The Workbook

The potential at a point is the work done per unit mass in bringing a small mass slowly from infinity to that point. The work is that of an external agent. Unit: J/kg. At the surface \(V = -gR = -10 \times 6.4 \times 10^{6} = -6.4 \times 10^{7}\) J/kg.

\(U = -\dfrac{Gm_1m_2}{r} = -\dfrac{6.67 \times 10^{-11} \times 6}{1} = -4 \times 10^{-10}\) J. To separate them to infinity (\(U = 0\)) an agent must supply \(+4 \times 10^{-10}\) J.

\(g = -\dfrac{dV}{dr}\). With \(V = -\dfrac{GM}{r}\): \(\dfrac{dV}{dr} = \dfrac{GM}{r^2}\), so \(g = -\dfrac{GM}{r^2}\): magnitude \(\dfrac{GM}{r^2}\), directed towards decreasing \(r\), that is towards the mass.

1. Outside, \(V = -\dfrac{GM}{r}\): \(V_{2R} = -\dfrac{GM}{2R}\), half the surface value in size.

2. At \(r = 0\): \(V_0 = -\dfrac{GM \times 3R^2}{2R^3} = -\dfrac{3GM}{2R}\).

3. The slope of \(V\) is \(-g\), and \(g\) is continuous at the surface (both formulas give \(\dfrac{GM}{R^2}\)). Equal slopes on both sides make the curve smooth there.

1. Inside, \(V\) equals the surface value: \(-\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 100}{1} = -6.7 \times 10^{-9}\) J/kg.

2. \(V = -\dfrac{GM}{2} = -3.3 \times 10^{-9}\) J/kg.

The field inside a shell is zero, so no work is done in moving a mass between any two inside points. The potential therefore cannot change inside; it stays at the value on the shell, \(-\dfrac{GM}{R}\).

\(V = -\dfrac{GM\left(3R^2 - R^2/4\right)}{2R^3} = -\dfrac{GM}{R}\cdot\dfrac{11/4}{2} = -\dfrac{11GM}{8R}\).

\(W = m(V_0 - V_s) = 1 \times \left(-\dfrac{3}{2}gR + gR\right) = -\dfrac{gR}{2} = -3.2 \times 10^{7}\) J. The work is negative: the potential falls towards the centre, so gravity pulls the mass inward and the agent must hold it back.

\(W = m(V_{2R} - V_R) = 1 \times \left(-\dfrac{gR}{2} + gR\right) = \dfrac{gR}{2} = 3.2 \times 10^{7}\) J. \(mgR = 6.4 \times 10^{7}\) J would be right only if \(g\) stayed \(10\) m/s\(^2\) all the way up; in fact it falls to \(\dfrac{g}{4}\) at the top.

1. The two pulls are equal and opposite, so the field at \(P\) is zero.

2. Potentials add as scalars: \(V = -2 \times \dfrac{G \times 10}{2} = -10G = -6.7 \times 10^{-10}\) J/kg.

3. \(W = mV = -6.7 \times 10^{-10}\) J. It is negative although the field at \(P\) is zero: work is decided by the potential, not by the field at the final point.

\(V = -4 \times \dfrac{G \times 1}{1} = -4G = -2.7 \times 10^{-10}\) J/kg. The field at the centre is zero by symmetry, but the potential is not.

Inside, \(V(r) = -\dfrac{GM(3R^2 - r^2)}{2R^3}\). At \(r = 0\): \(V_0 = -\dfrac{3GM}{2R}\). At \(r = R\): \(V_s = -\dfrac{GM \times 2R^2}{2R^3} = -\dfrac{GM}{R}\). So \(V_0 = \dfrac{3}{2}V_s\).

The self-energy is the work done by an external agent in assembling the body from matter spread out at infinity. Build the shell layer by layer. When it has mass \(m\), its potential is \(-\dfrac{Gm}{R}\), so bringing the next \(dm\) takes \(dW = -\dfrac{Gm\,dm}{R}\). \(U = \int_0^M -\dfrac{Gm\,dm}{R} = -\dfrac{GM^2}{2R}\).

Solid sphere: \(U = -\dfrac{3GM^2}{5R}\). Ratio \(\dfrac{U_{\text{shell}}}{U_{\text{sphere}}} = \dfrac{1/2}{3/5} = \dfrac{5}{6}\). The sphere is more tightly bound because its mass sits closer together.

1. The potential at the shell is \(-\dfrac{Gm}{R}\), so \(dW = V\,dm = -\dfrac{Gm\,dm}{R}\).

2. \(U = -\dfrac{G}{R}\int_0^M m\,dm = -\dfrac{G}{R}\cdot\dfrac{M^2}{2} = -\dfrac{GM^2}{2R}\).

\(U = -\dfrac{3GM^2}{5R} = -\dfrac{3 \times 6.67 \times 10^{-11} \times 36 \times 10^{48}}{5 \times 6.4 \times 10^{6}} \approx -2.3 \times 10^{32}\) J.

To scatter the matter to infinity the self-energy must be raised from \(-2.3 \times 10^{32}\) J to zero, so about \(2.3 \times 10^{32}\) J must be supplied. This is the Earth's gravitational binding energy.

As the pieces are brought together, gravity pulls them in and does positive work, so the agent does negative work. The assembled body has less energy than the scattered matter. A more negative self-energy means the body is more tightly bound and needs more energy to pull apart.

1. \(V_s - V_{2R} = -gR + \dfrac{gR}{2} = -\dfrac{gR}{2} = -3.2 \times 10^{7}\) J/kg.

2. \(\dfrac{1}{2}v^2 = 3.2 \times 10^{7}\), so \(v^2 = 6.4 \times 10^{7} = gR\) and \(v = 8000\) m/s \(= 8\) km/s.

Energy: \(K = U_1 - U_2 = -\dfrac{G}{1} + \dfrac{G}{0.5} = G = 6.7 \times 10^{-11}\) J in total. Momentum is conserved and the masses are equal, so each has \(\dfrac{G}{2}\): \(\dfrac{1}{2}v^2 = 3.3 \times 10^{-11}\), \(v \approx 8.2 \times 10^{-6}\) m/s.

\(V_\infty - V_s = 0 - (-gR) = gR = 1.6 \times 1.7 \times 10^{6} \approx 2.7 \times 10^{6}\) J/kg.

It must supply \(\dfrac{gR}{2}\) per kilogram: \(\dfrac{1}{2}v^2 = \dfrac{gR}{2}\), so \(v = \sqrt{gR} = 8\) km/s. Going up from the surface to \(R\) is the time reverse of falling from \(R\) to the surface. The same energy change gives the same speed.

\(W = m\,\Delta V = 1000 \times \dfrac{gR}{2} = 1000 \times 3.2 \times 10^{7} = 3.2 \times 10^{10}\) J.

\(W = m\left[V(R + h) - V(R)\right] = GMm\left(\dfrac{1}{R} - \dfrac{1}{R + h}\right) = \dfrac{GMmh}{R(R + h)}\). With \(GM = gR^2\): \(W = \dfrac{mghR}{R + h}\). For \(h \ll R\), \(R + h \approx R\) and \(W \approx mgh\).

Gravity is the only force and it must supply the centripetal force: \(\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}\). So \(v^2 = \dfrac{GM}{r}\) and \(v = \sqrt{\dfrac{GM}{r}}\).

1. \(g = \dfrac{GM}{R^2}\) gives \(GM = gR^2\), so \(v = \sqrt{\dfrac{gR^2}{R}} = \sqrt{gR} = \sqrt{10 \times 6.4 \times 10^{6}} = 8000\) m/s \(= 8\) km/s.

2. \(T = \dfrac{2\pi R}{v} = \dfrac{2\pi \times 6.4 \times 10^{6}}{8000} \approx 5000\) s, about \(84\) min.

\(r = 2R\): \(v = \sqrt{\dfrac{gR^2}{2R}} = \sqrt{\dfrac{gR}{2}} = \dfrac{8}{\sqrt2} = 4\sqrt2 \approx 5.7\) km/s. \(T = \dfrac{2\pi \times 2R}{v} = \dfrac{4\pi \times 6.4 \times 10^{6}}{5657} \approx 1.4 \times 10^{4}\) s, about \(4\) h. Check: \(T \propto r^{3/2}\), and \(2^{3/2} \times 84\) min \(\approx 238\) min.

No. The satellite's mass appears on both sides of \(\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}\) and cancels; the speed is fixed by \(M\) and \(r\) alone.

It falls towards the Earth continuously, but it also moves sideways very fast. In the time it falls a little, the ground has curved away by the same amount. The fall never closes the gap, so the path stays a circle.

\(v \propto \dfrac{1}{\sqrt r}\). Four times the radius gives half the speed.

\(T = \dfrac{2\pi r}{v} = 2\pi r\sqrt{\dfrac{r}{GM}} = 2\pi\sqrt{\dfrac{r^3}{GM}}\). Squaring: \(T^2 = \dfrac{4\pi^2}{GM}r^3\), so \(T^2 \propto r^3\), which is Kepler's third law.

1. \(T \propto r^{3/2}\): \(\dfrac{T_B}{T_A} = 4^{3/2} = 8\).

2. \(v \propto r^{-1/2}\): \(\dfrac{v_B}{v_A} = \dfrac{1}{\sqrt4} = \dfrac{1}{2}\).

\(\left(\dfrac{r}{R}\right)^{3/2} = 8\), so \(\dfrac{r}{R} = 8^{2/3} = 4\) and \(r = 4R\), a height of \(3R\).

\(M = \dfrac{4\pi^2r^3}{GT^2}\). \(T = 27.3 \times 86400 = 2.36 \times 10^{6}\) s. \(M = \dfrac{39.5 \times (3.84 \times 10^{8})^3}{6.67 \times 10^{-11} \times (2.36 \times 10^{6})^2} \approx 6 \times 10^{24}\) kg.

\(\omega = \dfrac{v}{R} = \dfrac{\sqrt{gR}}{R} = \sqrt{\dfrac{g}{R}} = \sqrt{\dfrac{10}{6.4 \times 10^{6}}} = 1.25 \times 10^{-3}\) rad/s.

Both have the same speed \(\sqrt{GM/r}\) and the same period \(2\pi\sqrt{r^3/GM}\), because the satellite's mass cancels in the force equation. Only their energies and momenta differ.

Its period must equal the Earth's rotation period, \(24\) h. It must orbit in the equatorial plane, in the same sense as the Earth turns, west to east. The orbit must be circular so that its angular speed is uniform.

1. \(r^3 = \dfrac{GMT^2}{4\pi^2} = \dfrac{4 \times 10^{14} \times (8.64 \times 10^{4})^2}{40} = \dfrac{4 \times 10^{14} \times 7.46 \times 10^{9}}{40} \approx 7.5 \times 10^{22}\) m\(^3\). \(r \approx 4.2 \times 10^{7}\) m, about \(6.6R\).

2. \(h = r - R \approx 42000 - 6400 \approx 36000\) km.

\(v = \dfrac{2\pi r}{T} = \dfrac{2\pi \times 4.2 \times 10^{7}}{86400} \approx 3.1 \times 10^{3}\) m/s \(\approx 3.1\) km/s.

The centre of every orbit is the centre of the Earth, so the orbit plane passes through the centre. A circle above latitude \(28^\circ\) would be centred on the axis, not on the Earth's centre, and gravity cannot hold a satellite on it. Only the equatorial plane both contains the centre and turns with the ground.

Three satellites spaced \(120^\circ\) apart around the geostationary ring can see almost the whole surface between them. The regions near the poles are left out: from there a geostationary satellite sits below the horizon.

1. \(r = 7200\) km \(= 7.2 \times 10^{6}\) m. \(T = 2\pi\sqrt{\dfrac{r^3}{GM}} = 2\pi\sqrt{\dfrac{3.73 \times 10^{20}}{4 \times 10^{14}}} = 2\pi \times 966 \approx 6100\) s, about \(100\) min.

2. The orbit plane stays fixed in space while the Earth turns beneath it. On successive passes the satellite flies over strips of ground further and further west, so it sweeps the whole globe.

\(g' = \dfrac{g}{4} = 2.5\) m/s\(^2\). In a circular orbit gravity is the only force, so the centripetal acceleration equals \(g'\): \(2.5\) m/s\(^2\).

The station and everyone in it fall freely with the same acceleration \(g'\). The floor does not push on the astronauts, so there is no normal force. Weightlessness is the absence of that push, not the absence of gravity.

1. At the orbit gravity per unit mass is \(g'\), and it must equal the centripetal acceleration: \(g' = \dfrac{v^2}{r}\), so \(v = \sqrt{g'r}\).

2. \(v = \sqrt{1.6 \times 1.6 \times 10^{7}} = \sqrt{2.56 \times 10^{7}} \approx 5.1 \times 10^{3}\) m/s.

\(v = \sqrt{gR} = \sqrt{1.6 \times 1.7 \times 10^{6}} = \sqrt{2.72 \times 10^{6}} \approx 1650\) m/s \(\approx 1.65\) km/s.

Same orbit, same speed. \(K = \dfrac{1}{2}mv^2 \propto m\): ratio \(1 : 2\). The period does not depend on mass: ratio \(1 : 1\).

\(T = 2\pi\sqrt{\dfrac{r^3}{GM}}\) with \(GM = gR^2\) gives \(T = 2\pi\sqrt{\dfrac{r^3}{gR^2}}\). For \(r = R\): \(T = 2\pi\sqrt{\dfrac{R^3}{gR^2}} = 2\pi\sqrt{\dfrac{R}{g}}\), the period found in Q2.

\(v^2 = \dfrac{GM}{r}\), so \(K = \dfrac{1}{2}mv^2 = \dfrac{GMm}{2r}\). \(U = -\dfrac{GMm}{r}\). \(E = K + U = -\dfrac{GMm}{2r}\). Comparing: \(E = -K\) and \(E = \dfrac{U}{2}\).

1. \(\dfrac{GMm}{r} = \dfrac{mgR^2}{2R} = \dfrac{mgR}{2} = 1000 \times 3.2 \times 10^{7} = 3.2 \times 10^{10}\) J. So \(U = -3.2 \times 10^{10}\) J, \(K = 1.6 \times 10^{10}\) J and \(E = -1.6 \times 10^{10}\) J.

2. \(K\) is the only positive curve. \(U\) is the lowest, twice as far below the axis as \(E\), and \(E\) lies exactly midway between \(U\) and the axis. All three shrink towards zero as \(r\) grows.

\(E < 0\) means the kinetic energy is less than the energy needed to climb to infinity, where \(U = 0\). The satellite is bound; it cannot get away. A positive total energy would mean it reaches infinity with kinetic energy to spare: it escapes.

\(r = R\): \(K = \dfrac{mgR}{2} = 3.2 \times 10^{10}\) J and \(E = -K = -3.2 \times 10^{10}\) J.

The binding energy is the least energy that must be supplied to carry the satellite to infinity and leave it at rest there: \(B = 0 - E = \dfrac{GMm}{2r}\). For Q2: \(1.6 \times 10^{10}\) J.

On the surface at rest: \(E_1 = -\dfrac{GMm}{R} = -mgR = -6.4 \times 10^{10}\) J. In orbit: \(E_2 = -1.6 \times 10^{10}\) J. Energy needed \(= E_2 - E_1 = 4.8 \times 10^{10}\) J, which is \(\dfrac{3}{4}mgR\).

\(K = \dfrac{GMm}{2r}\) halves, so it decreases. \(U = -\dfrac{GMm}{r}\) goes from \(-\dfrac{GMm}{r}\) to \(-\dfrac{GMm}{2r}\): it increases. \(E = -\dfrac{GMm}{2r}\) goes to \(-\dfrac{GMm}{4r}\): it increases. Work \(= \Delta E = \dfrac{GMm}{2r} - \dfrac{GMm}{4r} = \dfrac{GMm}{4r}\).

1. At \(R\): \(E = -\dfrac{mgR}{2} = -3.2 \times 10^{10}\) J. At \(2R\): \(E = -\dfrac{mgR}{4} = -1.6 \times 10^{10}\) J.

2. \(\Delta E = -1.6 \times 10^{10} + 3.2 \times 10^{10} = 1.6 \times 10^{10}\) J.

The burn adds kinetic energy, so the total energy rises and the satellite climbs to a larger orbit. While climbing, gravity does negative work and takes back more kinetic energy than the burn gave. In the new orbit \(v = \sqrt{GM/r}\) is smaller, though \(E\) is larger.

Drag lowers \(E\), so \(E = -\dfrac{GMm}{2r}\) becomes more negative and \(r\) shrinks. Since \(K = -E\), the kinetic energy grows as \(E\) falls. Gravity does positive work on the way in, more than drag removes, and the satellite speeds up.

\(B \propto \dfrac{1}{r}\): ratio \(\dfrac{B_R}{B_{2R}} = 2 : 1\). \(v \propto \dfrac{1}{\sqrt r}\): ratio \(\sqrt2 : 1\).

To just reach infinity with zero speed, the total energy must be zero: \(\dfrac{1}{2}mv_e^2 - \dfrac{GMm}{R} = 0\), so \(v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR} = \sqrt{1.28 \times 10^{8}} \approx 11.3 \times 10^{3}\) m/s.

1. Very far away \(U \to 0\), and the total energy is zero, so the kinetic energy and the speed also go to zero. \(E = 0\) throughout.

2. \(v_e^2 = 2gR\) and \(v_o^2 = gR\), so \(v_e^2 = 2v_o^2\) and \(v_e = \sqrt2\,v_o\).

\(v_e = \sqrt{2gR} = \sqrt{2 \times 1.6 \times 1.7 \times 10^{6}} = \sqrt{5.44 \times 10^{6}} \approx 2.3\) km/s.

\(v_e = \sqrt{\dfrac{2GM}{R}} \propto \sqrt{\dfrac{M}{R}}\). Doubling both leaves the ratio unchanged, so \(v_e = 11.3\) km/s, the same as for the Earth.

\(E = \dfrac{1}{2}m(2v_o)^2 - \dfrac{GMm}{R} = 2mv_o^2 - mv_o^2 = mv_o^2 > 0\), so it escapes. Far away all of \(E\) is kinetic: \(\dfrac{1}{2}mv_\infty^2 = mv_o^2\), so \(v_\infty = \sqrt2\,v_o \approx 11.3\) km/s.

1. Per unit mass: \(\dfrac{1}{2}\left(\dfrac{v_e}{2}\right)^2 - \dfrac{GM}{R} = 0 - \dfrac{GM}{R + h}\). With \(v_e^2 = 2gR\) and \(GM = gR^2\): \(\dfrac{gR}{4} - gR = -\dfrac{gR^2}{R + h}\).

2. \(-\dfrac{3gR}{4} = -\dfrac{gR^2}{R + h}\), so \(R + h = \dfrac{4R}{3}\) and \(h = \dfrac{R}{3}\).

No. The escape condition is \(E \ge 0\), and both \(\dfrac{1}{2}mv^2\) and \(U\) are scalars that do not depend on direction. Any direction that does not hit the ground works. Rotation and air would change the practical answer, which is why rockets launch eastward.

The rms speed is only an average; a good fraction of the molecules move much faster. On the Moon, \(1.9\) km/s is close to \(v_e = 2.3\) km/s, so the fast tail keeps escaping and over ages the gas is lost. On the Earth, oxygen at about \(0.5\) km/s is far below \(11.3\) km/s, so almost none escapes.

\(v_{\text{rms}} = \sqrt{\dfrac{3 \times 8.3 \times 300}{0.032}} = \sqrt{2.33 \times 10^{5}} \approx 480\) m/s \(\approx 0.48\) km/s, about \(\dfrac{1}{24}\) of the escape speed. The Earth keeps its oxygen.

\(T = \dfrac{Mv^2}{3R} = \dfrac{0.002 \times (1.13 \times 10^{4})^2}{3 \times 8.3} = \dfrac{0.002 \times 1.28 \times 10^{8}}{24.9} \approx 1.0 \times 10^{4}\) K.

\(E < 0\): the body is bound and moves in a closed orbit (circle or ellipse); it never reaches infinity. \(E = 0\): it just escapes, arriving at infinity with zero speed. \(E > 0\): it escapes and still has kinetic energy \(E\) when far away.

It must be given its binding energy, \(-E = \dfrac{GMm}{2r} = 1.6 \times 10^{10}\) J. Its total energy then becomes zero and it can just reach infinity.

The satellite already moves at \(v_o\). To escape from radius \(R\) it needs \(v_e = \sqrt2\,v_o\) in the same direction, so the extra speed is \(\sqrt2\,v_o - v_o = (\sqrt2 - 1)v_o \approx 0.414 \times 8 \approx 3.3\) km/s.

First: each planet moves on an ellipse with the Sun at one focus. Second: the line from the Sun to the planet sweeps equal areas in equal times. Third: the square of the period is proportional to the cube of the semi-major axis, \(T^2 \propto a^3\).

1. Angular momentum about the Sun. The Sun's pull is along the line to the Sun, so it has no torque about the Sun and cannot change \(L\).

2. At \(P\) and \(A\) the velocity is perpendicular to the radius: \(mv_Pr_P = mv_Ar_A\), so \(v_A = v_P\dfrac{r_P}{r_A} = \dfrac{30}{3} = 10\) km/s.

\(\dfrac{dA}{dt} = \dfrac{L}{2m} = \dfrac{2.7 \times 10^{40}}{2 \times 6 \times 10^{24}} = 2.25 \times 10^{15}\) m\(^2\)/s.

In time \(dt\) the radius vector turns through a small angle and sweeps a thin triangle of area \(dA = \dfrac{1}{2}r\,(v_\perp dt)\). Here \(v_\perp\) is the velocity component perpendicular to \(\vec r\). So \(\dfrac{dA}{dt} = \dfrac{1}{2}rv_\perp = \dfrac{L}{2m}\), since \(L = mrv_\perp\). Gravity is central, so \(L\) is constant and so is the areal velocity: Kepler's second law.

1. \(T = a^{3/2}\) in years when \(a\) is in AU: \(4^{3/2} = 8\) yr.

2. \(9^{3/2} = 27\) yr.

\(a = T^{2/3} = 27^{2/3} = 9\) AU. \(r_P + r_A = 2a = 18\) AU, so \(r_A = 17\) AU.

\(E = -\dfrac{GMm}{2a}\). Energy conservation gives \(\dfrac{1}{2}mv^2 - \dfrac{GMm}{r} = -\dfrac{GMm}{2a}\), so \(v^2 = GM\left(\dfrac{2}{r} - \dfrac{1}{a}\right)\). At \(r = a\): \(v^2 = \dfrac{GM}{a}\), exactly the circular-orbit speed at radius \(a\).

At \(P\) and \(A\) the velocity is perpendicular to \(\vec r\), so \(L = mv_Pr_P = mv_Ar_A\) and \(\dfrac{v_P}{v_A} = \dfrac{r_A}{r_P}\). Kinetic energy is largest at \(P\) (largest speed). Potential energy \(-\dfrac{GMm}{r}\) is largest, that is least negative, at \(A\) where \(r\) is largest.

\(a = \dfrac{0.6 + 35}{2} = 17.8\) AU. \(T = a^{3/2} = 17.8^{1.5} \approx 75\) yr, which matches its observed return every \(75\) to \(76\) years.

Angular momentum: \(L = mrv_\perp\) is fixed, so the smallest \(r\) goes with the largest speed. Energy: on the way in from aphelion, gravity pulls the planet inward along its motion and does positive work, turning potential energy into kinetic energy.

1. \(C\) divides \(d\) in the inverse ratio of the masses. The \(2M\) star is \(\dfrac{M}{3M}d = \dfrac{d}{3}\) from \(C\) and the \(M\) star is \(\dfrac{2d}{3}\) from \(C\).

2. The force on each star points at the other, through \(C\). Both stars must always lie on a line through \(C\), on opposite sides. So they turn through equal angles in equal times: same period, same \(\omega\).

3. \(v = \omega r\) with the same \(\omega\): \(\dfrac{v_1}{v_2} = \dfrac{d/3}{2d/3} = \dfrac{1}{2}\). The heavier star moves more slowly.

For star 1: \(\dfrac{Gm_1m_2}{d^2} = m_1\omega^2r_1\) with \(r_1 = \dfrac{m_2d}{m_1 + m_2}\). So \(\dfrac{Gm_2}{d^2} = \omega^2\dfrac{m_2d}{m_1 + m_2}\), giving \(\omega^2 = \dfrac{G(m_1 + m_2)}{d^3}\). Star 2 gives the same result.

\(T = 2\pi\sqrt{\dfrac{d^3}{G(m_1 + m_2)}} = 2\pi\sqrt{\dfrac{8 \times 10^{33}}{6.67 \times 10^{-11} \times 4 \times 10^{30}}} = 2\pi\sqrt{3.0 \times 10^{13}} \approx 2\pi \times 5.5 \times 10^{6} \approx 3.4 \times 10^{7}\) s, about \(1.1\) years.

Equal masses, so each is \(\dfrac{d}{2} = 10^{11}\) m from \(C\). \(v = \dfrac{2\pi \times 10^{11}}{3.4 \times 10^{7}} \approx 1.8 \times 10^{4}\) m/s \(\approx 18\) km/s.

\(\mu = \dfrac{m_1m_2}{m_1 + m_2}\). For \(2M\) and \(M\): \(\dfrac{2M^2}{3M} = \dfrac{2M}{3}\). For \(M\) and \(M\): \(\dfrac{M^2}{2M} = \dfrac{M}{2}\).

1. Each body accelerates towards the other: \(a_1 = \dfrac{F}{m_1}\) and \(a_2 = \dfrac{F}{m_2}\), in opposite directions. The separation therefore shrinks at the rate \(a_{\text{rel}} = a_1 + a_2 = F\left(\dfrac{1}{m_1} + \dfrac{1}{m_2}\right)\).

2. \(\dfrac{1}{m_1} + \dfrac{1}{m_2} = \dfrac{m_1 + m_2}{m_1m_2} = \dfrac{1}{\mu}\), so \(a_{\text{rel}} = \dfrac{F}{\mu}\), that is \(\mu\,a_{\text{rel}} = F\). The relative motion is that of a single body of mass \(\mu\) pulled by \(F\) towards a fixed centre.

\(E = -\dfrac{Gm_1m_2}{2d}\), the same form as for a satellite with \(d\) as the separation. \(E = -\dfrac{6.67 \times 10^{-11} \times 4 \times 10^{60}}{2 \times 2 \times 10^{11}} \approx -6.7 \times 10^{38}\) J.

The total kinetic energy about \(C\) equals \(\dfrac{1}{2}\mu v_{\text{rel}}^2\), where \(v_{\text{rel}} = \omega d\) is the relative speed. Using \(\omega^2 = \dfrac{G(m_1 + m_2)}{d^3}\): \(v_{\text{rel}}^2 = \dfrac{G(m_1 + m_2)}{d}\), so \(K = \dfrac{1}{2}\cdot\dfrac{m_1m_2}{m_1 + m_2}\cdot\dfrac{G(m_1 + m_2)}{d} = \dfrac{Gm_1m_2}{2d}\). Then \(E = K + U = \dfrac{Gm_1m_2}{2d} - \dfrac{Gm_1m_2}{d} = -\dfrac{Gm_1m_2}{2d}\), as in Q17.

From the Earth's centre: \(x = \dfrac{m_{\text{moon}}d}{M + m_{\text{moon}}} = \dfrac{384000}{82} \approx 4700\) km. This is less than the Earth's radius of \(6400\) km, so \(C\) lies inside the Earth, about \(1700\) km below the surface.

Both bodies circle \(C\) with the same period of \(27.3\) days, always on opposite sides of it. The Moon moves on a circle of radius about \(379000\) km. The Earth's centre moves on a small circle of radius \(4700\) km, a wobble that shows up in the Earth's motion about the Sun.

\(2a = 2R + 6R\), so \(a = 4R\). \(\dfrac{v_P}{v_A} = \dfrac{r_A}{r_P} = 3\). \(E = -\dfrac{GMm}{2a} = -\dfrac{GMm}{8R}\).

\(T^2 \propto a^3\), so \(\dfrac{a_A}{a_B} = \left(\dfrac{T_A}{T_B}\right)^{2/3} = 8^{2/3} = 4\).

\(T = 2\pi\sqrt{\dfrac{r^3}{GM}} \propto \dfrac{1}{\sqrt M}\). Doubling \(M\) divides the period by \(\sqrt2\): \(T = \dfrac{1}{\sqrt2}\) yr \(\approx 0.71\) yr \(\approx 258\) days.

\(T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{d^3}{G(m_1 + m_2)}}\). When \(m_2 \ll m_1\), \(m_1 + m_2 \approx m_1\) and \(T \approx 2\pi\sqrt{\dfrac{d^3}{Gm_1}}\), the ordinary satellite formula with the heavy body fixed at the centre.