JEE Corner — Advanced Gravitation

What JEE Adds to Gravitation

Sections 1 to 10 built this chapter properly, and that build is complete for the Board syllabus. But JEE — Main, and far more so Advanced — asks a family of gravitation questions the core never quite sets up. A rod pulls on a bead lying on its own axis. A stone is dropped down a hole bored clean through the Earth. A satellite is nudged out of one circle and has to arrive on another. Two stars circle a point in empty space between them.

None of it is harder physics. It is the same inverse-square law, asked from a place where you have to build the answer instead of quoting it.

Most of what follows sits outside the rationalised syllabus body text, but JEE Main and JEE Advanced ask it every year, so it is developed here from first principles.

The eight things this section teaches

#	Skill	Why it earns marks
1	Field and potential of an extended body by integration — rod, ring, disc	The examiner picks a body with no formula. You have to build one
2	The shell and the solid sphere, inside and out, with BOTH graphs	The single most drawn pair of graphs in the chapter
3	The tunnel through the Earth and the SHM it produces	One set-up, one period, and a beautiful coincidence
4	Elliptical orbits with energy AND angular momentum together	Two equations, two unknowns, every time
5	The vis-viva relation $v^2 = GM\left(\dfrac{2}{r}-\dfrac{1}{a}\right)$	One line that replaces both of the above
6	Orbit transfers and the Hohmann manoeuvre	Two burns, and the arithmetic is always the same
7	Binary stars and the two-body reduction	Both bodies move; the centre of mass never does
8	Gravitational self-energy, $-\dfrac{3GM^2}{5R}$	Assembling a body, not just moving one

Running underneath all eight is one habit: decide whether you are computing a field or a potential, an energy or an energy per kilogram, and write the sign down before the number. More marks are lost in this chapter to a dropped minus sign than to any integral.

Conventions, fixed now

Constants. Throughout this section $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$, $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^{6}$ m, so $GM_E = 3.982 \times 10^{14}$ m$^3$/s$^2$. Where a surface value of $g$ is needed it is 9.8 m/s$^2$. Every problem states which numbers it uses, and no problem mixes 9.8 with 10.

Notation. $V$ is the gravitational potential, measured in J/kg; $U$ is the gravitational potential energy of a body, measured in J, and $U = mV$. They are never the same object and the paper punishes anyone who treats them as one. $\vec{g}$ is the gravitational field (equivalently the acceleration a free particle would have), in N/kg or m/s$^2$. $a$ is a semi-major axis, $e$ an eccentricity, $r_p$ and $r_a$ the perigee and apogee distances.

The zero of potential is at infinity, always, in every line below. That is why $V$ and $U$ come out negative everywhere inside the universe of this chapter.

Key Point: Two relations connect everything in this section, and neither is new. $$V(r) = -\int_\infty^r \vec{g}\cdot d\vec{r} \qquad\text{and}\qquad \vec{g} = -\frac{dV}{dr}\,\hat{r}$$ Potential is the running total of the field; the field is the slope of the potential. If you can compute either one, you already have the other — and the potential is almost always easier, because potentials add as plain numbers while fields add as vectors.

[Exam Tip] Before writing a single line, ask three questions. Am I inside the body or outside it? Is the quantity asked for a scalar (potential, energy) or a vector (field, force)? Which radius belongs in the formula — the body's $R$, or the point's $r$? Answer those three and most of these problems are already half solved.

Field and Potential of an Extended Body, by Integration

A point mass has $\vec{g} = -\dfrac{GM}{r^2}\hat{r}$ and $V = -\dfrac{GM}{r}$, and that is the end of it. A rod, a ring or a disc is a continuous smear of point masses, and you have to add up their contributions.

The method, in four lines

1. Choose an element $dm$ every point of which is at the same distance from the field point $P$. Get this right and the integral is easy; get it wrong and it is impossible.
2. Express $dm$ through the density: $dm = \lambda\,dx$ for a wire, $\sigma\,dA$ for a lamina, $\rho\,dV$ for a solid.
3. Integrate the POTENTIAL first, because $dV = -\dfrac{G\,dm}{r}$ is a plain number with no direction to keep track of: $$V = -\int \frac{G\,dm}{r}$$
4. Differentiate once to get the field: $g = -\dfrac{dV}{dx}$ along the symmetry axis.

Integrating the field directly is legal, but then you must resolve every $d\vec{g}$ into components and argue that the transverse ones cancel. The potential route skips that entirely.

The rod, along its own axis

A thin rod of mass $M$ and length $L$ lies on a line. $P$ is on that same line, a distance $d$ from the near end. An element of length $dx$ at distance $x$ from $P$ has $dm = \dfrac{M}{L}dx$:

$$V = -\int_d^{d+L} \frac{G M}{L}\frac{dx}{x} = -\frac{GM}{L}\ln\!\frac{d+L}{d}$$

$$g = -\frac{dV}{dd} \quad\Longrightarrow\quad \boxed{\ g = \frac{GM}{d\,(d+L)}\ }$$

directed towards the rod. Notice that the answer is not $\dfrac{GM}{d^2}$ and not $\dfrac{GM}{(d + L/2)^2}$: the near end pulls harder than the far end, so the effective distance is the geometric mean $\sqrt{d(d+L)}$, not the arithmetic one. Push $d$ out to 500 times $L$ and $\dfrac{GM}{d(d+L)}$ sits 0.2% below $\dfrac{GM}{d^2}$, which is the point at which a distant rod finally counts as a point mass.

The ring, on its axis

Every point of a ring of radius $a$ is the same distance $\sqrt{a^2+x^2}$ from a point $P$ on the axis. So the potential needs no integration at all:

$$\boxed{\ V = -\frac{GM}{\sqrt{a^2+x^2}}\ } \qquad\Longrightarrow\qquad \boxed{\ g = \frac{GMx}{(a^2+x^2)^{3/2}}\ }$$

Read off three things at once.

• At the centre, $x = 0$: the field is zero by symmetry, but the potential is $-\dfrac{GM}{a}$, its most negative value. A ring is the cleanest example in the whole chapter of a place where $g = 0$ and $V \neq 0$.
• Far away, $x \gg a$: $g \to \dfrac{GM}{x^2}$ and $V \to -\dfrac{GM}{x}$, as any body must.
• In between, $g$ has a maximum. Setting $\dfrac{dg}{dx} = 0$ gives $$x = \frac{a}{\sqrt{2}} \qquad\text{and there}\qquad g_{max} = \frac{2GM}{3\sqrt{3}\,a^2} = 0.385\,\frac{GM}{a^2}$$

That maximum is asked directly, most years, in one form or another.

The disc, on its axis

A disc of mass $M$ and radius $a$ is a stack of rings. Take a ring of radius $s$ and width $ds$; its share of the mass is $dm = \dfrac{M}{\pi a^2}\,2\pi s\,ds = \dfrac{2M s\,ds}{a^2}$, and every point of it is $\sqrt{s^2+x^2}$ from $P$:

$$V = -\int_0^a \frac{G}{\sqrt{s^2+x^2}}\cdot\frac{2Ms\,ds}{a^2} = -\frac{2GM}{a^2}\left(\sqrt{a^2+x^2}-x\right)$$

$$\boxed{\ g = \frac{2GM}{a^2}\left(1 - \frac{x}{\sqrt{a^2+x^2}}\right)\ }$$

Two limits worth having on tap. Right at the centre of the face, $x \to 0$, the field is $\dfrac{2GM}{a^2}$ — finite, not infinite. And writing $\sigma = \dfrac{M}{\pi a^2}$ and letting $a \to \infty$ at fixed $\sigma$,

$$g \to 2\pi G\sigma$$

independent of $x$. An infinite sheet pulls with the same strength however far away you stand, exactly as a uniformly charged infinite sheet does in electrostatics. That parallel is worth carrying: every result in this block has an electrostatic twin with $\dfrac{1}{4\pi\epsilon_0}$ in place of $-G$.

The table, so you never have to rebuild these under time pressure

Body	Field point	$g$	$V$
Point mass $M$	distance $r$	$\dfrac{GM}{r^2}$	$-\dfrac{GM}{r}$
Rod, mass $M$, length $L$	on its axis, $d$ from the near end	$\dfrac{GM}{d(d+L)}$	$-\dfrac{GM}{L}\ln\dfrac{d+L}{d}$
Ring, mass $M$, radius $a$	on the axis, distance $x$	$\dfrac{GMx}{(a^2+x^2)^{3/2}}$	$-\dfrac{GM}{\sqrt{a^2+x^2}}$
Ring	at its centre	$0$	$-\dfrac{GM}{a}$
Disc, mass $M$, radius $a$	on the axis, distance $x$	$\dfrac{2GM}{a^2}\left(1-\dfrac{x}{\sqrt{a^2+x^2}}\right)$	$-\dfrac{2GM}{a^2}\left(\sqrt{a^2+x^2}-x\right)$
Infinite sheet, density $\sigma$	anywhere	$2\pi G\sigma$	not defined (no zero at infinity)

Key Point: In every one of these, $V$ came first and $g$ came from $-\dfrac{dV}{dx}$. Check any result you derive by differentiating your own potential — if the two disagree, you have made an algebra slip, and you will find it in thirty seconds instead of losing the question.

[Advanced] The rod result generalises to a broken rod or a rod with a gap by simple subtraction, because both $g$ and $V$ are linear in the mass distribution. A rod from $d$ to $d+L$ with a piece from $d_1$ to $d_2$ removed has $g = \dfrac{GM_{full}}{d(d+L)}$ minus the missing piece's contribution, computed with its own limits. The same "add and subtract" trick is what makes the cavity problem in the next block work.

The Shell and the Solid Sphere: Both Graphs, Side by Side

Section 2 stated the two shell theorems and Section 4 drew $g$ against $r$. What the paper actually asks for is both curves — the field and the potential — for both bodies, and the differences between them are exactly where the marks sit.

The four results for a uniform shell, mass $M$, radius $R$

$$r \ge R:\quad g = \frac{GM}{r^2},\quad V = -\frac{GM}{r} \qquad\qquad r < R:\quad g = 0,\quad V = -\frac{GM}{R}$$

Inside a shell the field vanishes but the potential does not. It sits at the constant value $-\dfrac{GM}{R}$, the same as on the surface. That is not a contradiction: a constant potential has zero slope, and zero slope is zero field. Flat means constant, not zero.

The four results for a uniform solid sphere, mass $M$, radius $R$

$$r \ge R:\quad g = \frac{GM}{r^2},\quad V = -\frac{GM}{r}$$ $$r < R:\quad g = \frac{GMr}{R^3},\quad V = -\frac{GM\left(3R^2 - r^2\right)}{2R^3}$$

Inside, only the mass within radius $r$ pulls, and that mass is $M\dfrac{r^3}{R^3}$, so the field grows linearly from zero at the centre to $\dfrac{GM}{R^2}$ at the surface. The potential is a downward parabola, deepest at the centre:

$$V(0) = -\frac{3GM}{2R} = 1.5 \times V(\text{surface})$$

Check the inside potential by differentiating it: $-\dfrac{dV}{dr} = -\dfrac{GMr}{R^3}$, whose magnitude is exactly the inside field. The two halves fit.

The whole thing in one table

Region	Shell: $g$	Shell: $V$	Solid sphere: $g$	Solid sphere: $V$
$r > R$	$\dfrac{GM}{r^2}$	$-\dfrac{GM}{r}$	$\dfrac{GM}{r^2}$	$-\dfrac{GM}{r}$
$r = R$	$\dfrac{GM}{R^2}$	$-\dfrac{GM}{R}$	$\dfrac{GM}{R^2}$	$-\dfrac{GM}{R}$
$r < R$	$0$	$-\dfrac{GM}{R}$	$\dfrac{GMr}{R^3}$	$-\dfrac{GM(3R^2-r^2)}{2R^3}$
$r = 0$	$0$	$-\dfrac{GM}{R}$	$0$	$-\dfrac{3GM}{2R}$

Key Point — read the graphs, do not memorise them.

• $g$ is discontinuous for a shell (it leaps from $0$ to $\dfrac{GM}{R^2}$ as you cross the surface) and continuous for a solid sphere.
• $V$ is continuous for both. A jump in potential would mean an infinite field, and there is none.
• $g$ is maximum at the surface of a solid sphere, never inside it and never outside it.
• $V$ is most negative at the centre of a solid sphere and reaches zero only at infinity.
• Both bodies look identical from outside. Nothing measured at $r > R$ can tell a shell from a solid sphere of the same mass.

Superposition: concentric bodies add, cavities subtract

Because both $\vec{g}$ and $V$ are linear in mass, a compound body is handled by adding the pieces — being careful that each piece is evaluated at the same field point, with its own inside-or-outside rule.

Worked, in two lines. A solid sphere of mass $M$ and radius $R$ sits at the centre of a concentric thin shell of mass $2M$ and radius $3R$.

At $r = 2R$ you are outside the sphere and inside the shell, so $$g = \frac{GM}{(2R)^2} + 0 = \frac{GM}{4R^2}, \qquad V = -\frac{GM}{2R} - \frac{2GM}{3R} = -\frac{7GM}{6R}$$

At the centre both fields vanish, but neither potential does: $$g = 0, \qquad V = -\frac{3GM}{2R} - \frac{2GM}{3R} = -\frac{13GM}{6R}$$

That second line is the one students get wrong. The shell contributes nothing to $g$ and a great deal to $V$.

The cavity. Scoop a sphere of radius $\dfrac{R}{2}$ out of a uniform sphere of radius $R$, with the hole's centre a distance $d$ from the big centre. Treat the hollow body as full sphere plus negative sphere. At any point $P$ inside the cavity, with $\vec{r}$ from the big centre and $\vec{r}^{\,\prime} = \vec{r} - \vec{d}$ from the hole's centre,

$$\vec{g} = -\frac{4}{3}\pi G\rho\,\vec{r} + \frac{4}{3}\pi G\rho\,\vec{r}^{\,\prime} = -\frac{4}{3}\pi G\rho\,\vec{d}$$

$\vec{r}$ has cancelled. The field everywhere inside the cavity is uniform, of magnitude $\dfrac{4}{3}\pi G\rho d = \dfrac{GMd}{R^3}$ (with $M$ the mass of the original, unhollowed sphere), pointing from the cavity's centre towards the sphere's centre. A genuinely surprising result, and a JEE favourite.

[Exam Tip] The most expensive error in this block is using the wrong radius. In $\dfrac{GM}{r^2}$ outside, $r$ is the distance of the field point. In $-\dfrac{GM}{R}$ inside a shell, $R$ is the shell's own radius and the field point's distance never appears. Write down which body you are inside before you write the formula, every single time.

The Tunnel Through the Earth

Bore a straight hole from one side of the Earth to the other, through the centre, evacuate it, and drop a stone in. What happens next is one of the most quoted results in the chapter — and, remarkably, one of the easiest to derive.

The force inside is proportional to the displacement

Treat the Earth as a uniform sphere of mass $M_E$ and radius $R_E$. At a distance $r$ from the centre, only the mass inside radius $r$ pulls, and the field is $g(r) = \dfrac{GM_E r}{R_E^3}$, directed inwards. So for a stone of mass $m$ at position $r$ measured from the centre,

$$F = -\frac{GM_E m}{R_E^3}\,r$$

That is $F = -kr$ with $k = \dfrac{GM_E m}{R_E^3}$: a linear restoring force, and therefore simple harmonic motion about the centre of the Earth.

$$\omega^2 = \frac{k}{m} = \frac{GM_E}{R_E^3} = \frac{g}{R_E} \qquad\Longrightarrow\qquad \boxed{\ T = 2\pi\sqrt{\frac{R_E}{g}}\ }$$

The numbers

With $R_E = 6.37 \times 10^6$ m and $g = 9.8$ m/s$^2$:

$$T = 2\pi\sqrt{\frac{6.37\times 10^{6}}{9.8}} = 2\pi(806.2) = 5.07 \times 10^{3}\ \text{s} = 84.4\ \text{minutes}$$

Event	Time
dropped at the surface, from rest	$0$
passing the centre, at maximum speed	$T/4 = 21.1$ min
momentarily at rest at the far surface	$T/2 = 42.2$ min
back where it started	$T = 84.4$ min

The maximum speed, at the centre, is the amplitude times $\omega$:

$$v_{max} = \omega R_E = \sqrt{gR_E} = \sqrt{(9.8)(6.37\times 10^{6})} = 7.90 \times 10^{3}\ \text{m/s} = 7.90\ \text{km/s}$$

The coincidence that is not a coincidence

Look at that number. 7.90 km/s is exactly the orbital speed of a satellite skimming the Earth's surface, and $2\pi\sqrt{R_E/g}$ is exactly its orbital period. The stone in the tunnel and the surface-skimming satellite keep the same time.

That is not luck. A circular orbit of radius $R_E$ projected onto any straight line through the centre is simple harmonic motion of amplitude $R_E$ and angular frequency $\sqrt{g/R_E}$. The satellite's shadow on the tunnel and the stone in the tunnel obey the same equation, so they arrive together.

Key Point: $T = 2\pi\sqrt{\dfrac{R_E}{g}} \approx 84$ minutes is the shortest period any object can have around the Earth, whether it goes through the planet or skims over it. Every higher orbit is slower. Remember the number: 84 minutes, or 5.07 kiloseconds.

The chord tunnel: the same 84 minutes

Now dig a tunnel that does not pass through the centre — a straight chord whose closest approach to the centre is $b$. Let $s$ be the displacement along the tunnel from its midpoint. At that point $r^2 = b^2 + s^2$, the field has magnitude $\dfrac{GM_E}{R_E^3}r$ pointing at the centre, and the component along the tunnel is that magnitude times $\dfrac{s}{r}$:

$$F_{\parallel} = -\frac{GM_E m}{R_E^3}\,r \cdot \frac{s}{r} = -\frac{GM_E m}{R_E^3}\,s$$

The $r$ cancels again. Same spring constant, same $\omega$, same period of 84.4 minutes — for every chord, however short. Only the amplitude changes: it is the half-length of the chord, $\sqrt{R_E^2 - b^2}$, so the maximum speed is $\omega\sqrt{R_E^2-b^2}$, smaller for a shorter tunnel.

[Advanced] This is the physics behind the old "gravity train" proposal: a frictionless tunnel between any two points on Earth is a 42-minute one-way journey, whether it links Delhi and Mumbai or Delhi and Buenos Aires. The trip time is the same because the period is the same. If a question tells you the tunnel misses the centre and expects a different period, the question is wrong — but if it asks for the maximum speed, that genuinely does depend on the chord.

[Exam Tip] Two traps. First, $T$ depends only on $R_E$ and $g$ — not on the mass of the stone, not on the length of the tunnel. Second, the SHM picture uses the uniform-density Earth. The real Earth has a dense core, so the real field does not rise linearly from the centre and the real period would differ; every exam question assumes uniform density, and you should say so in one line if you are writing a full solution.

Elliptical Orbits: Two Conservation Laws, Used Together

A circular orbit has one unknown speed and one equation. An elliptical orbit has two speeds you care about — at perigee and at apogee — and needs two equations. Those two equations are always the same pair.

The pair

Angular momentum. Gravity is a central force, so its torque about the focus is zero and $\vec{L}$ is constant. At perigee and apogee the velocity is perpendicular to the radius, so

$$\boxed{\ m v_p r_p = m v_a r_a \quad\Longrightarrow\quad \frac{v_p}{v_a} = \frac{r_a}{r_p}\ }$$

Energy. Gravity is conservative, so

$$\boxed{\ \frac{1}{2}v_p^2 - \frac{GM}{r_p} = \frac{1}{2}v_a^2 - \frac{GM}{r_a}\ }$$

Two equations, two unknowns. Solving them together, with $r_p = a(1-e)$ and $r_a = a(1+e)$:

$$v_p = \sqrt{\frac{GM}{a}\cdot\frac{1+e}{1-e}}, \qquad v_a = \sqrt{\frac{GM}{a}\cdot\frac{1-e}{1+e}}, \qquad v_p v_a = \frac{GM}{a}$$

That last relation is a lovely, and rarely quoted, check: the product of the perigee and apogee speeds equals the circular speed for radius $a$, squared.

The geometry you must be able to write down cold

Quantity	In terms of $a$ and $e$	In terms of $r_p$ and $r_a$
perigee distance	$r_p = a(1-e)$	—
apogee distance	$r_a = a(1+e)$	—
semi-major axis	$a$	$a = \dfrac{r_p+r_a}{2}$
eccentricity	$e$	$e = \dfrac{r_a-r_p}{r_a+r_p}$
semi-minor axis	$b = a\sqrt{1-e^2}$	$b = \sqrt{r_p r_a}$
total energy	$E = -\dfrac{GMm}{2a}$	same, with $a$ from the sum
period	$T = 2\pi\sqrt{\dfrac{a^3}{GM}}$	same

Key Point: $E$ and $T$ depend on $a$ alone — not on $e$. Two satellites with the same semi-major axis have the same total energy and the same period, however differently shaped their orbits are. A circle is just the special case $e = 0$, $a = r$, and every formula above collapses to the circular one when you put $e=0$.

Vis-viva: the one line that does both jobs

Combine the two conservation laws and eliminate everything but $r$:

$$\boxed{\ v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)\ }$$

This is the vis-viva equation, and it is the single most useful formula in orbital mechanics. It gives the speed at any point of any orbit — circle, ellipse, parabola or hyperbola — from just two numbers.

Feed it the special cases and watch them fall out:

Put in	You get	Meaning
$a = r$	$v = \sqrt{\dfrac{GM}{r}}$	the circular orbit speed
$a \to \infty$	$v = \sqrt{\dfrac{2GM}{r}}$	the escape speed — a parabola is an "ellipse" with $a$ infinite
$r = r_p = a(1-e)$	$v_p = \sqrt{\dfrac{GM}{a}\dfrac{1+e}{1-e}}$	fastest point
$r = r_a = a(1+e)$	$v_a = \sqrt{\dfrac{GM}{a}\dfrac{1-e}{1+e}}$	slowest point
$r = a$ (ends of the minor axis)	$v = \sqrt{\dfrac{GM}{a}}$	the same speed a circular orbit of radius $a$ would have

That last row is worth a moment. At the two points where the satellite is exactly $a$ from the focus, it is moving at precisely the circular speed for that distance — but not in a circle, because its velocity is not perpendicular to the radius there.

Areal velocity, in the form that gets asked

Kepler's second law is angular-momentum conservation wearing a hat:

$$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt} = \frac{L}{2m} = \text{constant}$$

so the time to sweep any region equals its area divided by $\dfrac{L}{2m}$. Sweeping the whole ellipse takes $T$, and the whole area is $\pi a b$, which gives $T = \dfrac{2\pi a b m}{L}$ — another route to Kepler's third law if you ever need it.

[Exam Tip] When a problem gives you one speed and one distance on an ellipse and asks for another, do not set up the two conservation laws from scratch. Get $a$ from vis-viva at the point you know, then use vis-viva again at the point you want. Two substitutions, no simultaneous equations.

Orbit Transfers, the Hohmann Manoeuvre, and Catching a Satellite

A spacecraft in a circular orbit of radius $r_1$ has to end up in a circular orbit of radius $r_2$. It cannot simply "move outwards" — every path it takes is itself an orbit. The cheapest route, and the only one you are asked about, is the Hohmann transfer: half an ellipse that touches the inner circle at its perigee and the outer circle at its apogee.

The transfer ellipse

$$r_p = r_1, \qquad r_a = r_2, \qquad \boxed{\ a_t = \frac{r_1+r_2}{2}\ }$$

Burn 1, at $r_1$: speed up from the circular speed to the transfer ellipse's perigee speed. $$v_1 = \sqrt{\frac{GM}{r_1}}, \qquad v_p = \sqrt{GM\left(\frac{2}{r_1}-\frac{1}{a_t}\right)} = \sqrt{\frac{2GM r_2}{r_1(r_1+r_2)}}$$ $$\boxed{\ \Delta v_1 = v_p - v_1 = \sqrt{\frac{GM}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}} - 1\right)\ }$$

Coast half an ellipse. That takes half the transfer orbit's period: $$\boxed{\ t_{transfer} = \pi\sqrt{\frac{a_t^{3}}{GM}}\ }$$

Burn 2, at $r_2$: the craft arrives at apogee moving too slowly to hold a circular orbit there, so it must speed up again. $$v_a = \sqrt{\frac{2GM r_1}{r_2(r_1+r_2)}}, \qquad v_2 = \sqrt{\frac{GM}{r_2}}$$ $$\boxed{\ \Delta v_2 = v_2 - v_a = \sqrt{\frac{GM}{r_2}}\left(1 - \sqrt{\frac{2r_1}{r_1+r_2}}\right)\ }$$

Key Point: Both burns are prograde — both speed the craft up — yet the craft ends up SLOWER than it started. A low orbit is a fast orbit. The engine adds kinetic energy twice, and the orbit's total energy rises from $-\dfrac{GMm}{2r_1}$ to $-\dfrac{GMm}{2r_2}$, which is less negative; but the speed $\sqrt{GM/r}$ has fallen. Energy up, speed down. Section 8 called this the orbit-raising paradox; here is the manoeuvre that actually does it.

The ledger, in one worked line

Going from a 7000 km orbit to the geostationary radius of 42 200 km, with $GM_E = 3.982\times10^{14}$:

Step	Value
circular speed at $r_1$	7.54 km/s
perigee speed on the transfer ellipse	9.88 km/s
$\Delta v_1$	2.34 km/s
apogee arrival speed	1.64 km/s
circular speed at $r_2$	3.07 km/s
$\Delta v_2$	1.43 km/s
total $\Delta v$	3.77 km/s
coast time	19 200 s = 5.34 h

Those are real launch numbers: getting a communications satellite from a parking orbit to geostationary really does cost a little under 4 km/s of velocity change and a little over five hours of coasting.

Two satellites in the same orbit: the docking paradox

Here is the set-up that catches everybody. Two satellites are in the same circular orbit, and one is a few degrees behind the other. How does the chaser catch up?

Not by firing its engine forward. Speeding up raises the orbit, which lengthens the period, which makes the chaser fall further behind.

$$\text{fire forward} \;\Rightarrow\; \text{higher orbit} \;\Rightarrow\; \text{longer period} \;\Rightarrow\; \text{you lose ground}$$

The chaser must fire backwards. That drops it into a slightly smaller ellipse whose period is shorter, so it gains a little angle on every lap. After enough laps the gap is closed, and a second forward burn puts it back on the original circle, now alongside the target. This is called a phasing orbit, and it is exactly how a spacecraft docks with a space station.

$$T_{phasing} = T\left(1 - \frac{\Delta\theta}{2\pi N}\right), \qquad a_{phasing} = \left(\frac{GM\,T_{phasing}^2}{4\pi^2}\right)^{1/3}$$

where $\Delta\theta$ is the angle to be made up and $N$ the number of laps you are prepared to spend. Fewer laps means a bigger drop and a bigger $\Delta v$; the manoeuvre is cheap only if you are patient.

Two satellites in different orbits: when do they line up?

If two satellites have periods $T_1$ and $T_2$, their angular speeds differ by $\left\lvert \frac{2\pi}{T_1} - \frac{2\pi}{T_2}\right\rvert$, so they return to the same relative position after

$$\boxed{\ \frac{1}{T_{syn}} = \left\lvert\frac{1}{T_1} - \frac{1}{T_2}\right\rvert\ }$$

the synodic period. If they orbit in opposite senses, the angular speeds add instead: $$\frac{1}{T_{meet}} = \frac{1}{T_1} + \frac{1}{T_2}$$

[Exam Tip] Read the sense of rotation before you pick a formula. "Same direction" gives a difference, "opposite directions" gives a sum, and swapping them is the single most common slip in this family of questions.

Binary Stars, Self-Energy, and the Traps That Cost Marks

Two bodies, one centre of mass

Everything so far assumed a tiny satellite around a huge fixed planet. Drop that assumption and both bodies move — around their common centre of mass, which stays put because no external force acts.

Two stars of masses $m_1$ and $m_2$, separated by a distance $d$, in circular orbits. The centre of mass divides the separation in the inverse ratio of the masses:

$$r_1 = \frac{m_2\,d}{m_1+m_2}, \qquad r_2 = \frac{m_1\,d}{m_1+m_2}, \qquad r_1 + r_2 = d$$

Both stars sweep the same angle in the same time, so they share one $\omega$ and one period. For star 1, the gravitational pull supplies its centripetal need:

$$\frac{Gm_1m_2}{d^2} = m_1\omega^2 r_1 = m_1\omega^2\frac{m_2 d}{m_1+m_2} \quad\Longrightarrow\quad \boxed{\ \omega^2 = \frac{G(m_1+m_2)}{d^3}\ }$$

$$\boxed{\ T = 2\pi\sqrt{\frac{d^3}{G(m_1+m_2)}}\ }$$

which is Kepler's third law with the total mass in the denominator. Everything else follows:

Quantity	Result	The sentence that goes with it
speeds	$v_1 = \omega r_1$, $v_2 = \omega r_2$	$\dfrac{v_1}{v_2} = \dfrac{m_2}{m_1}$: the heavier star moves slower
momenta	$m_1 v_1 = m_2 v_2$	equal and opposite; total momentum is zero
total KE	$\dfrac{1}{2}\mu v_{rel}^2$ with $v_{rel} = v_1+v_2$	one particle again
total energy	$E = -\dfrac{Gm_1m_2}{2a}$	$a$ is the semi-major axis of the relative orbit; $a = d$ if circular

The reduced mass

The relative position $\vec{r} = \vec{r}_2 - \vec{r}_1$ obeys

$$\mu\,\frac{d^2\vec{r}}{dt^2} = -\frac{Gm_1m_2}{r^2}\hat{r}, \qquad\text{where}\qquad \boxed{\ \mu = \frac{m_1m_2}{m_1+m_2}\ }$$

So the two-body problem is exactly a one-body problem: a single particle of mass $\mu$ orbiting a fixed centre that exerts the same force. Every orbital result you know — vis-viva, $T^2 \propto a^3$, $E = -\dfrac{Gm_1m_2}{2a}$ — applies to the relative orbit unchanged.

Two sanity checks on $\mu$. If $m_2 \gg m_1$ then $\mu \to m_1$, and the light body orbits a fixed heavy one, which is where we started. If $m_1 = m_2 = m$ then $\mu = \dfrac{m}{2}$, and the two stars sit diametrically opposite at $\dfrac{d}{2}$ each, moving at $v = \sqrt{\dfrac{Gm}{2d}}$.

Gravitational self-energy

Everything so far has been the energy of a body in a field. Self-energy is different: it is the work done by gravity in assembling the body from dust scattered at infinity, or equivalently minus the work you must do to pull it apart and scatter it again.

Build a uniform sphere shell by shell. When the sphere already has radius $r$, its mass is $m = M\dfrac{r^3}{R^3}$, and the next shell of thickness $dr$ carries $dm = \dfrac{3Mr^2}{R^3}dr$. Bringing that shell from infinity onto a sphere of mass $m$ releases

$$dU = -\frac{G\,m\,dm}{r} = -\frac{3GM^2 r^4}{R^6}\,dr$$

Integrate from $0$ to $R$:

$$\boxed{\ U_{self} = -\frac{3}{5}\frac{GM^2}{R}\ }$$

For a thin shell, where every scrap of mass ends up at the same radius $R$, the same argument gives $U_{self} = -\dfrac{GM^2}{2R}$.

The Earth's number. With $M_E = 5.97\times10^{24}$ kg and $R_E = 6.37\times10^{6}$ m,

$$U_{self} = -\frac{3(6.67\times10^{-11})(5.97\times10^{24})^2}{5(6.37\times10^{6})} = -2.24\times10^{32}\ \text{J}$$

so blowing the Earth apart and dispersing it to infinity would cost $+2.24\times10^{32}$ J — about seven days of the Sun's entire energy output, in every direction at once.

Key Point: Do not confuse the three energies. $U = -\dfrac{GM_Em}{r}$ is the energy of a body $m$ in the Earth's field. $E = -\dfrac{GM_Em}{2r}$ is the total energy of that body in orbit. $U_{self} = -\dfrac{3GM_E^2}{5R_E}$ is the energy stored in the Earth itself, and it involves $M_E^2$, not $M_E m$. Look at which masses appear before you pick a formula.

The seven traps, collected

1. Potential is not potential energy. $V$ is J/kg, $U = mV$ is J. A question asking for "the potential at the centre of the sphere" wants $-\dfrac{3GM}{2R}$, not $-\dfrac{3GMm}{2R}$.
2. Zero field is not zero potential. Inside a shell, at the centre of a ring, at the centre of a two-mass system: $g = 0$ and $V \neq 0$ at all three.
3. Superposition of fields is a VECTOR sum. Two equal masses on opposite sides of a point give zero field and a doubled (more negative) potential. Adding their magnitudes is the classic sign error.
4. The wrong radius. $-\dfrac{GM}{R}$ inside a shell uses the shell's radius; $\dfrac{GM}{r^2}$ outside uses the field point's distance; $\dfrac{GM_E}{(R_E+h)^2}$ uses the sum. Three different radii, three different formulas.
5. $T$ and $E$ of an ellipse depend on $a$, not on $r$. Do not put the current distance where the semi-major axis belongs.
6. Both Hohmann burns speed you up, and you end up slower. If your working has a retro-burn in a raising manoeuvre, you have the sign backwards.
7. Self-energy carries $M^2$. If your answer to a self-energy question is linear in $M$, it is wrong before you check the number.

[Advanced] One more relation worth carrying. For any bound orbit under an inverse-square force, the time-averaged kinetic and potential energies satisfy $\langle U\rangle = -2\langle K\rangle$, so $E = -\langle K\rangle = \dfrac{1}{2}\langle U\rangle$. That is the virial theorem, and for a circle it is exactly the $K : U : E = 1 : -2 : -1$ ratio Section 8 established. For an ellipse the instantaneous values wander, but the averages over one full period obey it exactly.

Solved Examples

Twelve problems at Advanced level. Try each on paper before reading the solution — the value here is in the set-up, not the arithmetic. Where numbers are used, $G = 6.67\times10^{-11}$, $M_E = 5.97\times10^{24}$ kg, $R_E = 6.37\times10^{6}$ m and $g = 9.8$ m/s$^2$.

Example 1: A ring, on its own axis

A uniform ring of mass $M$ and radius $a$ lies in the $yz$-plane with its centre at the origin. For a point $P$ on the $x$-axis a distance $x$ from the centre, find (a) the potential $V$, (b) the field $g$, (c) the value of $x$ at which $g$ is largest and the value of $g$ there, and (d) the speed with which a particle released from rest at $x = \sqrt{3}\,a$ passes through the centre of the ring.

Solution:

1. (a) The potential needs no integral. Every element of the ring is the same distance $\sqrt{a^2+x^2}$ from $P$, so $$V = -\frac{G}{\sqrt{a^2+x^2}}\int dm = -\frac{GM}{\sqrt{a^2+x^2}}$$

2. (b) Differentiate once. $$g = -\frac{dV}{dx} = -\frac{d}{dx}\left[-GM\left(a^2+x^2\right)^{-1/2}\right] = -\frac{GMx}{(a^2+x^2)^{3/2}}$$ The minus sign says the field points back towards the ring. Its magnitude is $\dfrac{GMx}{(a^2+x^2)^{3/2}}$, which is zero at the centre and zero at infinity.

3. (c) The maximum. Set the derivative to zero: $$\frac{d}{dx}\left[\frac{x}{(a^2+x^2)^{3/2}}\right]= \frac{(a^2+x^2)^{3/2} - x\cdot 3x(a^2+x^2)^{1/2}}{(a^2+x^2)^{3}} = 0 \ \Longrightarrow\ a^2 + x^2 = 3x^2$$ $$x = \frac{a}{\sqrt{2}}, \qquad g_{max} = \frac{GM\,(a/\sqrt{2})}{\left(\tfrac{3}{2}a^2\right)^{3/2}} = \frac{2GM}{3\sqrt{3}\,a^2} = 0.385\,\frac{GM}{a^2}$$

4. (d) Energy conservation, with potentials. At $x = \sqrt{3}a$, $\sqrt{a^2+3a^2} = 2a$, so $$V_i = -\frac{GM}{2a}, \qquad V_f = -\frac{GM}{a}$$ $$\frac{1}{2}v^2 = V_i - V_f = -\frac{GM}{2a}+\frac{GM}{a} = \frac{GM}{2a} \quad\Longrightarrow\quad v = \sqrt{\frac{GM}{a}}$$

Final Answer: (a) $-\dfrac{GM}{\sqrt{a^2+x^2}}$; (b) $\dfrac{GMx}{(a^2+x^2)^{3/2}}$ towards the ring; (c) $x = \dfrac{a}{\sqrt{2}}$, $g_{max} = \dfrac{2GM}{3\sqrt{3}a^2}$; (d) $v = \sqrt{\dfrac{GM}{a}}$.

Takeaway: Integrate the potential, then differentiate. The ring is the one body where the potential integral is free, and everything else follows from one derivative. And note part (d): the particle speeds up all the way to the centre even though the field is zero there — the field being zero says nothing about the potential.

Example 2: A rod, and the distance that is not the average

A thin uniform rod of mass $M$ and length $L$ lies along the $x$-axis. A point $P$ lies on the same axis, a distance $d$ from the nearer end. Find (a) the potential at $P$, (b) the field at $P$, (c) both at $d = \dfrac{L}{2}$, and (d) the percentage by which the exact field differs from the point-mass estimate $\dfrac{GM}{d^2}$ when $d = 500L$.

Solution:

1. (a) One integral. Take an element $dx$ at distance $x$ from $P$, with $dm = \dfrac{M}{L}dx$. Every point of it is at $x$, so $$V = -\int_d^{d+L}\frac{G}{x}\cdot\frac{M}{L}dx = -\frac{GM}{L}\left[\ln x\right]_d^{d+L} = -\frac{GM}{L}\ln\!\frac{d+L}{d}$$

2. (b) Differentiate with respect to $d$. $$\frac{dV}{dd} = -\frac{GM}{L}\left(\frac{1}{d+L}-\frac{1}{d}\right) = \frac{GM}{L}\cdot\frac{L}{d(d+L)} = \frac{GM}{d(d+L)}$$ so the field has magnitude $\dfrac{GM}{d(d+L)}$, directed towards the rod. (Getting the same answer by integrating $\dfrac{G\,dm}{x^2}$ directly is a useful check: $\displaystyle\int_d^{d+L}\frac{GM}{L}\frac{dx}{x^2} = \frac{GM}{L}\left(\frac{1}{d}-\frac{1}{d+L}\right)$, the same thing.)

3. (c) Put $d = L/2$. $$g = \frac{GM}{\left(\frac{L}{2}\right)\left(\frac{3L}{2}\right)} = \frac{4GM}{3L^2} = 1.333\,\frac{GM}{L^2}$$ $$V = -\frac{GM}{L}\ln 3 = -1.099\,\frac{GM}{L}$$

4. (d) The far-field check. With $d = 500L$, $$g_{exact} = \frac{GM}{500L\cdot 501L} = \frac{GM}{250\,500\,L^2}, \qquad g_{point} = \frac{GM}{250\,000\,L^2}$$ $$\frac{g_{point}}{g_{exact}} = \frac{250\,500}{250\,000} = 1.002$$ The point-mass estimate is 0.2% too big, even at five hundred rod-lengths away.

Final Answer: (a) $-\dfrac{GM}{L}\ln\dfrac{d+L}{d}$; (b) $\dfrac{GM}{d(d+L)}$; (c) $\dfrac{4GM}{3L^2}$ and $-\dfrac{GM}{L}\ln 3$; (d) 0.2% high.

Takeaway: The effective distance to a rod on its own axis is the geometric mean $\sqrt{d(d+L)}$, not the arithmetic mean $d + \dfrac{L}{2}$. Every extended body approaches a point mass eventually — but "eventually" is further away than students expect.

Example 3: A disc, and the sheet it becomes

A uniform disc of mass $M$ and radius $a$ has a point $P$ on its axis at distance $x$ from the centre. (a) Derive $V$ and $g$ at $P$. (b) Evaluate both at $x = 0.75a$. (c) Show that as $a \to \infty$ at fixed surface density $\sigma$, the field becomes independent of $x$, and state its value.

Solution:

1. (a) Slice into rings. A ring of radius $s$ and width $ds$ has area $2\pi s\,ds$ and therefore mass $$dm = \frac{M}{\pi a^2}\,2\pi s\,ds = \frac{2Ms\,ds}{a^2}$$ Every point of it is $\sqrt{s^2+x^2}$ from $P$, so it contributes $dV = -\dfrac{G\,dm}{\sqrt{s^2+x^2}}$ and $$V = -\frac{2GM}{a^2}\int_0^a \frac{s\,ds}{\sqrt{s^2+x^2}} = -\frac{2GM}{a^2}\left[\sqrt{s^2+x^2}\right]_0^a = -\frac{2GM}{a^2}\left(\sqrt{a^2+x^2}-x\right)$$ $$g = -\frac{dV}{dx} = -\frac{2GM}{a^2}\left(\frac{x}{\sqrt{a^2+x^2}} - 1\right) \quad\Longrightarrow\quad \lvert g\rvert = \frac{2GM}{a^2}\left(1-\frac{x}{\sqrt{a^2+x^2}}\right)$$

2. (b) At $x = 0.75a$. Then $\sqrt{a^2+0.5625a^2} = 1.25a$, so $$g = \frac{2GM}{a^2}\left(1-\frac{0.75}{1.25}\right) = \frac{2GM}{a^2}(0.40) = 0.800\,\frac{GM}{a^2}$$ $$V = -\frac{2GM}{a^2}\left(1.25a - 0.75a\right) = -\frac{2GM}{a^2}(0.5a) = -\frac{GM}{a}$$

3. (c) The infinite sheet. Write $M = \sigma\pi a^2$, so $\dfrac{2GM}{a^2} = 2\pi G\sigma$: $$g = 2\pi G\sigma\left(1 - \frac{x}{\sqrt{a^2+x^2}}\right) \xrightarrow[\ a\to\infty\ ]{} 2\pi G\sigma\left(1-0\right) = 2\pi G\sigma$$ The $x$ has vanished. An infinite sheet pulls equally hard at every distance.

Final Answer: (a) as boxed above; (b) $g = 0.800\dfrac{GM}{a^2}$, $V = -\dfrac{GM}{a}$; (c) $g \to 2\pi G\sigma$, independent of $x$.

Takeaway: A disc is a stack of rings, and each ring is already solved. Note also what happens at $x=0$: the field on the face of a disc is a finite $\dfrac{2GM}{a^2}$, not infinite — the mass right under your feet is spread out, and its contributions cancel sideways.