Summary and Quick Revision

How to Use This Section

This is the last section of the chapter, and it is built for one job: to be read the night before the paper, and again in the queue outside the hall.

Nothing new is taught here. Every card below is a compression of something the earlier sections worked through properly, in the same notation and with the same results. So if a line here surprises you, that is not a line to memorise — it is a signal to go back and reread the section that owns it.

Ten formula cards, two graphs, one solar-system table, one mistake checklist, one 60-second panic list and one fast self-test. Screenshot the three figures.

Four notation reminders.

• $M_E$ and $R_E$ are the Earth's mass and radius. Plenty of coaching material writes plain $M$ and $R$ for exactly the same things. Read both without blinking.
• $U$ is gravitational potential ENERGY, in joules; $V$ is gravitational POTENTIAL, in J/kg. They are related by $U = mV$ and they are not interchangeable. This is the single most examined confusion in the chapter, so the two symbols are kept rigorously apart on every card below.
• $v_e$ is escape speed, $v_o$ is orbital speed, $a$ is the semi-major axis, $e$ is the eccentricity, and $g_h$, $g_d$, $g_\lambda$ are the values of $g$ at height $h$, at depth $d$ and at latitude $\lambda$.
• Every card that puts a number on the page uses $g = 9.8$ m/s$^2$, $R_E = 6.4\times10^{6}$ m and $G = 6.67\times10^{-11}$ N m$^2$/kg$^2$. Those choices give $v_e = 11.2$ km/s exactly. Some worked examples elsewhere use $R_E = 6.37\times10^6$ m; the difference is well under 1%, but never mix two values inside one problem.

Four of the topics on these cards — the variation of $g$ with latitude, geostationary and polar satellites, weightlessness and binding energy — sit outside the rationalised syllabus body text, and so does the comparison of inertial and gravitational mass. Board papers, JEE Main, JEE Advanced and NEET ask all five every year, so they are on these cards in full.

Card 1 — Kepler's Three Laws

The statements, in the words the paper wants

Key Point — the law of ORBITS (first law): Every planet moves in an ellipse, with the Sun at one of its two foci. Not at the centre, and the other focus is empty. A circle is the special case $e = 0$, where the two foci merge at the centre.

Key Point — the law of AREAS (second law): The line joining a planet to the Sun sweeps out equal areas in equal intervals of time. Equivalently, the areal velocity $\frac{dA}{dt}$ is constant.

Key Point — the law of PERIODS (third law): $$T^2 \propto a^3, \qquad\text{that is}\qquad T^2 = \frac{4\pi^2}{GM}a^3$$ where $a$ is the semi-major axis — the average of the nearest and farthest distances, not either one of them.

The ellipse, in the four numbers you actually need

What the second law really is

$$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$$

The area law is conservation of angular momentum. Gravity is a central force — it acts along the line joining the two bodies — so its torque about the Sun is $\vec{r}\times\vec{F} = \vec{0}$ and $\vec{L}$ cannot change. Two consequences get asked directly:

• Fast when close, slow when far, with $v_pr_p = v_ar_a$ at the two apsides, so $\dfrac{v_p}{v_a} = \dfrac{r_a}{r_p}$.
• On any planetary orbit two quantities are conserved: angular momentum and total mechanical energy. Speed, kinetic energy and potential energy are each conserved only for a circle.

Reading the third law

In astronomical units and years the constant becomes 1, so $T^2 = a^3$ and $a = T^{2/3}$. The constant depends only on the mass of the central body — the same number for every planet round the Sun, a different number for every moon round the Earth, and never anything to do with the orbiting body's own mass.

[NEET Important] Two eccentric orbits with the same semi-major axis have the same period, however different their shapes. The third law does not see eccentricity at all.

Card 2 — The Universal Law, Superposition and the Two Shell Theorems

The law

Key Point: $$F = G\frac{m_1m_2}{r^2}, \qquad \vec{F}_{12} = -\,G\frac{m_1m_2}{r^2}\hat{r}_{12}$$ Always attractive, always along the line joining the two bodies, and always an action-reaction pair of equal magnitude — the Earth pulls the apple exactly as hard as the apple pulls the Earth. $G = 6.67\times10^{-11}$ N m$^2$/kg$^2$, with dimensions $\mathrm{M^{-1}L^{3}T^{-2}}$. The minus sign and $\hat{r}_{12}$ together say the force on 1 points back towards 2.

The law as written is for point masses. It applies to real bodies only when they are far apart compared with their sizes, or when they are uniform spheres — for which the shell theorems make it exact.

Superposition

Key Point: The force on one body due to several others is the vector sum of the pairwise forces, each computed as though the others were not there: $$\vec{F}_1 = \sum_{i \ne 1}\vec{F}_{1i}$$ Gravity cannot be shielded. Placing a lead wall, a planet or a hollow shell between two masses changes the force between them by exactly zero.

The scalar partner is easier and gets used more: potentials simply add, with no directions to worry about, $$V = \sum_i \left(-\frac{Gm_i}{r_i}\right)$$

The two shell theorems

Key Point — theorem 1 (outside): A uniform spherical shell attracts a particle outside it as though the shell's entire mass were concentrated at its centre.

Key Point — theorem 2 (inside): A uniform spherical shell exerts no gravitational force at all on a particle placed anywhere inside it. The force is zero at every interior point, not just at the centre.

The second theorem is why the field inside a solid sphere depends only on the mass beneath you: every shell further out contributes nothing.

$$g_{\text{inside}}(r) = \frac{GM r}{R^3} \propto r, \qquad g_{\text{outside}}(r) = \frac{GM}{r^2} \propto \frac{1}{r^2}$$

Key Point — the trap built on theorem 2: Zero field does not mean zero potential. Inside the shell $V$ is a constant $-\frac{GM}{R}$ everywhere, and $g = -\frac{dV}{dr} = 0$ precisely because $V$ is flat, not because it is zero.

[JEE Tip] The "remove one from a symmetric set" trick: if $n$ equal masses spaced round a circle give zero field at the centre, then removing one leaves a field of $\frac{Gm}{R^2}$ pointing away from the gap — because the survivors must supply minus whatever the missing one used to. The answer never contains $n$.

Card 3 — $g$ at the Surface, and Two Kinds of Mass

Key Point: $$\boxed{g = \frac{GM_E}{R_E^2}} \qquad\text{and equivalently}\qquad \boxed{g = \frac{4}{3}\pi G\rho R_E}$$ Use the first when the question gives a mass, the second when it gives a density. $g$ depends on the planet alone — never on the mass, shape or material of the falling body, which is why a coin and a feather fall together in a vacuum.

Rearranged, the first form gives $M_E = \frac{gR_E^2}{G} = 5.97\times10^{24}$ kg, which is what "Cavendish weighed the Earth" means: measuring $G$ in a laboratory is what turns a known $g$ into a known planetary mass. The mean density follows as about 5500 kg/m$^3$, roughly twice that of surface rock — the evidence for a dense iron core.

Scaling, the way it is asked

$$g \propto \frac{M}{R^2} \qquad\text{at fixed density,}\qquad g \propto \rho R$$

so half the radius at fixed mass gives $4g$, while half the radius at fixed density gives $\frac{g}{2}$. Deciding which quantity the question holds fixed is the whole exercise.

Inertial and gravitational mass

Key Point: Inertial mass is defined by $m_i = \frac{F}{a}$ — a body's resistance to being accelerated by any force at all. Gravitational mass is defined by $m_g = \frac{FR^2}{GM}$ — the strength with which a body responds to gravity. They are logically independent definitions, and experiment finds them equal to better than one part in $10^{12}$.

That equality is exactly why $g$ is the same for every body: $a = \frac{m_g}{m_i}g = g$. It is the observational foundation of the general theory of relativity.

[Board Important] Mass and weight: mass is the same everywhere, measured in kilograms by a beam balance; weight is $mg$, measured in newtons by a spring balance, and it changes from planet to planet, with latitude, with altitude and with depth. A 60 kg astronaut has a mass of 60 kg on the Moon and a weight of only 97 N there.

Card 4 — The Variation of $g$: Altitude, Depth, Latitude and Shape

Panel (a) is the $g$-versus-$r$ graph this card is about; panel (b) is the $V$-versus-$r$ graph, which belongs to the next card. They are drawn together because the paper likes to ask about both at once.

Going up: altitude

Key Point — exact, always valid: $$g_h = \frac{GM_E}{\left(R_E+h\right)^2} = g\left(\frac{R_E}{R_E+h}\right)^{2} = g\left(1+\frac{h}{R_E}\right)^{-2}$$ Approximate, and only for $h \ll R_E$: $$g_h \approx g\left(1 - \frac{2h}{R_E}\right)$$ The approximation is the first two terms of a binomial expansion. It is within 1% up to about 360 km and 8% low at 1000 km, and at $h = R_E$ it returns a negative value, which is the algebra telling you it has been pushed far past its range.

Going down: depth

Key Point — exact for a uniform Earth: $$g_d = g\left(1 - \frac{d}{R_E}\right)$$ This is the shell theorem in action: only the sphere of radius $R_E - d$ beneath you pulls, and everything in the shell above you contributes exactly nothing. $g$ falls linearly with depth and reaches zero at the centre of the Earth.

Two comparisons the paper enjoys:

• Near the surface, $g$ falls twice as fast going up as going down — $\frac{2h}{R_E}$ against $\frac{d}{R_E}$ — so for small changes the depth at which $g$ matches its value at a height $h$ is $d = 2h$. That shortcut fails once $d$ or $h$ is a serious fraction of $R_E$, where the exact formulae must be used.
• The maximum value of $g$ is at the surface. It falls off on both sides.

Rotation: latitude

Key Point: $$g_\lambda = g - \omega^2 R_E\cos^2\lambda$$ A body at latitude $\lambda$ moves in a circle of radius $R_E\cos\lambda$, so part of the gravitational pull is spent supplying its centripetal acceleration and the apparent weight is smaller by $m\omega^2R_E\cos^2\lambda$.

• At the equator ($\lambda = 0$) the correction is largest: $\omega^2R_E = \left(7.29\times10^{-5}\right)^2\left(6.4\times10^6\right) = 0.034$ m/s$^2$, which is 0.35% of $g$.
• At the poles ($\lambda = 90°$) the correction is zero — a body there sits on the axis, traces no circle, and needs no centripetal force.
• For the equator to become weightless, $\omega$ would have to grow about 17 times, shortening the day to roughly 1.4 hours.

Shape: oblateness

The Earth is not a sphere. Its equatorial radius exceeds its polar radius by about 21 km, so a pole is nearer the centre and $g$ is larger there for that reason too. Rotation and shape together give roughly

$$g_{\text{pole}} \approx 9.83\ \text{m/s}^2, \qquad g_{\text{equator}} \approx 9.78\ \text{m/s}^2$$

[NEET Important] Ranking questions recur every year. From largest to smallest: at the poles, then at the equator, then at a modest depth, then at a modest height, and zero at the centre. And a beam balance compares masses, so it reads the same everywhere; only a spring balance notices the change.

Card 5 — Gravitational Potential Energy and Gravitational Potential

The two definitions, kept apart

Key Point — potential ENERGY, in joules: $$\boxed{U = -\frac{GMm}{r}}$$ the work done by an external agent in bringing the mass $m$ from infinity to a distance $r$ from $M$, both at rest. It belongs to the pair of bodies, not to either one of them.

Key Point — POTENTIAL, in J/kg: $$\boxed{V = -\frac{GM}{r}}$$ the potential energy per unit mass at that point. It belongs to the point, and it exists whether or not you put anything there.

$$\boxed{U = mV}$$

The two-second test: does the question hand you a mass in kilograms? If yes, it wants $U$ and an answer in joules. If not, it wants $V$ and an answer in J/kg.

The signs, and the zero

Key Point: With the zero taken at infinity — the universal convention — both $U$ and $V$ are negative everywhere, and both approach zero only as $r \to \infty$. Moving a body away from the Earth makes $U$ less negative, that is, it increases $U$. There is no point in a purely gravitational problem where $V$ is positive.

Recovering $mgh$

For a rise from $R_E$ to $R_E + h$, $$\Delta U = \frac{GM_Emh}{R_E\left(R_E+h\right)} \;\xrightarrow{\;h \ll R_E\;}\; \frac{GM_Emh}{R_E^2} = mgh$$

$mgh$ is an approximation, valid only near the surface. The exact result for $h = R_E$ is $\frac{mgR_E}{2}$, whereas $mgh$ would claim $mgR_E$ — twice too much.

Systems of masses

$$U_{\text{system}} = -G\sum_{\text{pairs}}\frac{m_im_j}{r_{ij}}$$

Three particles give three pairs, four particles give six. For three equal masses $m$ at the corners of an equilateral triangle of side $a$, $$U = -\frac{3Gm^2}{a}$$ and the work an external agent must do to separate them completely is $+\frac{3Gm^2}{a}$ — positive, because pulling gravitating bodies apart always raises $U$ towards zero.

$V$ for the standard bodies

At the centre of a solid sphere $V = -\frac{3GM}{2R}$, one and a half times as deep as at the surface. At the centre of a shell $V = -\frac{GM}{R}$, the same as at its surface.

[JEE Tip] Fields cancel; potentials do not. At the null point between two masses the field is zero and the potential is a definite negative number. Inside a shell the field is zero everywhere and $V = -\frac{GM}{R}$ everywhere. Any option offering "zero potential because the field is zero" is wrong by construction.

Card 6 — Escape Speed

Key Point: $$\boxed{v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}} \qquad\text{and for the Earth}\qquad v_e = 11.2\ \text{km/s}$$ It comes from setting the launch kinetic energy equal to the binding energy at the surface: $\frac{1}{2}mv_e^2 = \frac{GMm}{R}$. The mass $m$ cancels from both sides, which is the whole reason for the next box.

Key Point — what escape speed does NOT depend on: the mass of the projected body, its shape, its composition, and the direction of projection. A pebble and a rocket need the same 11.2 km/s. Launching at $45°$ works exactly as well as launching straight up, because energy is a scalar and the potential depends only on distance. It does depend on the planet's mass and radius, and on the launch height.

The forms worth carrying

$$v_e = \sqrt{2gR} = \sqrt{2}\,v_o = R\sqrt{\frac{8\pi G\rho}{3}}, \qquad v_e(\text{from height } h) = \sqrt{\frac{2GM}{R+h}}$$ $$v_\infty^2 = v_0^2 - v_e^2 \qquad\text{(speed left over far away, after launching at } v_0 > v_e)$$ $$h_{\max} = \frac{k^2R}{1-k^2} \qquad\text{(launched at } k\,v_e \text{ with } k<1)$$

Speeds never subtract linearly. Launched at 15 km/s, a body reaches infinity at $\sqrt{15^2 - 11.2^2} = 10.0$ km/s, not at 3.8 km/s.

Why the Moon has no atmosphere

Gas molecules have a spread of speeds, and any molecule faster than the escape speed simply leaves. The Moon's $v_e$ is only 2.4 km/s, comparable with the thermal speeds of light molecules, so over geological time it has lost everything. The Earth's 11.2 km/s is far above the thermal speed of nitrogen and oxygen, which is why the air is still here — and why hydrogen and helium, being light and therefore fast, have largely escaped even from the Earth.

Card 7 — Orbital Speed and Time Period

Key Point: Gravity supplies the centripetal force, $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$, so $$\boxed{v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM_E}{R_E+h}}} \qquad\qquad \boxed{T = 2\pi\sqrt{\frac{r^3}{GM}}}$$ with $r$ measured from the centre of the planet, never from the ground. Neither expression contains the satellite's mass.

Squaring the period gives $T^2 = \frac{4\pi^2}{GM}r^3$ — Kepler's third law derived rather than observed, with the constant depending only on the central mass. That is how a planet is weighed from a moon's orbit: $M = \frac{4\pi^2r^3}{GT^2}$.

The scaling family, all for a circular orbit

$$v_o \propto r^{-1/2}, \qquad T \propto r^{3/2}, \qquad L \propto r^{1/2}, \qquad E \propto -\,r^{-1}$$

Higher means slower and longer. A satellite at $4R_E$ moves at half the surface-skimming speed and takes eight times as long. Every satellite-scaling question in this chapter is one of those four proportionalities.

The two anchor numbers

Anchor to those two and any other Earth orbit is one application of $T \propto r^{3/2}$ away.

For a satellite skimming a planet of density $\rho$ the period contains no radius at all: $$T = \sqrt{\frac{3\pi}{G\rho}}$$ so two planets of the same density have surface-skimming satellites of identical period, whatever their sizes.

Escape speed against orbital speed

Key Point: At the same radius, $$\frac{v_e}{v_o} = \sqrt{2} \approx 1.41$$ always, for every body in the solar system. To turn a circular orbit into an escape you must raise the speed by 41.4%, which costs an extra energy exactly equal to the satellite's present kinetic energy.

[NEET Important] Read the table for the pattern, not the digits: $v_e$ is always $\sqrt{2}$ times $v_o$, and both are set by $g$ and $R$ alone. Notice that Saturn's surface gravity barely exceeds the Earth's while its escape speed is over three times as large — because $v_e = \sqrt{2gR}$ and Saturn is enormous.

Card 8 — The Energy of an Orbiting Satellite, and Binding Energy

Key Point — the triple, for a circular orbit of radius $r$: $$K = +\frac{GMm}{2r}, \qquad U = -\frac{GMm}{r}, \qquad E = K + U = -\frac{GMm}{2r}$$ and therefore $$\boxed{E = -K} \qquad\qquad \boxed{U = 2E} \qquad\qquad \boxed{U = -2K}$$ Compute any one and the other two are free. $K$ comes from the force balance $v^2 = \frac{GM}{r}$; nothing here is a separate thing to memorise.

Key Point — why $E$ is negative: a negative total energy is what "bound" means. The satellite has less energy than it would need to sit at rest at infinity, where $E = 0$, so it cannot get there. Raise $E$ to zero and it escapes; a positive $E$ is an unbound, hyperbolic path.

Binding energy

Key Point: $$B = -E = +\frac{GMm}{2r}$$ the energy that must be supplied to take the satellite from its orbit to rest at infinity. It is positive, and for a circular orbit it is numerically equal to the kinetic energy.

Two cases with the same name, differing by a factor of two — read which one the question means:

So a body on the ground needs 11.2 km/s to leave, while a satellite already circling at 7.9 km/s needs only 3.3 km/s more.

Raising an orbit

Moving from radius $r_1$ to $r_2 > r_1$ needs $$\Delta E = \frac{GM_Em}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right) > 0$$

and while that happens $U$ rises by twice as much as $K$ falls. The satellite in the higher orbit moves more slowly, even though energy was spent putting it there — the tidiest counter-intuitive result in the chapter. Air drag runs the same argument backwards: a satellite losing energy to drag spirals inward and speeds up.

Card 9 — Geostationary Satellites, Polar Satellites and Weightlessness

The geostationary conditions

Key Point: A satellite appears fixed above one point of the ground only if all four conditions hold:

1. its period is exactly one sidereal day, 24 hours (23 h 56 min to be precise);
2. its orbit lies in the equatorial plane;
3. it moves west to east, the same sense as the Earth's rotation;
4. the orbit is circular.

Its mass is irrelevant — that is never one of the conditions.

Put $T = 24$ h into $T = 2\pi\sqrt{\frac{r^3}{GM_E}}$ and the radius follows:

Because the orbit must be equatorial, no geostationary satellite can hang over New Delhi — or over any place away from the equator. It can only sit over a point on the equator at the same longitude. Three satellites $120°$ apart cover essentially the whole globe except the polar caps.

Polar satellites

Key Point: A polar satellite orbits low — typically 500 to 800 km up — in a plane passing close to both poles, with a period of roughly 100 minutes. Each orbit crosses a strip of ground shifted about $25°$ of longitude west of the previous one, because the Earth turns underneath. After enough orbits it has viewed the whole globe, poles included.

Low means quick and close, so the resolution is high. Polar satellites are used for remote sensing, mapping, resource survey and weather; geostationary satellites are used for communications, broadcasting and continuous weather watch over one region. The two designs are chosen for opposite reasons.

Weightlessness

Key Point: An astronaut in an orbiting spacecraft is not beyond the Earth's gravity. At 400 km the local $g$ is still 8.7 m/s$^2$, about 89% of its surface value, and it is exactly that gravity which holds the spacecraft in orbit. What has vanished is the contact force: the astronaut, the spacecraft and everything in it fall together with the same acceleration, so nothing presses on anything.

$$mg_{\text{local}} - N = ma = mg_{\text{local}} \qquad\Longrightarrow\qquad N = 0$$

So a spring balance reads zero, a beam balance is useless, a pendulum has no restoring force and does not oscillate at all, and mercury will not stay in a barometer. Mass is entirely unchanged.

[NEET Important] Two situations both give zero reading on a spring balance and they are physically different. At the centre of the Earth the true $g$ really is zero. In an orbiting spacecraft $g$ is large and the body is in free fall. Options that conflate the two are set every year.

Card 10 — The Mistakes That Cost the Most Marks

Every one of these was flagged somewhere in the earlier sections. They are ordered roughly by how often they actually turn up in answer scripts.

1. Confusing gravitational potential with gravitational potential energy. The one that costs the most, by a distance. $V = -\frac{GM}{r}$ is in J/kg and belongs to a point; $U = -\frac{GMm}{r}$ is in joules and belongs to a pair of bodies; $U = mV$. If the question gives you a mass in kilograms and wants joules, it wants $U$. If it names a point and wants J/kg, it wants $V$. Copying one across as the other loses the whole mark.

2. Losing a sign. With the zero at infinity, $U$ and $V$ are negative everywhere, the total energy of any bound orbit is negative, and binding energy is positive. Moving a body outward makes $U$ less negative, so $\Delta U > 0$. If a bound satellite comes out with positive total energy, or the work to separate two masses comes out negative, you have subtracted the wrong way round.

3. Using the wrong radius. $R_E$ is the surface. A point at height $h$ is at $R_E + h$ from the centre; a point at depth $d$ is at $R_E - d$. The $r$ in every orbital formula is measured from the centre of the Earth, never from the ground. "At a height equal to twice the Earth's radius" means $r = 3R_E$, giving $\frac{g}{9}$ — not $\frac{g}{4}$.

4. Using $mgh$ far from the surface. $mgh$ is the first term of the exact expression and is honest only while $h \ll R_E$. At $h = R_E$ the true change in potential energy is $\frac{mgR_E}{2}$, while $mgh$ claims $mgR_E$ — a 100% error. The same warning applies to $g_h \approx g\left(1-\frac{2h}{R_E}\right)$, which returns a negative $g$ at $h = R_E$ and is 8% low even at 1000 km.

5. Thinking astronauts are beyond gravity. Weightlessness in orbit is free fall, not the absence of gravity. At the height of a space station $g$ is about 89% of its surface value. What is zero is the normal reaction, not the gravitational field, and the astronaut's mass is unchanged.

6. Concluding that zero field means zero potential. Inside a uniform shell the field is zero everywhere and $V = -\frac{GM}{R}$ everywhere. At the null point between two masses the field cancels and the potential does not — potentials are scalars and both contributions are negative, so they add. $\vec{g} = -\frac{dV}{dr}$ measures the slope of $V$, not its value.

7. Mixing up the depth formula with the height formula. Down: $g_d = g\left(1-\frac{d}{R_E}\right)$, linear, exact for a uniform Earth, zero at the centre. Up: $g_h = \frac{GM_E}{(R_E+h)^2}$, inverse square, zero only at infinity. They agree only at the surface, where $g$ takes its largest value.

8. Getting the $\cos^2\lambda$ backwards. $g_\lambda = g - \omega^2R_E\cos^2\lambda$. The correction is largest at the equator, where $\lambda = 0$ and $\cos^2\lambda = 1$, and zero at the poles, where the body sits on the axis and traces no circle. So $g$ is greatest at the poles.

9. Losing the factor of two in the energy triple. $K = \frac{GMm}{2r}$, $U = -\frac{GMm}{r}$, $E = -\frac{GMm}{2r}$. Quoting $|U|$ when the question asked for the binding energy doubles the answer. And a body at rest on the surface needs $mgR_E$ to escape, while a satellite already in a surface-skimming orbit needs only half that.

10. Feeding a perigee distance into Kepler's third law. $T^2 \propto a^3$ needs the semi-major axis, which is $\frac{r_p + r_a}{2}$. For a circle they coincide; for an ellipse they do not, and the question that hands you both apsides is testing exactly this.

11. Forgetting that speeds do not subtract. $v_\infty^2 = v_0^2 - v_e^2$. Launched at 15 km/s a body reaches infinity at 10.0 km/s, not at $15 - 11.2 = 3.8$ km/s.

12. Treating gravity as shieldable, or as depending on the falling body. No material blocks gravity, and $g$ is the same for a coin and a feather in a vacuum. The escape speed does not depend on the projectile's mass, though the escape energy certainly does.

Key Point: Three more that cost single marks each — using $g \propto \frac{1}{R^2}$ when the question fixed the density rather than the mass, quoting the escape speed where the orbital speed was wanted (they differ by $\sqrt{2}$), and mixing $g = 9.8$ with $g = 10$ inside one problem. Pick one value of $g$, write it at the top of your working, and use it everywhere.

The 60-Second Revision

You are in the queue outside the hall. This is the irreducible minimum.

Kepler. Ellipse, Sun at a focus. Equal areas in equal times, which is $L$ conserved, so $v_pr_p = v_ar_a$. $T^2 \propto a^3$ with $a$ the semi-major axis, and $r_p + r_a = 2a$, $e = \frac{r_a-r_p}{r_a+r_p}$.

The law. $F = G\frac{m_1m_2}{r^2}$, attractive, along the join, action-reaction, $G = 6.67\times10^{-11}$. Superposition: add the pairwise vectors for forces, the plain scalars for potentials. Shell theorems: mass at the centre for an outside point, zero force for an inside point.

$g$. $g = \frac{GM_E}{R_E^2} = \frac{4}{3}\pi G\rho R_E = 9.8$ m/s$^2$. Up: $g_h = \frac{GM_E}{(R_E+h)^2}$, or $g\left(1-\frac{2h}{R_E}\right)$ for small $h$. Down: $g_d = g\left(1-\frac{d}{R_E}\right)$, zero at the centre. Latitude: $g_\lambda = g - \omega^2R_E\cos^2\lambda$, largest at the poles.

$U$ and $V$. $U = -\frac{GMm}{r}$ in J; $V = -\frac{GM}{r}$ in J/kg; $U = mV$; $\vec{g} = -\frac{dV}{dr}\hat{r}$. Both negative, both zero only at infinity. $mgh$ only near the surface. Inside a shell $V = -\frac{GM}{R}$, constant. At the centre of a solid sphere $V = -\frac{3GM}{2R}$.

Escape. $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = 11.2$ km/s. Independent of the body's mass, shape and launch direction. $v_\infty^2 = v_0^2 - v_e^2$.

Orbit. $v_o = \sqrt{\frac{GM}{r}}$, $T = 2\pi\sqrt{\frac{r^3}{GM}}$, $v_e = \sqrt2\,v_o$. Higher is slower and longer. Floor: 7.9 km/s and 84.6 minutes.

Energy. $K = \frac{GMm}{2r}$, $U = -\frac{GMm}{r}$, $E = -\frac{GMm}{2r}$, so $E = -K$ and $U = 2E$. Binding energy $= -E$, positive. $mgR_E$ from the ground, $\frac{mgR_E}{2}$ from a skimming orbit.

Satellites. Geostationary: 24 h, equatorial, west to east, circular, $r = 4.2\times10^{7}$ m $= 6.6R_E$, about 36 000 km up, 3.1 km/s. Polar: low, about 100 minutes, sweeps the globe, used for remote sensing. Weightless means free fall, not zero gravity.

Habits. Name the radius. Decide $U$ or $V$. Check the sign. Ask whether the approximation is allowed. Pick one value of $g$ and keep it.

The Fast Self-Test

Cover the answers. Fourteen questions, five minutes. Anything you miss tells you which card to reopen tonight.

1. Where does the Sun sit in a planet's elliptical orbit?
2. Kepler's second law is a statement about the conservation of what?
3. Which length does Kepler's third law use — the perihelion distance, the aphelion distance, or the semi-major axis?
4. What is the gravitational force on a particle placed anywhere inside a uniform spherical shell?
5. Write $g$ at the Earth's surface in terms of $G$, $M_E$ and $R_E$, and again in terms of $G$, $\rho$ and $R_E$.
6. At a height equal to twice the Earth's radius, what fraction of $g$ survives?
7. At a depth equal to half the Earth's radius, what fraction of $g$ survives?
8. At which latitude is the rotational reduction in $g$ zero, and why?
9. State the SI units of $V$ and of $U$, and the relation between them.
10. Is the gravitational potential inside a uniform shell zero? What is it?
11. Write the escape speed in two forms, and name three things it does not depend on.
12. A satellite's orbital radius is doubled. What happens to its speed and to its period?
13. For a circular orbit, express $E$ in terms of $K$, and $U$ in terms of $E$.
14. Why does an astronaut in an orbiting spacecraft feel weightless?

Answers. 1. At one focus, not the centre. 2. Angular momentum. 3. The semi-major axis $a$. 4. Exactly zero, everywhere inside. 5. $g = \frac{GM_E}{R_E^2} = \frac{4}{3}\pi G\rho R_E$. 6. One ninth, since $r = 3R_E$. 7. One half. 8. At the poles, $\lambda = 90°$, because a body there lies on the axis of rotation and moves in no circle. 9. J/kg and J, with $U = mV$. 10. No — it is a constant $-\frac{GM}{R}$; the field is zero, which is why $V$ is flat. 11. $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$; it does not depend on the body's mass, its shape, or the direction of projection. 12. Speed falls by $\sqrt2$; period grows by $2\sqrt2 \approx 2.83$. 13. $E = -K$ and $U = 2E$. 14. Because the astronaut and the spacecraft fall together with the same acceleration, so the normal reaction is zero — gravity there is still about 89% of its surface value.

That is the whole chapter. Go and get the marks.