Solved Examples

How to Use This Problem Bank

Gravitation is one of those chapters where every idea eventually talks to every other one. A satellite question is a circular-motion question and an energy question at the same time. A "how deep is the mine" question is a shell-theorem question. A comet question is Kepler plus angular momentum plus energy. These 44 problems are laid out so the early ones drill a single idea and the later ones make two or three of them work together.

Work them with a pen. Cover the solution, try it, then compare — including the checks at the end of each solution, because the checks are where marks are usually lost.

The five questions to ask before you write anything

1. Which distance goes in the formula? Almost always the distance between centres, never between surfaces, and never the height above the ground unless the formula is specifically written in terms of $h$.
2. Am I being asked for $U$ or for $V$? Gravitational potential energy $U$ belongs to a pair of bodies and is measured in joules. Gravitational potential $V$ belongs to a point in space and is measured in joules per kilogram. Mixing them is the single most expensive error in this chapter.
3. Is anything a vector here? Force and field add as vectors and can cancel. Potential energy and potential are scalars and, for ordinary masses, are all negative — so they can only pile up.
4. What is my zero? Everywhere below, the zero of both $U$ and $V$ is at infinity. That is what makes both of them negative near a mass.
5. Does the sign make sense? $U$ negative, $V$ negative, the total energy of a bound orbit negative, binding energy positive. If a "binding energy" comes out negative you have dropped a minus sign.

Key Point: $\vec{g}$, the gravitational field, and $g$, the acceleration due to gravity, are the same quantity in the same units. A satellite in orbit has an acceleration exactly equal to the local $g$, because gravity is the only force on it.

The constants, and the value of $g$

Unless a problem states its own data, everything below uses

Quantity	Symbol	Value
Gravitational constant	$G$	$6.67 \times 10^{-11}$ N m$^2$/kg$^2$
Mass of the Earth	$M_E$	$5.97 \times 10^{24}$ kg
Radius of the Earth	$R_E$	$6.37 \times 10^{6}$ m
Product, worth memorising	$GM_E$	$3.982 \times 10^{14}$ m$^3$/s$^2$
Surface gravity	$g$	$9.8$ m/s$^2$
Mass of the Sun	$M_S$	about $2 \times 10^{30}$ kg
One astronomical unit	AU	$1.5 \times 10^{11}$ m

Each problem states the data it uses in its own statement, and no problem mixes $g = 9.8$ with $g = 10$. Several problems deliberately quote slightly different values of $M_E$ and $R_E$ (such as $6.0 \times 10^{24}$ kg and $6.4 \times 10^{6}$ m) — when that happens the statement says so, and the answer is worked with those numbers throughout.

[Board Important] Every solution below writes the formula on its own line before any number goes into it. Do the same in the exam. A correct formula with an arithmetic slip still earns most of the marks; a bare number earns none.

Solved Examples

Part 1: Kepler's Three Laws — Planets, Moons and Comets

Example 1: A planet that laps the Earth

Suppose a planet existed that went round the Sun exactly twice as fast as the Earth does — that is, its year is half of ours. How would the size of its orbit compare with the size of the Earth's?

Solution:

1. Quote the harmonic law. For any two bodies orbiting the same Sun, $$\frac{T_p^{2}}{T_E^{2}} = \frac{a_p^{3}}{a_E^{3}}$$ where $a$ is the semi-major axis. For a nearly circular orbit, $a$ is just the radius.

2. Put in the period ratio, $\frac{T_p}{T_E} = \frac{1}{2}$: $$\frac{a_p^{3}}{a_E^{3}} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4}$$

3. Take the cube root. $$\frac{a_p}{a_E} = \left(\frac{1}{4}\right)^{1/3} = 4^{-1/3} = 0.630$$

4. Read it in kilometres. With $a_E = 1.5 \times 10^{11}$ m, the planet would sit at $$a_p = 0.630 \times 1.5 \times 10^{11} = 9.45 \times 10^{10} \text{ m}$$ which is between the orbits of Venus and the Earth.

5. Check the speed too. Orbital speed goes as $\frac{1}{\sqrt{a}}$, so the planet moves $\frac{1}{\sqrt{0.630}} = 1.26$ times as fast as the Earth. Half the period on a $0.63$-sized orbit means $\frac{2\pi a}{T}$ has gone up by $\frac{0.63}{0.5} = 1.26$. Consistent.

Final Answer: its orbit would be $0.63$ times the size of the Earth's, about $9.45 \times 10^{10}$ m, and it would travel $1.26$ times as fast.

Takeaway: Halving the period does not halve the radius — it shrinks it only to $0.63$ of the original. The two-thirds power in $a \propto T^{2/3}$ makes orbital size respond lazily to changes in period.

Example 2: How far out is Saturn?

A year on Saturn lasts 29.5 Earth years. The Earth's orbit has a mean radius of $1.50 \times 10^{8}$ km. How far is Saturn from the Sun?

Solution:

1. Quote the law and rearrange it for $a$. $$\frac{T_S^{2}}{T_E^{2}} = \frac{a_S^{3}}{a_E^{3}} \qquad \Longrightarrow \qquad a_S = a_E\left(\frac{T_S}{T_E}\right)^{2/3}$$

2. Substitute $\frac{T_S}{T_E} = 29.5$: $$\left(29.5\right)^{2/3} = 9.55$$

3. Multiply out. $$a_S = 9.55 \times 1.50 \times 10^{8} = 1.43 \times 10^{9} \text{ km} = 1.43 \times 10^{12} \text{ m}$$

4. Sanity check. That is $9.55$ AU. Saturn is indeed a little under ten times as far from the Sun as we are, and takes about thirty times as long to get round. The three-halves relation between those two numbers ($9.55^{3/2} = 29.5$) is Kepler's third law in one line.

Final Answer: about $1.43 \times 10^{9}$ km from the Sun, or $9.55$ AU.

Takeaway: You never need $G$ or the Sun's mass to compare two orbits round the same star. Ratios are enough, and ratios are much less error-prone than absolute values.

Example 3: Weighing Jupiter with one of its moons

Io, one of Jupiter's moons, has an orbital period of $1.769$ days and an orbital radius of $4.22 \times 10^{8}$ m. Find Jupiter's mass, and compare it with the Sun's mass of $2 \times 10^{30}$ kg. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. The period rule for a satellite of a body of mass $M$. $$T = 2\pi\sqrt{\frac{r^{3}}{GM}} \qquad \Longrightarrow \qquad M = \frac{4\pi^{2}r^{3}}{GT^{2}}$$

2. Convert the period to seconds. $$T = 1.769 \times 24 \times 3600 = 1.528 \times 10^{5} \text{ s}$$

3. Work out the two big pieces. $$r^{3} = \left(4.22 \times 10^{8}\right)^{3} = 7.515 \times 10^{25} \text{ m}^3$$ $$T^{2} = \left(1.528 \times 10^{5}\right)^{2} = 2.336 \times 10^{10} \text{ s}^2$$

4. Assemble. $$M_J = \frac{4\pi^{2} \times 7.515 \times 10^{25}}{6.67 \times 10^{-11} \times 2.336 \times 10^{10}} = \frac{2.967 \times 10^{27}}{1.558} = 1.90 \times 10^{27} \text{ kg}$$

5. Compare with the Sun. $$\frac{M_J}{M_S} = \frac{1.90 \times 10^{27}}{2 \times 10^{30}} = 9.5 \times 10^{-4} \approx \frac{1}{1050}$$

Final Answer: $M_J = 1.90 \times 10^{27}$ kg, which is about one thousandth of the Sun's mass.

Takeaway: Watch anything go round a body and you have weighed that body. The orbiting object's own mass never appears — Io's mass is nowhere in this calculation, and that is why one moon is enough.

Example 4: How long is a galactic year?

Take our galaxy to contain $2.5 \times 10^{11}$ stars, each of one solar mass ($2 \times 10^{30}$ kg), and suppose that mass acts as though it were concentrated at the galactic centre. How long does a star $50\,000$ light-years out take to complete one revolution? Take one light-year as $9.46 \times 10^{15}$ m and $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. The total mass. $$M = 2.5 \times 10^{11} \times 2 \times 10^{30} = 5.0 \times 10^{41} \text{ kg}$$

2. The orbital radius in metres. $$r = 5.0 \times 10^{4} \times 9.46 \times 10^{15} = 4.73 \times 10^{20} \text{ m}$$

3. Quote the period formula and substitute. $$T = 2\pi\sqrt{\frac{r^{3}}{GM}}$$ $$r^{3} = \left(4.73 \times 10^{20}\right)^{3} = 1.058 \times 10^{62} \text{ m}^3$$ $$GM = 6.67 \times 10^{-11} \times 5.0 \times 10^{41} = 3.335 \times 10^{31} \text{ m}^3\text{/s}^2$$

4. Evaluate. $$T = 2\pi\sqrt{\frac{1.058 \times 10^{62}}{3.335 \times 10^{31}}} = 2\pi\sqrt{3.17 \times 10^{30}} = 2\pi \times 1.78 \times 10^{15}$$ $$T = 1.12 \times 10^{16} \text{ s}$$

5. In years. One year is $3.156 \times 10^{7}$ s, so $$T = \frac{1.12 \times 10^{16}}{3.156 \times 10^{7}} = 3.55 \times 10^{8} \text{ years}$$

Final Answer: about $1.12 \times 10^{16}$ s, which is roughly $355$ million years.

Takeaway: A galactic year is about 355 million years, so the Sun has gone round perhaps twenty times since it formed. The model here is crude — real galactic mass is spread out, not lumped at the centre — but the order of magnitude is right, and the method is exactly the method used on Jupiter two problems ago.

Example 5: What stays constant as a comet goes round

A comet moves round the Sun on a very elongated ellipse. Neglecting any mass it loses when it comes close to the Sun, state whether each of the following stays constant through one orbit: (a) linear speed, (b) angular speed, (c) angular momentum about the Sun, (d) kinetic energy, (e) potential energy, (f) total energy.

Solution:

1. The two things that genuinely do not change. The only force on the comet is the Sun's pull. It acts along the line joining them, so its torque about the Sun is zero, so angular momentum about the Sun is constant. That force is also conservative, so total energy is constant.

2. (a) Linear speed — not constant. At perihelion the comet is deep in the Sun's potential well and moves fastest; at aphelion it crawls. For an orbit with perihelion and aphelion distances $r_p$ and $r_a$, conservation of angular momentum gives $$v_pr_p = v_ar_a \qquad \Longrightarrow \qquad \frac{v_p}{v_a} = \frac{r_a}{r_p}$$ For a comet with $r_a = 50r_p$, the speed changes by a factor of fifty.

3. (b) Angular speed — not constant. Constant areal velocity means $r^{2}\omega$ is constant, so $\omega \propto \frac{1}{r^{2}}$. Between the two ends of that same orbit, $\omega$ changes by a factor of $50^{2} = 2500$.

4. (c) Angular momentum — constant. As argued in step 1. This is exactly the content of the equal-areas law, since the areal velocity is $\frac{L}{2m}$.

5. (d) Kinetic energy — not constant. It follows the speed, so it varies enormously.

6. (e) Potential energy — not constant. $U = -\frac{GM_Sm}{r}$ changes as $r$ changes; it is most negative at perihelion.

7. (f) Total energy — constant. $K$ and $U$ trade back and forth, but their sum does not move.

Quantity	Constant?	Why
linear speed	no	fastest at perihelion, slowest at aphelion
angular speed	no	$\omega \propto \frac{1}{r^{2}}$
angular momentum	yes	the Sun's pull exerts no torque about the Sun
kinetic energy	no	it follows the speed
potential energy	no	it follows $\frac{1}{r}$
total energy	yes	gravity is a conservative force

Final Answer: only the angular momentum about the Sun and the total energy are constant. The other four all vary through the orbit.

Takeaway: Two conserved quantities, four that change. Nearly every elliptical-orbit numerical in this chapter is solved by writing down those same two conservation statements and nothing else.

Example 6: Two moons of Saturn

Titan orbits Saturn at a mean distance of $1.222 \times 10^{9}$ m with a period of $15.95$ days. Rhea orbits the same planet at $5.27 \times 10^{8}$ m. Find (a) Rhea's period and (b) the mass of Saturn. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. (a) Two satellites of the same planet, so use the ratio form. $$\frac{T_R}{T_T} = \left(\frac{r_R}{r_T}\right)^{3/2}$$

2. The radius ratio. $$\frac{r_R}{r_T} = \frac{5.27 \times 10^{8}}{1.222 \times 10^{9}} = 0.4313$$ $$\left(0.4313\right)^{3/2} = 0.4313 \times \sqrt{0.4313} = 0.4313 \times 0.6567 = 0.2832$$

3. So Rhea's period is $$T_R = 0.2832 \times 15.95 = 4.52 \text{ days} = 3.90 \times 10^{5} \text{ s}$$

4. (b) For the mass, now we do need $G$. Use Titan, converting its period: $$T_T = 15.95 \times 86\,400 = 1.378 \times 10^{6} \text{ s}$$ $$M = \frac{4\pi^{2}r_T^{3}}{GT_T^{2}} = \frac{4\pi^{2}\left(1.222 \times 10^{9}\right)^{3}}{6.67 \times 10^{-11} \times \left(1.378 \times 10^{6}\right)^{2}}$$

5. Evaluate. $$r_T^{3} = 1.825 \times 10^{27} \text{ m}^3, \qquad T_T^{2} = 1.899 \times 10^{12} \text{ s}^2$$ $$M = \frac{7.205 \times 10^{28}}{126.7} = 5.69 \times 10^{26} \text{ kg}$$

6. Cross-check with Rhea. Using $r_R$ and the $T_R$ we just predicted must give the same mass, because that is precisely the statement $\frac{r^{3}}{T^{2}}$ is the same for both. It does.

Final Answer: Rhea's period is about $4.52$ days, and Saturn's mass is $5.69 \times 10^{26}$ kg — roughly 95 Earth masses.

Takeaway: Part (a) needed no constants at all; part (b) needed $G$. Ratios of orbits round one body are free; absolute masses always cost you a $G$.

Part 2: Adding Up Forces — Two, Three and Four Masses, and Null Points

Example 7: Two heavy spheres on a table

Two spheres, each of mass 100 kg and radius $0.10$ m, are placed on a horizontal table with their centres $1.0$ m apart. Find the gravitational force and the gravitational potential at the mid-point of the line joining the centres. Is a small object placed there in equilibrium? If so, is it stable? Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. First, the shell theorem lets us treat each sphere as a point. Each sphere is uniform and the test point lies outside both of them, so each behaves exactly as though its whole 100 kg sat at its centre. The radius $0.10$ m plays no part except to confirm the point is outside.

2. The force. The mid-point is $0.50$ m from each centre. The two pulls have equal magnitude $$F_1 = F_2 = \frac{Gm}{r^{2}} \times m_{test} = \frac{6.67 \times 10^{-11} \times 100}{0.25}\,m_{test}$$ but they point in exactly opposite directions along the line, so $$\vec{F}_{net} = 0$$

3. The potential. Potential is a scalar, so there is no cancellation — the two contributions add. $$V = -\frac{Gm}{r} - \frac{Gm}{r} = -\frac{2 \times 6.67 \times 10^{-11} \times 100}{0.50}$$ $$V = -2.67 \times 10^{-8} \text{ J/kg}$$ with the zero taken at infinity.

4. Is it equilibrium? Yes — zero net force is exactly what equilibrium means.

5. Is it stable? Displace the object a little along the line, towards one sphere. It is now nearer that sphere and further from the other, so the near pull grows and the far pull shrinks, and the net force drags it further along. That is the signature of instability. (Displace it sideways instead and the two pulls do develop a small restoring component — but a point that is unstable in even one direction is an unstable equilibrium.)

Final Answer: the net force is zero, the potential is $-2.67 \times 10^{-8}$ J/kg, and the object is in unstable equilibrium.

Takeaway: Zero field does not mean zero potential. Vectors cancel; scalars from ordinary masses cannot, because they all carry the same negative sign.

Example 8: Three masses in a row

Masses of 2 kg, 4 kg and 6 kg lie along a straight line at $x = 0$, $x = 1.0$ m and $x = 3.0$ m. Find the net gravitational force on the 4 kg mass. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. List the two pulls on the 4 kg, each from Newton's law $F = \frac{Gm_1m_2}{r^{2}}$.

2. From the 2 kg, a distance $1.0$ m away, pulling in the $-x$ direction: $$F_1 = \frac{G \times 2 \times 4}{1.0^{2}} = 8G$$

3. From the 6 kg, a distance $2.0$ m away, pulling in the $+x$ direction: $$F_2 = \frac{G \times 4 \times 6}{2.0^{2}} = \frac{24G}{4} = 6G$$

4. Add them as vectors. They are anti-parallel, so subtract magnitudes and keep the sign of the larger: $$F_{net} = 8G - 6G = 2G = 2 \times 6.67 \times 10^{-11} = 1.33 \times 10^{-10} \text{ N}$$ in the $-x$ direction, that is, towards the 2 kg mass.

5. The point of the problem. The 6 kg is the heavier neighbour, yet it loses. Trebling the mass multiplies the pull by 3; doubling the distance divides it by 4. Distance wins.

Final Answer: $1.33 \times 10^{-10}$ N directed towards the 2 kg mass, along the $-x$ direction.

Takeaway: In an inverse-square law, distance is a stronger lever than mass. Always compare $\frac{m}{r^{2}}$, never $m$ alone.

Example 9: Four masses on a square, and the pull on one corner

Four equal masses $m$ sit at the corners of a square of side $a$. Find the net gravitational force on any one of them due to the other three. Then evaluate it for $m = 20$ kg and $a = 0.50$ m, with $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. Name the corners. Call the corner we are interested in $A$, the two next to it $B$ and $D$, and the far one $C$. Then $AB = AD = a$ and $AC = a\sqrt{2}$.

2. The two side pulls. Each has magnitude $$F_{side} = \frac{Gm^{2}}{a^{2}}$$ and they are at right angles to each other, so their resultant is $$\sqrt{2}\,\frac{Gm^{2}}{a^{2}} = 1.414\,\frac{Gm^{2}}{a^{2}}$$ directed along the diagonal $AC$ — which is just the diagonal of the square built on the two equal arrows.

3. The diagonal pull. $C$ is $a\sqrt{2}$ away, so $$F_{diag} = \frac{Gm^{2}}{\left(a\sqrt{2}\right)^{2}} = \frac{Gm^{2}}{2a^{2}} = 0.500\,\frac{Gm^{2}}{a^{2}}$$ and it points along $AC$ as well.

4. They are parallel, so simply add. $$F = \left(\sqrt{2} + \frac{1}{2}\right)\frac{Gm^{2}}{a^{2}} = 1.914\,\frac{Gm^{2}}{a^{2}}$$ directed along the diagonal, towards the centre of the square.

5. Put the numbers in. $$\frac{Gm^{2}}{a^{2}} = \frac{6.67 \times 10^{-11} \times 400}{0.25} = 1.067 \times 10^{-7} \text{ N}$$ $$F = 1.914 \times 1.067 \times 10^{-7} = 2.04 \times 10^{-7} \text{ N}$$

6. A symmetry check. By the same argument, every corner feels a pull of the same size directed at the centre. Four inward pulls, arranged symmetrically — so the whole set would collapse inwards on itself, which is exactly what an isolated four-mass square would do.

Final Answer: $F = 1.914\,\frac{Gm^{2}}{a^{2}}$ along the diagonal towards the centre; for the given numbers, $2.04 \times 10^{-7}$ N.

Takeaway: Do the two perpendicular pulls first, then notice the third one is already lined up with their resultant. Choosing that order turns a three-vector problem into one right-angled triangle and one addition.

Example 10: Firing a rocket at the Sun

A rocket is launched from the Earth straight towards the Sun. At what distance from the Earth's centre does the net gravitational force on it fall to zero? Take the Sun's mass as $2 \times 10^{30}$ kg, the Earth's as $6 \times 10^{24}$ kg, and the Earth-Sun distance as $1.5 \times 10^{11}$ m. Neglect all other bodies.

Solution:

1. Set up. Let the null point be a distance $x$ from the Earth's centre, so it is $d - x$ from the Sun's centre, with $d = 1.5 \times 10^{11}$ m. The two pulls act in opposite directions there, so setting them equal: $$\frac{GM_Em_r}{x^{2}} = \frac{GM_Sm_r}{(d-x)^{2}}$$

2. The rocket's mass $m_r$ cancels, and so does $G$. Take the square root of both sides: $$\frac{d-x}{x} = \sqrt{\frac{M_S}{M_E}}$$

3. The mass ratio. $$\frac{M_S}{M_E} = \frac{2 \times 10^{30}}{6 \times 10^{24}} = 3.333 \times 10^{5} \qquad \Longrightarrow \qquad \sqrt{\frac{M_S}{M_E}} = 577.4$$

4. Solve for $x$. $$d - x = 577.4\,x \qquad \Longrightarrow \qquad d = 578.4\,x \qquad \Longrightarrow \qquad x = \frac{1.5 \times 10^{11}}{578.4}$$ $$x = 2.59 \times 10^{8} \text{ m}$$

5. Read it. That is about $2.6 \times 10^{5}$ km from the Earth's centre — only $0.17\%$ of the way to the Sun, and about two-thirds of the way out to the Moon, whose orbit sits at $3.84 \times 10^{8}$ m.

6. Why so close to the Earth? The Sun is $333\,000$ times heavier, so to balance it the rocket must be $\sqrt{333\,000} = 577$ times closer to the Earth than to the Sun.

Final Answer: about $2.59 \times 10^{8}$ m from the Earth's centre, on the line towards the Sun.

Takeaway: The null point always sits far nearer the lighter body, and the ratio of distances is the square root of the ratio of masses. Take that square root early — the numbers get much friendlier.

Example 11: An unequal triangle, done properly with components

Masses of 1 kg, 2 kg and 3 kg sit at the corners $A$, $B$ and $C$ of an equilateral triangle of side $1.0$ m. Find the magnitude and direction of the net gravitational force on the 3 kg mass. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. Set up coordinates. Put $A$ at the origin, $B$ at $(1, 0)$, and $C$ at $\left(0.5,\ 0.866\right)$, all in metres, since the height of an equilateral triangle of side 1 is $\frac{\sqrt{3}}{2} = 0.866$.

2. The two pulls on the 3 kg at $C$. Both have $r = 1.0$ m. $$F_{CA} = \frac{G \times 3 \times 1}{1^{2}} = 3G, \qquad F_{CB} = \frac{G \times 3 \times 2}{1^{2}} = 6G$$ $F_{CA}$ points from $C$ towards $A$, and $F_{CB}$ from $C$ towards $B$.

3. The unit vectors. $$\hat{u}_{CA} = (-0.5,\ -0.866), \qquad \hat{u}_{CB} = (+0.5,\ -0.866)$$

4. Add components. $$F_x = 3G(-0.5) + 6G(+0.5) = -1.5G + 3.0G = +1.5G$$ $$F_y = 3G(-0.866) + 6G(-0.866) = -9G \times 0.866 = -7.794G$$

5. Magnitude. $$F = G\sqrt{1.5^{2} + 7.794^{2}} = G\sqrt{2.25 + 60.75} = G\sqrt{63} = 7.937G$$ $$F = 7.937 \times 6.67 \times 10^{-11} = 5.29 \times 10^{-10} \text{ N}$$

6. Direction. Taking the angle from the $+x$ axis, $$\theta = \tan^{-1}\left(\frac{-7.794}{+1.5}\right) = -79.1°$$ so the force points downward and slightly to the right: $79.1°$ below the $+x$ axis, which is $10.9°$ away from straight down, tilted towards $B$.

7. Check it with the cosine rule. The angle between the two pulls is $60°$, so $$F = G\sqrt{3^{2} + 6^{2} + 2(3)(6)\cos 60°} = G\sqrt{9 + 36 + 18} = G\sqrt{63}$$ The same $\sqrt{63}$. Good.

Final Answer: $5.29 \times 10^{-10}$ N, directed $79.1°$ below the $AB$ direction — that is, mostly straight down towards the side $AB$, leaning $10.9°$ towards the heavier neighbour $B$.

Takeaway: Resolve into components and let the signs do the work. Two arrows and a cosine rule is faster here, but components generalise to four masses, five masses and any awkward angle, and they never let you lose a direction.

Example 12: The Sun pulls harder, the Moon raises the bigger tide

Compare (a) the gravitational force the Sun exerts on the Earth with that exerted by the Moon, and (b) the difference in that pull between the near side and the far side of the Earth for each of them. Use $M_S = 2 \times 10^{30}$ kg at $1.5 \times 10^{11}$ m, $M_M = 7.35 \times 10^{22}$ kg at $3.84 \times 10^{8}$ m, $M_E = 5.97 \times 10^{24}$ kg and $R_E = 6.37 \times 10^{6}$ m.

Solution:

1. (a) The two pulls. $$F_S = \frac{GM_SM_E}{r_{SE}^{2}} = \frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 5.97 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^{2}} = 3.54 \times 10^{22} \text{ N}$$ $$F_M = \frac{GM_MM_E}{r_{ME}^{2}} = \frac{6.67 \times 10^{-11} \times 7.35 \times 10^{22} \times 5.97 \times 10^{24}}{\left(3.84 \times 10^{8}\right)^{2}} = 1.98 \times 10^{20} \text{ N}$$ $$\frac{F_S}{F_M} = \frac{3.54 \times 10^{22}}{1.98 \times 10^{20}} = 178$$

2. So the Sun wins comfortably — by a factor of nearly 180.

3. (b) But a tide is not caused by the pull. A tide is caused by the pull being stronger on the near side than on the far side, which stretches the oceans. The relevant quantity is therefore the difference $$\Delta g = GM\left[\frac{1}{(r - R_E)^{2}} - \frac{1}{(r + R_E)^{2}}\right] \approx \frac{4GMR_E}{r^{3}}$$

4. Take the ratio of those differences. The $4GR_E$ cancels: $$\frac{\Delta g_M}{\Delta g_S} = \frac{M_M}{M_S}\left(\frac{r_{SE}}{r_{ME}}\right)^{3} = \frac{7.35 \times 10^{22}}{2 \times 10^{30}} \times \left(\frac{1.5 \times 10^{11}}{3.84 \times 10^{8}}\right)^{3}$$

5. Evaluate. $$= 3.675 \times 10^{-8} \times (390.6)^{3} = 3.675 \times 10^{-8} \times 5.96 \times 10^{7} = 2.19$$

6. Read it. The Moon's tidal stretch is about $2.2$ times the Sun's, even though the Moon's pull is 178 times weaker.

Final Answer: the Sun pulls 178 times harder, but the Moon's tidal effect is about $2.2$ times the Sun's.

Takeaway: The pull goes as $\frac{M}{r^{2}}$, but the tide goes as $\frac{M}{r^{3}}$. That extra power of $r$ is the whole story: being close beats being massive when what matters is how much the field changes across an object.

Part 3: Inside and Outside a Sphere, Weighing Worlds, and the Variations of $g$

Example 13: Can you hide from gravity inside a hollow shell?

A charge can be shielded from an electric field by putting it inside a hollow conductor. Can a body be shielded from the gravitational influence of nearby matter by putting it inside a hollow sphere, or by any other means? Support the answer with a calculation. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. What the shell itself does. A uniform spherical shell exerts no net force on any body placed anywhere inside it. That is genuine and exact: for any point inside, the extra mass of the far part of the shell is precisely compensated by its extra distance, so $$\vec{g}_{shell,\ inside} = 0$$

2. So far, so encouraging — but that is not shielding. Shielding would mean the shell blocking the field of something else. It does not.

3. A number to prove it. Put a hollow shell of mass 1000 kg and radius $1.0$ m around a small body, and place a 100 kg mass $2.0$ m from the centre, outside the shell. The field at the centre from that external mass is $$g_{ext} = \frac{GM}{r^{2}} = \frac{6.67 \times 10^{-11} \times 100}{\left(2.0\right)^{2}} = 1.67 \times 10^{-9} \text{ m/s}^2$$ Exactly what it would have been with no shell at all. The shell contributes zero, and it removes nothing.

4. Why electricity is different. A conductor shields because it has mobile charges of both signs, which rearrange until they cancel the outside field within the metal. Mass comes in only one sign, so nothing can rearrange to cancel anything. There is no negative mass to work with.

Final Answer: No. A hollow sphere contributes zero field inside itself, but it does not screen the field of any external body, and no arrangement of matter can. Gravitational shielding does not exist.

Takeaway: "Zero field from the shell" and "shielded from everything" are different statements. Inside a shell you are free of the shell — you are not free of the Earth, the Sun or the person standing next to you.

Example 14: A solid sphere, from three different places

A uniform solid sphere has mass 1000 kg and radius $0.50$ m. Find the gravitational force it exerts on a $1.0$ kg particle placed (a) $1.0$ m from the centre, (b) exactly on its surface, and (c) $0.25$ m from the centre. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. The two rules for a uniform sphere. Outside, the sphere behaves as a point mass at its centre. Inside, only the mass within the particle's radius counts, and it too acts as though it were at the centre: $$r \ge R:\quad F = \frac{GMm}{r^{2}} \qquad\qquad r \le R:\quad F = \frac{GMmr}{R^{3}}$$

2. (a) At $r = 1.0$ m, outside. $$F = \frac{6.67 \times 10^{-11} \times 1000 \times 1.0}{1.0^{2}} = 6.67 \times 10^{-8} \text{ N}$$

3. (b) On the surface, $r = R = 0.50$ m. Both formulas agree here, which is a useful check: $$F = \frac{6.67 \times 10^{-11} \times 1000 \times 1.0}{0.25} = 2.67 \times 10^{-7} \text{ N}$$

4. (c) At $r = 0.25$ m, inside. Only the inner sphere of radius $0.25$ m pulls, and it holds $\left(\frac{0.25}{0.50}\right)^{3} = \frac{1}{8}$ of the mass, that is $125$ kg, at a distance of $0.25$ m: $$F = \frac{6.67 \times 10^{-11} \times 125 \times 1.0}{\left(0.25\right)^{2}} = \frac{8.34 \times 10^{-9}}{0.0625} = 1.33 \times 10^{-7} \text{ N}$$

5. Check against the linear rule. At half the radius the inside formula predicts exactly half the surface value: $\frac{2.67 \times 10^{-7}}{2} = 1.33 \times 10^{-7}$ N. It agrees.

Final Answer: (a) $6.67 \times 10^{-8}$ N, (b) $2.67 \times 10^{-7}$ N, (c) $1.33 \times 10^{-7}$ N.

Takeaway: The maximum pull of a uniform sphere is at its surface. Move outwards and it falls as $\frac{1}{r^{2}}$; move inwards and it falls linearly to zero at the centre. Two different laws meeting at one point.

Example 15: A 250 N body, taken down and taken up

A body weighs 250 N at the Earth's surface. Assuming the Earth to be a sphere of uniform density, find its weight (a) half way down to the centre and (b) at a height equal to half the Earth's radius.

Solution:

1. (a) Depth: use the linear law. With $d = \frac{R_E}{2}$, $$g_d = g\left(1 - \frac{d}{R_E}\right) = g\left(1 - \frac{1}{2}\right) = \frac{g}{2}$$

2. So the weight halves. $$W_d = \frac{250}{2} = 125 \text{ N}$$

3. (b) Height: use the inverse square law, with $r = R_E + \frac{R_E}{2} = \frac{3R_E}{2}$: $$g_h = g\left(\frac{R_E}{r}\right)^{2} = g\left(\frac{2}{3}\right)^{2} = \frac{4g}{9}$$

4. So the weight becomes $$W_h = \frac{4}{9} \times 250 = 111 \text{ N}$$

5. Compare the two. The same displacement of half an Earth radius costs $50\%$ of the weight going down but $56\%$ going up. Going up is worse — because the inverse square is a steeper law than the straight line.

Final Answer: (a) $125$ N half way to the centre; (b) $111$ N at a height of half an Earth radius.

Takeaway: Down is linear, up is inverse-square, and they are not symmetric. The only place where both give the same value is the surface itself.

Example 16: A tunnel to the centre

Imagine a narrow tunnel bored from the surface of the Earth to its centre, and take the Earth to be a uniform sphere with $g = 9.8$ m/s$^2$ at the surface and $R_E = 6.37 \times 10^{6}$ m. (a) Write the acceleration of a body at a distance $r$ from the centre. (b) Find $g$ at a depth of $1000$ km. (c) Where along the tunnel is the pull greatest, and where is it least?

Solution:

1. (a) Only the sphere below you counts. At a distance $r$ from the centre, the mass inside is $M_r = M_E\left(\frac{r}{R_E}\right)^{3}$, and the shells above contribute nothing. So $$g_r = \frac{GM_r}{r^{2}} = \frac{GM_E}{R_E^{3}}\,r = g\,\frac{r}{R_E}$$ The pull is directly proportional to the distance from the centre.

2. In terms of depth $d = R_E - r$, the same statement reads $$g_d = g\left(1 - \frac{d}{R_E}\right)$$

3. (b) At $d = 1000$ km $= 1.0 \times 10^{6}$ m. $$\frac{d}{R_E} = \frac{1.0 \times 10^{6}}{6.37 \times 10^{6}} = 0.157$$ $$g_d = 9.8 \times (1 - 0.157) = 9.8 \times 0.843 = 8.26 \text{ m/s}^2$$ A drop of about $16\%$ for a thousand kilometres of descent.

4. (c) Greatest and least. Since $g_r$ rises steadily with $r$ all the way out to the surface, and then falls as $\frac{1}{r^{2}}$ beyond it, the maximum is at the surface itself, $9.8$ m/s$^2$. The minimum along the tunnel is at the centre, where $r = 0$ and so $g = 0$: a body released at the centre would float, because it is pulled equally in every direction.

5. A caution about the real Earth. The Earth is not uniform; the iron core is roughly four times denser than the crust. As a result the real $g$ actually rises slightly for the first few thousand kilometres of depth before falling. The uniform-density model is the one intended in every exam question, and it is the one used here.

Final Answer: (a) $g_r = g\frac{r}{R_E}$; (b) $8.26$ m/s$^2$ at a depth of $1000$ km; (c) greatest at the surface, zero at the centre.

Takeaway: The centre of the Earth is the one place where a body has weight zero because the field is zero. Remember that later, when weightlessness in orbit turns out to be a completely different phenomenon with the same instrument reading.

Example 17: Weighing the Earth from a stone dropped down a well

A stone dropped from rest into a dry well falls $20.0$ m in $2.02$ s. Using only this measurement and $R_E = 6.37 \times 10^{6}$ m, find the mass of the Earth and its mean density. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$ and ignore air resistance.

Solution:

1. Get $g$ from the fall. From rest, $h = \frac{1}{2}gt^{2}$, so $$g = \frac{2h}{t^{2}} = \frac{2 \times 20.0}{\left(2.02\right)^{2}} = \frac{40.0}{4.080} = 9.80 \text{ m/s}^2$$

2. Quote the surface-gravity relation and invert it. $$g = \frac{GM_E}{R_E^{2}} \qquad \Longrightarrow \qquad M_E = \frac{gR_E^{2}}{G}$$

3. Substitute. $$R_E^{2} = \left(6.37 \times 10^{6}\right)^{2} = 4.058 \times 10^{13} \text{ m}^2$$ $$M_E = \frac{9.80 \times 4.058 \times 10^{13}}{6.67 \times 10^{-11}} = \frac{3.977 \times 10^{14}}{6.67 \times 10^{-11}} = 5.96 \times 10^{24} \text{ kg}$$

4. The mean density. $$\rho = \frac{M_E}{\frac{4}{3}\pi R_E^{3}} = \frac{5.96 \times 10^{24}}{\frac{4}{3}\pi \left(6.37 \times 10^{6}\right)^{3}} = \frac{5.96 \times 10^{24}}{1.083 \times 10^{21}} = 5.51 \times 10^{3} \text{ kg/m}^3$$

5. What that tells you about the inside. Surface rock has a density near $2700$ kg/m$^3$, only half of this. So the interior must be far denser than the crust — which is exactly how we know there is an iron core down there, without ever having been near it.

Final Answer: $M_E \approx 5.96 \times 10^{24}$ kg, with a mean density of about $5510$ kg/m$^3$, or $5.51$ g/cm$^3$.

Takeaway: A tape measure, a stopwatch and the value of $G$ are enough to weigh the planet you are standing on. This is why Cavendish's experiment mattered so much — before $G$ was known, $gR_E^2$ was all anyone could measure.

Example 18: Weighing the Sun, twice over

Estimate the Sun's mass (a) from the Earth's orbit, of mean radius $1.50 \times 10^{11}$ m and period one year, and (b) from Jupiter's orbit, of mean radius $7.78 \times 10^{11}$ m and period $11.86$ years. Comment on the agreement. Take one year as $3.156 \times 10^{7}$ s and $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. The idea. The Sun's pull supplies the centripetal force for a planet in a circular orbit: $$\frac{GM_Sm}{r^{2}} = \frac{mv^{2}}{r} = \frac{4\pi^{2}mr}{T^{2}} \qquad \Longrightarrow \qquad M_S = \frac{4\pi^{2}r^{3}}{GT^{2}}$$ The planet's mass $m$ has cancelled, which is why any planet will do.

2. (a) From the Earth. $$r^{3} = \left(1.50 \times 10^{11}\right)^{3} = 3.375 \times 10^{33} \text{ m}^3, \qquad T^{2} = \left(3.156 \times 10^{7}\right)^{2} = 9.96 \times 10^{14} \text{ s}^2$$ $$M_S = \frac{4\pi^{2} \times 3.375 \times 10^{33}}{6.67 \times 10^{-11} \times 9.96 \times 10^{14}} = \frac{1.332 \times 10^{35}}{6.643 \times 10^{4}} = 2.01 \times 10^{30} \text{ kg}$$

3. (b) From Jupiter. $$T = 11.86 \times 3.156 \times 10^{7} = 3.743 \times 10^{8} \text{ s}$$ $$r^{3} = \left(7.78 \times 10^{11}\right)^{3} = 4.709 \times 10^{35} \text{ m}^3, \qquad T^{2} = 1.401 \times 10^{17} \text{ s}^2$$ $$M_S = \frac{4\pi^{2} \times 4.709 \times 10^{35}}{6.67 \times 10^{-11} \times 1.401 \times 10^{17}} = \frac{1.859 \times 10^{37}}{9.345 \times 10^{6}} = 1.99 \times 10^{30} \text{ kg}$$

4. The comparison. The two answers differ by less than $1\%$, even though Jupiter's orbit is five times wider and its year twelve times longer. That agreement is Kepler's third law being confirmed and used at the same time.

Final Answer: about $2.0 \times 10^{30}$ kg either way — $2.01 \times 10^{30}$ kg from the Earth and $1.99 \times 10^{30}$ kg from Jupiter.

Takeaway: Every planet gives the same answer for the Sun's mass, and that is the real test of the theory. If different planets gave different answers, the inverse-square law would be wrong.

Example 19: The tower and the mine

Take $g = 9.8$ m/s$^2$ at the Earth's surface and $R_E = 6.37 \times 10^{6}$ m, and treat the Earth as uniform. Find $g$ (a) at the top of a tower $828$ m high and (b) at the bottom of a mine $3.9$ km deep. In part (a), compare the exact inverse-square answer with the binomial shortcut and say how big the shortcut's error is.

Solution:

1. (a) The exact form. $$g_h = g\left(\frac{R_E}{R_E+h}\right)^{2} = g\left(1 + \frac{h}{R_E}\right)^{-2}$$ $$\frac{h}{R_E} = \frac{828}{6.37 \times 10^{6}} = 1.300 \times 10^{-4}$$ $$g_h = 9.8 \times \left(1.00013\right)^{-2} = 9.79745 \text{ m/s}^2$$ so the exact drop is $9.8 - 9.79745 = 2.5472 \times 10^{-3}$ m/s$^2$.

2. The binomial shortcut, valid when $h \ll R_E$: $$g_h \approx g\left(1 - \frac{2h}{R_E}\right) \qquad \Longrightarrow \qquad \text{drop} \approx \frac{2gh}{R_E} = 2.5477 \times 10^{-3} \text{ m/s}^2$$

3. How wrong is the shortcut? $$\text{error} = \frac{2.5477 - 2.5472}{2.5472} = 1.95 \times 10^{-4} = 0.02\%$$ Two parts in ten thousand — utterly negligible. (The general result is that the binomial form overstates the drop by a fraction $\frac{3h}{2R_E}$, which for any tower on Earth is tiny.)

4. (b) The mine, using the depth law. $$g_d = g\left(1 - \frac{d}{R_E}\right) = 9.8\left(1 - \frac{3900}{6.37 \times 10^{6}}\right) = 9.8\left(1 - 6.122 \times 10^{-4}\right)$$ $$g_d = 9.8 - 6.00 \times 10^{-3} = 9.7940 \text{ m/s}^2$$

5. Compare the two losses. Going up $828$ m costs $2.55 \times 10^{-3}$ m/s$^2$; going down $3900$ m costs $6.00 \times 10^{-3}$ m/s$^2$ — about $2.4$ times as much, for $4.7$ times the displacement.

6. Why the ratio is not 4.7. Per metre, altitude costs $\frac{2g}{R_E}$ and depth costs only $\frac{g}{R_E}$ — exactly half as much. So $4.7$ metres of depth is worth only $2.35$ metres of height, and $\frac{4.71}{2} = 2.36$, which is the ratio we found.

Final Answer: (a) $9.79745$ m/s$^2$ at the top of the tower, with the binomial shortcut in error by only $0.02\%$; (b) $9.7940$ m/s$^2$ at the bottom of the mine.

Takeaway: Near the surface, one metre of depth is worth exactly half a metre of height. And the binomial approximation is not a guess — its error is $\frac{3h}{2R_E}$, which you can quote and defend.

Example 20: How much of your weight does the Earth's spin steal?

The Earth turns once relative to the stars in $8.6164 \times 10^{4}$ s. Taking $R_E = 6.37 \times 10^{6}$ m and $g = 9.8$ m/s$^2$ in the absence of rotation, find (a) the reduction in apparent $g$ at the equator, as a percentage, (b) the latitude at which the reduction is exactly half its equatorial value, and (c) how much lighter a 60 kg person is at the equator than at the pole, on this account alone.

Solution:

1. Quote the latitude formula. At latitude $\lambda$, the body moves on a circle of radius $R_E\cos\lambda$, and only the component of the centripetal requirement along the local vertical shows up in a spring balance: $$g_\lambda = g - \omega^{2}R_E\cos^{2}\lambda$$

2. The angular speed. $$\omega = \frac{2\pi}{T} = \frac{2\pi}{8.6164 \times 10^{4}} = 7.292 \times 10^{-5} \text{ rad/s}$$

3. (a) At the equator, $\lambda = 0$, so $\cos^{2}\lambda = 1$: $$\omega^{2}R_E = \left(7.292 \times 10^{-5}\right)^{2} \times 6.37 \times 10^{6} = 5.317 \times 10^{-9} \times 6.37 \times 10^{6} = 3.387 \times 10^{-2} \text{ m/s}^2$$ $$\frac{\omega^{2}R_E}{g} = \frac{0.03387}{9.8} = 3.46 \times 10^{-3} = 0.35\%$$

4. (b) Half of that reduction requires $\cos^{2}\lambda = \frac{1}{2}$, so $$\cos\lambda = \frac{1}{\sqrt{2}} \qquad \Longrightarrow \qquad \lambda = 45°$$ and there $$g_{45} = 9.8 - 0.03387 \times 0.5 = 9.783 \text{ m/s}^2$$

5. (c) The 60 kg person. At the pole the spring balance reads $60 \times 9.8 = 588$ N. At the equator it reads $$60 \times (9.8 - 0.03387) = 60 \times 9.766 = 585.97 \text{ N}$$ a loss of $$60 \times 0.03387 = 2.03 \text{ N}$$

6. A note on the real Earth. The measured difference between pole and equator is about $0.05$ m/s$^2$, larger than the $0.034$ m/s$^2$ found here, because the Earth is also flattened — the equator is about $21$ km further from the centre. Rotation accounts for roughly two-thirds of the observed pole-to-equator difference, and the bulge for the rest.

Final Answer: (a) $0.35\%$; (b) at latitude $45°$, where $g_\lambda = 9.783$ m/s$^2$; (c) about $2.0$ N lighter, which is roughly 200 grams on a kitchen scale.

Takeaway: Rotation robs you of a third of a percent at the equator and nothing at all at the poles, with a $\cos^{2}\lambda$ in between. Note that this is an apparent change: the Earth's actual pull on you has not altered by one newton.

Part 4: Potential Energy, Potential, and Bodies Set Free

A reminder before you start this part. $U$ is a property of a system of masses and is measured in joules. $V$ is a property of a point in space and is measured in joules per kilogram. Everywhere below, the zero of both is at infinity, which is why both come out negative.

Example 21: A right-angled triangle of masses

Masses of 3 kg, 4 kg and 5 kg are placed at the corners of a right-angled triangle whose sides are $3$ m, $4$ m and $5$ m: the 3 kg at the right-angle corner, the 4 kg at the far end of the $4$ m side, and the 5 kg at the far end of the $3$ m side. Find (a) the gravitational potential energy of the system, and (b) the gravitational potential and the gravitational field at the mid-point of the hypotenuse. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. Set coordinates. Put the 3 kg at the origin $A(0, 0)$, the 4 kg at $B(4, 0)$ and the 5 kg at $C(0, 3)$, all in metres. Then $AB = 4$ m, $AC = 3$ m and $BC = 5$ m, which is the hypotenuse.

2. (a) Three particles give three pairs. $$U = -G\left(\frac{m_Am_B}{AB} + \frac{m_Am_C}{AC} + \frac{m_Bm_C}{BC}\right)$$ $$U = -G\left(\frac{3 \times 4}{4} + \frac{3 \times 5}{3} + \frac{4 \times 5}{5}\right) = -G\left(3 + 5 + 4\right) = -12G$$ $$U = -12 \times 6.67 \times 10^{-11} = -8.00 \times 10^{-10} \text{ J}$$

3. (b) The mid-point of the hypotenuse is $M(2,\ 1.5)$. Now a small piece of geometry pays off: in a right-angled triangle the mid-point of the hypotenuse is equidistant from all three vertices. Check it: $$MA = \sqrt{2^{2} + 1.5^{2}} = 2.5 \text{ m}, \quad MB = \sqrt{2^{2} + 1.5^{2}} = 2.5 \text{ m}, \quad MC = \sqrt{2^{2} + 1.5^{2}} = 2.5 \text{ m}$$

4. The potential is then a one-line scalar sum. $$V_M = -\frac{G}{2.5}\left(3 + 4 + 5\right) = -\frac{12G}{2.5} = -4.8G = -3.20 \times 10^{-10} \text{ J/kg}$$

5. The field is not a one-liner — it needs components. Each mass pulls towards itself with $\frac{Gm}{2.5^{2}} = \frac{Gm}{6.25}$. The unit vectors from $M$ are $$\hat{u}_{MA} = (-0.8,\ -0.6), \qquad \hat{u}_{MB} = (+0.8,\ -0.6), \qquad \hat{u}_{MC} = (-0.8,\ +0.6)$$ $$g_x = \frac{G}{6.25}\left[3(-0.8) + 4(+0.8) + 5(-0.8)\right] = \frac{G}{6.25}(-3.2) = -0.512G$$ $$g_y = \frac{G}{6.25}\left[3(-0.6) + 4(-0.6) + 5(+0.6)\right] = \frac{G}{6.25}(-1.2) = -0.192G$$

6. Magnitude and direction. $$\lvert \vec{g} \rvert = G\sqrt{0.512^{2} + 0.192^{2}} = 0.547G = 3.65 \times 10^{-11} \text{ m/s}^2$$ $$\theta = \tan^{-1}\left(\frac{-0.192}{-0.512}\right) \text{ in the third quadrant } = 180° + 20.6° = 200.6°$$ that is, pointing down and to the left, $20.6°$ below the $-x$ direction.

Final Answer: (a) $U = -8.00 \times 10^{-10}$ J; (b) $V_M = -3.20 \times 10^{-10}$ J/kg and $\lvert \vec{g} \rvert = 3.65 \times 10^{-11}$ m/s$^2$ directed $20.6°$ below the $-x$ axis.

Takeaway: One picture, three questions, three different amounts of work. $U$ needed three pairs, $V$ needed one division, and $\vec{g}$ needed six components. Learn to spot which one is actually being asked.

Example 22: Two particles released in deep space

Two particles of masses $1.0$ kg and $2.0$ kg are held $1.0$ m apart in deep space, far from everything else, and released from rest. Find their speeds when their separation has fallen to $0.50$ m. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. Two conservation laws, because there is no external force. $$\text{momentum:}\quad m_1v_1 = m_2v_2 \qquad\qquad \text{energy:}\quad K_f + U_f = K_i + U_i$$

2. Momentum first. They start at rest, so the total momentum is zero and stays zero. With $m_1 = 1$ and $m_2 = 2$, $$1 \times v_1 = 2 \times v_2 \qquad \Longrightarrow \qquad v_1 = 2v_2$$ The light one moves twice as fast, as expected.

3. The energy released. $$U_i = -\frac{Gm_1m_2}{1.0} = -1.334 \times 10^{-10} \text{ J}, \qquad U_f = -\frac{Gm_1m_2}{0.50} = -2.668 \times 10^{-10} \text{ J}$$ $$K_f = U_i - U_f = -1.334 \times 10^{-10} + 2.668 \times 10^{-10} = 1.334 \times 10^{-10} \text{ J}$$

4. Share that kinetic energy out. $$K_f = \frac{1}{2}(1)(2v_2)^{2} + \frac{1}{2}(2)v_2^{2} = 2v_2^{2} + v_2^{2} = 3v_2^{2}$$ $$3v_2^{2} = 1.334 \times 10^{-10} \qquad \Longrightarrow \qquad v_2^{2} = 4.447 \times 10^{-11}$$ $$v_2 = 6.67 \times 10^{-6} \text{ m/s}, \qquad v_1 = 1.33 \times 10^{-5} \text{ m/s}$$

5. The speed of approach. They close on each other at $v_1 + v_2 = 2.00 \times 10^{-5}$ m/s — about $20$ micrometres per second. Gravity between kilogram masses really is that feeble.

Final Answer: the 1 kg mass moves at $1.33 \times 10^{-5}$ m/s and the 2 kg mass at $6.67 \times 10^{-6}$ m/s, closing at $2.00 \times 10^{-5}$ m/s.

Takeaway: In deep space, momentum conservation is not optional. Writing $\frac{1}{2}mv^{2} = \Delta U$ for one particle alone is the standard mistake here — both bodies move, and both carry kinetic energy.

Example 23: The field at the centre of a hemispherical bowl

A uniform hemispherical shell — a thin bowl — has mass $M$ and radius $R$. Find the direction and magnitude of the gravitational field at $O$, the centre of its circular rim, and also the gravitational potential there. Then evaluate both for $M = 100$ kg and $R = 0.40$ m, with $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. The direction, by symmetry alone. Slice the bowl into thin rings, each centred on the axis of symmetry. Every ring pulls $O$ towards itself, and by symmetry the sideways contributions from opposite sides of a ring cancel exactly. Only the component along the axis survives, and it points from $O$ into the bowl. So the field at $O$ is directed straight down the axis, towards the material.

2. The magnitude, by adding up the rings. Take the ring at polar angle $\theta$ from the axis. Every point of it is exactly $R$ from $O$, so it contributes $\frac{G\,dm}{R^{2}}$, of which the surviving axial part is $\frac{G\,dm}{R^{2}}\cos\theta$.

3. Write $dm$ for that ring. With surface density $\sigma = \frac{M}{2\pi R^{2}}$, a ring of angular width $d\theta$ has circumference $2\pi R\sin\theta$ and width $R\,d\theta$, so $$dm = \sigma\,2\pi R^{2}\sin\theta\,d\theta = M\sin\theta\,d\theta$$

4. Integrate over the hemisphere, $\theta$ from $0$ to $\frac{\pi}{2}$: $$g_O = \int_0^{\pi/2}\frac{GM\sin\theta\cos\theta}{R^{2}}\,d\theta = \frac{GM}{R^{2}}\left[\frac{\sin^{2}\theta}{2}\right]_0^{\pi/2} = \frac{GM}{2R^{2}}$$

5. The potential is far easier. Every element of the bowl is exactly $R$ from $O$, and potential is a scalar, so $$V_O = -\frac{GM}{R}$$ which is the same as for a complete sphere of mass $M$ and radius $R$.

6. The numbers. $$g_O = \frac{6.67 \times 10^{-11} \times 100}{2 \times \left(0.40\right)^{2}} = \frac{6.67 \times 10^{-9}}{0.32} = 2.08 \times 10^{-8} \text{ m/s}^2$$ $$V_O = -\frac{6.67 \times 10^{-11} \times 100}{0.40} = -1.67 \times 10^{-8} \text{ J/kg}$$

Final Answer: $\vec{g}$ at $O$ has magnitude $\frac{GM}{2R^{2}} = 2.08 \times 10^{-8}$ m/s$^2$ directed along the axis into the bowl, while $V_O = -\frac{GM}{R} = -1.67 \times 10^{-8}$ J/kg.

Takeaway: Half a sphere gives the full sphere's potential but only half the "point-mass" field, and the field is no longer zero. Closing the bowl into a complete shell would drive the field to zero while leaving $V$ untouched — a clean demonstration that $V$ knows nothing about direction.

Example 24: Three spheres in a row, and one of them removed

Three small spheres, each of mass $5.0$ kg, lie on a straight line at $x = 0$, $x = 1.0$ m and $x = 2.0$ m. Find (a) the gravitational potential energy of the system, (b) the work an external agent must do to take the middle sphere away to infinity while the other two stay put, and (c) the work needed to scatter all three to infinity. Take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. (a) Three particles, three pairs. Two pairs are $1.0$ m apart and one pair is $2.0$ m apart: $$U_i = -G\left(\frac{25}{1.0} + \frac{25}{1.0} + \frac{25}{2.0}\right) = -G\left(25 + 25 + 12.5\right) = -62.5G$$ $$U_i = -62.5 \times 6.67 \times 10^{-11} = -4.17 \times 10^{-9} \text{ J}$$

2. (b) Remove the middle one. Two of the three pairs disappear; the outer pair, $2.0$ m apart, survives: $$U_f = -\frac{25G}{2.0} = -12.5G = -8.34 \times 10^{-10} \text{ J}$$

3. The work is the change in energy. $$W = U_f - U_i = -12.5G - (-62.5G) = 50G = 3.34 \times 10^{-9} \text{ J}$$

4. (c) Scatter all three. Now the final state has no pairs left at all, so $U_f = 0$: $$W_{all} = 0 - U_i = 62.5G = 4.17 \times 10^{-9} \text{ J}$$

5. A quick sanity comparison. Taking away one sphere costs $80\%$ of the price of taking away all three. That is because the middle sphere is involved in two of the three bonds, and in the two tightest ones.

Final Answer: (a) $-4.17 \times 10^{-9}$ J; (b) $3.34 \times 10^{-9}$ J; (c) $4.17 \times 10^{-9}$ J.

Takeaway: The work needed to dismantle a system is $+\lvert U \rvert$ only if you dismantle it completely. Remove one particle and you must recompute $U$ for what is left — never assume the whole thing goes to zero.

Example 25: A rocket fired straight up at 5 km/s

A rocket is fired vertically from the Earth's surface with a speed of $5.0$ km/s. Neglecting air resistance and the Earth's rotation, how far from the Earth's centre does it get before falling back? Take the Earth's mass as $6.0 \times 10^{24}$ kg, its radius as $6.4 \times 10^{6}$ m and $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. Note first that $5$ km/s is well under the escape speed, which for this data is $\sqrt{\frac{2GM_E}{R_E}} = 11.2$ km/s. So the rocket must come back, and it has a definite highest point.

2. Energy conservation, per kilogram, with the exact potential energy. At the highest point the speed is zero: $$\frac{1}{2}v^{2} - \frac{GM_E}{R_E} = 0 - \frac{GM_E}{r_{max}}$$

3. Evaluate the two terms on the left. $$GM_E = 6.67 \times 10^{-11} \times 6.0 \times 10^{24} = 4.002 \times 10^{14} \text{ m}^3\text{/s}^2$$ $$\frac{1}{2}v^{2} = \frac{1}{2}\left(5000\right)^{2} = 1.250 \times 10^{7} \text{ J/kg}$$ $$\frac{GM_E}{R_E} = \frac{4.002 \times 10^{14}}{6.4 \times 10^{6}} = 6.253 \times 10^{7} \text{ J/kg}$$

4. So the total energy per kilogram is $$E = 1.250 \times 10^{7} - 6.253 \times 10^{7} = -5.003 \times 10^{7} \text{ J/kg}$$ Negative, confirming that the rocket is bound.

5. Find $r_{max}$. $$-\frac{GM_E}{r_{max}} = -5.003 \times 10^{7} \qquad \Longrightarrow \qquad r_{max} = \frac{4.002 \times 10^{14}}{5.003 \times 10^{7}} = 8.0 \times 10^{6} \text{ m}$$

6. The height above the surface. $$h = 8.0 \times 10^{6} - 6.4 \times 10^{6} = 1.6 \times 10^{6} \text{ m} = 1600 \text{ km}$$

7. What a constant-$g$ estimate would have said. With $g = \frac{GM_E}{R_E^2} = 9.77$ m/s$^2$ held fixed, $h = \frac{v^2}{2g} = 1280$ km — a $20\%$ underestimate, because $g$ has already fallen to $6.3$ m/s$^2$ by the top of the climb.

Final Answer: the rocket reaches $8.0 \times 10^{6}$ m from the Earth's centre, that is $1600$ km above the surface.

Takeaway: Once the climb is a sizeable fraction of $R_E$, $\frac{v^2}{2g}$ is no longer good enough. Write energy conservation with $-\frac{GMm}{r}$ and let the algebra hand you $r_{max}$ directly.

Example 26: Two stars falling together

Two stars, each of one solar mass ($2 \times 10^{30}$ kg) and radius $10^{4}$ km, are $10^{9}$ km apart and approaching each other head-on with negligible speed. With what speed do they collide? Assume they stay undistorted until they touch, and take $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

Solution:

1. Convert to metres. Initial separation $d_i = 10^{9}$ km $= 1.0 \times 10^{12}$ m. Each radius is $10^{4}$ km $= 1.0 \times 10^{7}$ m, so at the moment of contact their centres are $$d_f = 2 \times 1.0 \times 10^{7} = 2.0 \times 10^{7} \text{ m}$$ apart. Note that the collision separation is centre-to-centre, not surface-to-surface.

2. Symmetry does the momentum bookkeeping. Equal masses starting at rest means each ends with the same speed $v$, in opposite directions.

3. Energy conservation for the pair. $$0 - \frac{GM^{2}}{d_i} = 2\left(\frac{1}{2}Mv^{2}\right) - \frac{GM^{2}}{d_f}$$ $$Mv^{2} = GM^{2}\left(\frac{1}{d_f} - \frac{1}{d_i}\right)$$

4. Substitute. $$GM^{2} = 6.67 \times 10^{-11} \times \left(2 \times 10^{30}\right)^{2} = 2.668 \times 10^{50}$$ $$\frac{1}{d_f} - \frac{1}{d_i} = \frac{1}{2.0 \times 10^{7}} - \frac{1}{1.0 \times 10^{12}} = 5.0 \times 10^{-8} - 1.0 \times 10^{-12} \approx 5.0 \times 10^{-8}$$ The starting separation contributes almost nothing — the stars might as well have begun infinitely far apart.

5. Solve for $v$. $$Mv^{2} = 2.668 \times 10^{50} \times 5.0 \times 10^{-8} = 1.334 \times 10^{43}$$ $$v^{2} = \frac{1.334 \times 10^{43}}{2 \times 10^{30}} = 6.67 \times 10^{12} \qquad \Longrightarrow \qquad v = 2.58 \times 10^{6} \text{ m/s}$$

Final Answer: each star is moving at about $2.6 \times 10^{6}$ m/s when they touch, so they close on each other at some $5.2 \times 10^{6}$ m/s.

Takeaway: When the starting separation is enormous compared with the finishing one, the $\frac{1}{d_i}$ term is dead weight — but write it down anyway. Checking that it is negligible is a legitimate step; assuming it is, is not.