\pi r$ and the time is the period $T$ :

$v_o = \frac{2\pi r}{T} \qquad \text{and} \qquad v_o = \sqrt{\frac{G M_E}{r}}$

Setting these equal, squaring, and rearranging for $r$ :

$\frac{4\pi^{2} r^{2}}{T^{2}} = \frac{G M_E}{r} \qquad \Longrightarrow \qquad r = \left( \frac{G M_E T^{2}}{4\pi^{2}} \right)^{\!1/3}$

where $r$ is the orbital radius from Earth's centre (m), $T$ is the period (s), $M_E$ is Earth's mass (kg). Putting in $T = 86400~\text{s}$ gives $r \approx 4.23 \times 10^{7}~\text{m}$ , so the height above the surface is $h = r - R_E \approx 3.59 \times 10^{4}~\text{km}$ , and the orbital speed there is about $3.07~\text{km s}^{-1}$ . Learn these three numbers — they come up directly.

Where students lose marks

The formula gives $r$ , the radius from Earth's centre. The question often asks for height above the surface — you must then subtract $R_E$ . Read carefully whether "radius" or "height" is wanted; they differ by 6400 km.

Worked Example 4 — Radius of a synchronous satellite of Venus HARD

Given	$M = 4.867 \times 10^{24}~\text{kg}$ · $T = 243~\text{days} = 2.093 \times 10^{7}~\text{s}$
Required	Orbital radius $r$ (m) of a synchronous satellite
Formula	$r = \left( \dfrac{G M T^{2}}{4\pi^{2}} \right)^{1/3}$
Substitution	$r = \left( \dfrac{6.67 \times 10^{-11} \times 4.867 \times 10^{24} \times (2.093 \times 10^{7})^{2}}{4 \times (3.14)^{2}} \right)^{1/3}$
Calculation	$r = 1.53 \times 10^{9}~\text{m} = 1.53 \times 10^{6}~\text{km}$

∴

r \approx 1.53 \times 10^{9}~\text{m}

Examiner tip: the cube root is the whole exam skill here. Compute the bracket fully first, then take the one-third power last — never before.

5. Gravitational Potential and Potential Energy

Near the ground you have used $\Delta \text{P.E.} = mgh$ . But that assumes $g$ is constant, which only holds close to the surface. Out in space $g$ weakens with distance, so we need a formula built directly from Newton's law. This is the conceptual heart of the unit.

Definition — gravitational potential

The gravitational potential $V$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point:

$V = \frac{U}{m} = -\frac{G M_E}{r}$

where $V$ is in $\text{J kg}^{-1}$ , $M_E$ is the source mass (kg) and $r$ is the distance from its centre (m).

Definition — gravitational potential energy

The gravitational potential energy $U$ of a mass $m$ at distance $r$ from a mass $M_E$ is:

$U = -\frac{G M_E\, m}{r}$

where $U$ is in joules (J). The reference point of zero energy is taken at infinity.

The minus sign is the whole idea and examiners test it directly. We set potential energy to zero at infinity, where gravity's influence vanishes. Any finite distance is a place the mass has "fallen into" the field, so its energy is below zero — hence negative. As $r$ increases toward infinity, $U$ increases toward zero.

Fig. 4 — Gravitational P.E. is negative and rises toward zero as distance grows. Reproduce this shape in the exam.

Closer to home

Put Earth's numbers into $V = -GM_E/R_E$ and you get about $-62.5~\text{MJ kg}^{-1}$ at the surface. That means it would take 62.5 million joules to drag just one kilogram all the way out to infinity — a clean measure of how deep Earth's gravitational "well" really is.

Worked Example 5 — Potential and P.E. of Earth in the Sun's field HARD

Given	$M_S = 1.99 \times 10^{30}~\text{kg}$ · $M_E = 6 \times 10^{24}~\text{kg}$ · $r = 1.5 \times 10^{11}~\text{m}$
Required	Gravitational P.E. $U$ (J) and potential $V$ (J kg $^{-1}$ )
Formula	$U = -\dfrac{G M_S M_E}{r}$ and $V = -\dfrac{G M_S}{r}$
Substitution	$U = -\dfrac{6.67 \times 10^{-11} \times 1.99 \times 10^{30} \times 6 \times 10^{24}}{1.5 \times 10^{11}}$
Calculation	$U = -5.31 \times 10^{33}~\text{J}$ ; $V = -\dfrac{6.67 \times 10^{-11} \times 1.99 \times 10^{30}}{1.5 \times 10^{11}} = -8.85 \times 10^{8}~\text{J kg}^{-1}$

∴

U \approx -5.31 \times 10^{33}~\text{J}

and

V \approx -885~\text{MJ kg}^{-1}

Examiner tip: both answers are negative — if you write a positive value you have lost the concept mark.

V

is just

U

divided by the orbiting mass, so

V = U/M_E

also checks your arithmetic.

6. Key Terminology and Quick Reference

Table 6.1 — Every formula, its symbols and the mistake that costs marks
Quantity	Formula	Common exam mistake
Law of gravitation	$F = \dfrac{G m_1 m_2}{r^{2}}$	Using surface-to-surface, not centre-to-centre $r$
Field strength (planet)	$g = \dfrac{G M}{r^{2}}$	Forgetting to square $r$
g at altitude	$g_h = \dfrac{g R_E^{2}}{(R_E + h)^{2}}$	Writing $h$ instead of $R_E + h$
Orbital velocity	$v_o = \sqrt{\dfrac{G M}{r}}$	Thinking it depends on satellite mass
Geostationary radius	$r = \left(\dfrac{G M T^{2}}{4\pi^{2}}\right)^{1/3}$	Taking the cube root too early
Gravitational P.E.	$U = -\dfrac{G M m}{r}$	Dropping the minus sign
Gravitational potential	$V = -\dfrac{G M}{r}$	Confusing $V$ (per kg) with $U$ (total)

The three "it cancels" facts examiners love

Learn why a mass disappears in each derivation: in $g = GM/r^{2}$ the falling body's mass $m_0$ cancels (so all bodies fall alike); in $v_o = \sqrt{GM/r}$ the satellite mass cancels (so orbital speed is mass-independent). These one-line reasons are worth easy marks in every long question.

7. Sixty-Minute Revision Checklist

State Newton's law of universal gravitation word for word.
Build $F = Gm_1m_2/r^{2}$ from the two proportionalities.
Quote $G = 6.67 \times 10^{-11}~\text{N m}^{2}\,\text{kg}^{-2}$ with its unit.
Explain why gravitation is consistent with Newton's third law.
Explain why the apple falls but Earth doesn't visibly move.
Define gravitational field strength and give its unit $\text{N kg}^{-1}$ .
Derive $g = GM_E/R_E^{2}$ and state why $m_0$ cancels.
Write $g_h = gR_E^{2}/(R_E+h)^{2}$ and explain why $g$ is ~constant near the surface.
Sketch radial field lines with arrows pointing inward (Fig. 2).
State the uniform-sphere rule for a point outside the sphere.
Derive $v_o = \sqrt{GM_E/r}$ from gravity = centripetal force.
Explain why higher satellites move more slowly.
Draw the orbit diagram: gravity inward, velocity tangent (Fig. 3).
Define a geostationary satellite (24 h, equatorial, west→east).
Derive $r = (GM_ET^{2}/4\pi^{2})^{1/3}$ and know $r \approx 4.23\times10^{7}$ m.
Convert geostationary radius to height above the surface.
Define gravitational potential and gravitational P.E.
Explain why $U$ and $V$ are negative (zero taken at infinity).
Sketch the $U$ vs $r$ curve rising to zero (Fig. 4).
Re-work the ISS orbital-velocity example without looking.
Re-work the Sun–Earth potential-energy example without looking.

Final Word from Athens Academy

Gravitation rewards preparation more than almost any other unit. There are only seven formulae, four definitions to memorise word-for-word, and four diagrams to reproduce — master those and you can walk into the exam confident of the marks.

Two habits will carry you: always measure $r$ from the centre, and never lose the minus sign on potential and potential energy. Those two slips account for most of the marks dropped in this unit. Fix them now, in practice, so they never happen under exam pressure.

Learn the derivations by writing them out, not just reading them. Best of luck — Exam March 2027.

Athens Academy Islamabad · Physics · Grade 12 · Unit 15 — Gravitation · FBISE · Exam March 2027 · Sourced from National Book Foundation Physics Grade 12.

🔬 Chapter 15 · Gravitation