Chapter 7: Gravitation

At the perihelion P and the aphelion A, the velocity vector is perpendicular to the radius vector (these are the turning points of the radial motion).

The gravitational force on the planet always passes through the Sun S, so the torque of this force about S is zero. Hence the angular momentum of the planet about the Sun is conserved:

$$L = m\,r_P v_P = m\,r_A v_A$$

Therefore

$$\frac{v_P}{v_A} = \frac{r_A}{r_P}$$

Since $$r_P < r_A$$, we get $$v_P > v_A$$: the planet moves fastest at the perihelion and slowest at the aphelion.

By Kepler's second law, the line joining the Sun to the planet sweeps out equal areas in equal times. The area of the sector SBAC (which contains the aphelion side) is larger than the area of the sector SCPB (which contains the perihelion side), because the planet is on average farther from the Sun on the aphelion side. Equal areas in equal times therefore requires the planet to take a longer time on the arc BAC than on the arc CPB.

So the planet does not take equal times to traverse BAC and CPB.

(a) All three vertex masses equal to $$m$$.

Each vertex mass $$m$$ exerts a gravitational pull on the centroid mass $$2m$$ of magnitude

$$F = \frac{G\,(m)(2m)}{(1\,\mathrm{m})^2} = 2Gm^2 \,\mathrm{N}$$

directed from G towards the respective vertex. The position vectors GA, GB, GC have equal magnitudes and are mutually inclined at $$120^{\circ}$$. Their vector sum is therefore zero:

$$\vec{F}_A + \vec{F}_B + \vec{F}_C = \vec{0}$$

Hence the net force on the mass $$2m$$ at G is zero.

(b) Mass at vertex A doubled to $$2m$$.

The forces from B and C are unchanged:

$$F_B = F_C = 2Gm^2 \,\mathrm{N}$$, directed from G along GB and GC respectively.

From (a), $$\vec{F}_A + \vec{F}_B + \vec{F}_C = \vec{0}$$ when all three vertex masses are $$m$$. So

$$\vec{F}_B + \vec{F}_C = -\vec{F}_A^{(\text{old})}$$

which is a vector of magnitude $$2Gm^2$$ directed from G in the direction opposite to GA.

The new force from the doubled mass at A is

$$F_A^{\text{new}} = \frac{G\,(2m)(2m)}{(1)^2} = 4Gm^2\,\mathrm{N}$$

directed from G towards A. Adding,

$$F_{\text{net}} = 4Gm^2 - 2Gm^2 = 2Gm^2\,\mathrm{N}$$

directed from G towards A.

Label the four equal point masses (each of mass $$m$$) at the corners of a square of side $$l$$. The number of distinct pairs is $$\binom{4}{2}=6$$. Of these:

• 4 pairs lie along the sides of the square, separated by $$l$$.
• 2 pairs lie along the diagonals, separated by $$l\sqrt{2}$$.

The gravitational potential energy of a pair of point masses $$m$$ separated by $$r$$ is $$U_{12} = -Gm^2/r$$. Summing over all 6 pairs:

$$U = -4\,\frac{Gm^2}{l} - 2\,\frac{Gm^2}{l\sqrt{2}}$$

$$U = -\frac{Gm^2}{l}\left(4 + \frac{2}{\sqrt{2}}\right) = -\frac{Gm^2}{l}\left(4 + \sqrt{2}\right)$$

$$U \approx -5.41\,\frac{Gm^2}{l}$$

Potential at the centre of the square. The distance from the centre to each vertex is half the diagonal:

$$d = \frac{l\sqrt{2}}{2} = \frac{l}{\sqrt{2}}$$

The gravitational potential is the sum of contributions from each mass (a scalar):

$$V = 4 \times \left(-\frac{Gm}{d}\right) = -\frac{4 Gm}{l/\sqrt{2}} = -\frac{4\sqrt{2}\,Gm}{l}$$

Let $$O_1$$, $$O_2$$ be the centres of the spheres of mass $$M$$ and $$4M$$ respectively, with $$O_1O_2 = 6R$$. The projectile is launched from the surface of the $$M$$-sphere (distance $$R$$ from $$O_1$$, distance $$5R$$ from $$O_2$$) directly towards $$O_2$$.

Step 1 — Locate the neutral point N. Let N be on the line $$O_1O_2$$ at distance $$x$$ from $$O_1$$. The two gravitational pulls balance when

$$\frac{GMm}{x^2} = \frac{G(4M)m}{(6R - x)^2}$$

$$(6R - x)^2 = 4 x^2 \;\Rightarrow\; 6R - x = 2x \;\Rightarrow\; x = 2R$$

So N is $$2R$$ from $$O_1$$ (i.e. $$4R$$ from $$O_2$$). Between the launch point and N the net force is directed back towards $$O_1$$ (retarding); beyond N it is directed towards $$O_2$$ (accelerating). The projectile needs just enough kinetic energy to reach N — once it crosses N, the second sphere pulls it the rest of the way.

Step 2 — Energy conservation from the launch point to N. For the minimum speed, the projectile reaches N with zero kinetic energy.

Initial gravitational PE (using point masses at the centres, valid because the projectile is outside both spheres):

$$U_i = -\frac{GMm}{R} - \frac{G(4M)m}{5R}$$

PE at N (distances $$2R$$ from $$O_1$$ and $$4R$$ from $$O_2$$):

$$U_N = -\frac{GMm}{2R} - \frac{G(4M)m}{4R} = -\frac{GMm}{2R} - \frac{GMm}{R}$$

Setting total energy at the launch point equal to total energy at N:

$$\tfrac12 m v^2 + U_i = 0 + U_N$$

$$\tfrac12 v^2 = U_N - U_i = \left(-\frac{GM}{2R} - \frac{GM}{R}\right) - \left(-\frac{GM}{R} - \frac{4GM}{5R}\right)$$

$$\tfrac12 v^2 = -\frac{GM}{2R} + \frac{4GM}{5R} = \frac{GM}{R}\left(\frac{4}{5} - \frac{1}{2}\right) = \frac{3\,GM}{10\,R}$$

$$v_{\min} = \sqrt{\frac{3\,GM}{5\,R}}$$

(i) Mass of Mars from Phobos's orbit.

$$T = 7\,\mathrm{h}\,39\,\mathrm{min} = (7 \times 3600 + 39 \times 60)\,\mathrm{s} = 27540\,\mathrm{s}$$

$$r = 9.4 \times 10^3\,\mathrm{km} = 9.4 \times 10^6\,\mathrm{m}$$

For a circular orbit, gravitational attraction provides the centripetal force:

$$\frac{G M_{\mathrm{Mars}}\,m}{r^2} = m\left(\frac{2\pi}{T}\right)^2 r \;\Rightarrow\; M_{\mathrm{Mars}} = \frac{4\pi^2 r^3}{G T^2}$$

$$M_{\mathrm{Mars}} = \frac{4\pi^2 (9.4 \times 10^6)^3}{(6.67 \times 10^{-11})(27540)^2}$$

$$M_{\mathrm{Mars}} = \frac{4\pi^2 \times 8.31 \times 10^{20}}{(6.67 \times 10^{-11})(7.58 \times 10^{8})} \approx 6.48 \times 10^{23}\,\mathrm{kg}$$

(ii) Length of the Martian year.

Kepler's third law applied to Earth and Mars (both orbiting the Sun):

$$\left(\frac{T_M}{T_E}\right)^2 = \left(\frac{r_M}{r_E}\right)^3 = (1.52)^3$$

$$T_M = T_E \,(1.52)^{3/2} = 365\,\mathrm{d} \times 1.874 \approx 684\,\mathrm{days}$$

Method 1 — from the value of $$g$$ at the Earth's surface.

At the surface, the acceleration due to gravity satisfies

$$g = \frac{GM_E}{R_E^2} \;\Rightarrow\; M_E = \frac{g R_E^2}{G}$$

$$M_E = \frac{(9.81)(6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = \frac{(9.81)(4.058 \times 10^{13})}{6.67 \times 10^{-11}}$$

$$M_E \approx 5.97 \times 10^{24}\,\mathrm{kg}$$

Method 2 — from the orbital motion of the Moon.

The Moon moves in a (nearly) circular orbit around the Earth of radius $$R$$ and period $$T$$. Gravity provides the centripetal force:

$$\frac{GM_E m}{R^2} = m\left(\frac{2\pi}{T}\right)^2 R \;\Rightarrow\; M_E = \frac{4\pi^2 R^3}{G T^2}$$

Convert $$T = 27.3$$ days $$= 27.3 \times 86400 \,\mathrm{s} \approx 2.36 \times 10^6$$ s. Then

$$M_E = \frac{4\pi^2 (3.84 \times 10^8)^3}{(6.67 \times 10^{-11})(2.36 \times 10^6)^2}$$

$$M_E = \frac{4\pi^2 \times 5.66 \times 10^{25}}{(6.67 \times 10^{-11})(5.57 \times 10^{12})} \approx 6.02 \times 10^{24}\,\mathrm{kg}$$

The two independent estimates agree to within $$\sim 1\%$$, both giving $$M_E \approx 6 \times 10^{24}\,\mathrm{kg}$$.

The constant $$k$$ comes from Kepler's third law, $$T^2 = k\,r^3$$. We are given

$$k = 10^{-13}\,\mathrm{s^2\,m^{-3}}$$

Convert seconds to days and metres to kilometres:

$$1\,\mathrm{s} = \frac{1}{86400}\,\mathrm{day}, \qquad 1\,\mathrm{m} = 10^{-3}\,\mathrm{km}$$

So $$1\,\mathrm{s^2\,m^{-3}} = \dfrac{1}{(86400)^2}\,\mathrm{day^2} \times \dfrac{1}{(10^{-3})^3\,\mathrm{km^3}} = \dfrac{10^9}{(86400)^2}\,\mathrm{day^2\,km^{-3}}$$

$$k = 10^{-13} \times \dfrac{10^9}{(8.64 \times 10^4)^2}\,\mathrm{day^2\,km^{-3}}$$

$$k = \dfrac{10^{-4}}{7.464 \times 10^9}\,\mathrm{day^2\,km^{-3}} \approx 1.33 \times 10^{-14}\,\mathrm{day^2\,km^{-3}}$$

Period of the Moon. With $$r = 3.84 \times 10^5$$ km,

$$T^2 = k\,r^3 = (1.33 \times 10^{-14})\,(3.84 \times 10^5)^3$$

$$T^2 = (1.33 \times 10^{-14})(5.66 \times 10^{16}) \approx 752.8\,\mathrm{day^2}$$

$$T \approx 27.4\,\mathrm{days}$$

For a satellite of mass $$m$$ in a circular orbit of radius $$r$$ about the Earth, the centripetal-force condition $$GM_E m/r^2 = mv^2/r$$ gives

$$KE(r) = \tfrac12 m v^2 = \dfrac{GM_E m}{2r}, \qquad PE(r) = -\dfrac{GM_E m}{r}$$

$$E(r) = KE + PE = -\dfrac{GM_E m}{2r}$$

Use $$GM_E = g R_E^2$$. With $$r_1 = 2R_E$$, $$r_2 = 4R_E$$ and $$m = 400$$ kg:

$$E_1 = -\dfrac{g R_E^2 m}{4 R_E} = -\dfrac{g R_E m}{4}, \qquad E_2 = -\dfrac{g R_E m}{8}$$

Energy required to raise the orbit:

$$\Delta E = E_2 - E_1 = -\dfrac{g R_E m}{8} + \dfrac{g R_E m}{4} = \dfrac{g R_E m}{8}$$

$$\Delta E = \dfrac{(9.81)(6.37 \times 10^6)(400)}{8} \approx 3.13 \times 10^{9}\,\mathrm{J}$$

Change in kinetic energy:

$$\Delta KE = KE_2 - KE_1 = \dfrac{g R_E m}{8} - \dfrac{g R_E m}{4} = -\dfrac{g R_E m}{8} \approx -3.13 \times 10^{9}\,\mathrm{J}$$

The KE decreases: the higher orbit is the slower one.

Change in potential energy:

$$\Delta PE = PE_2 - PE_1 = -\dfrac{g R_E m}{4} + \dfrac{g R_E m}{2} = +\dfrac{g R_E m}{4} \approx +6.25 \times 10^{9}\,\mathrm{J}$$

Consistency: $$\Delta E = \Delta KE + \Delta PE = -3.13 \times 10^{9} + 6.25 \times 10^{9} = +3.13 \times 10^{9}\,\mathrm{J}$$ ✓

Let $$T_E$$, $$R_E$$ denote the period and the orbital radius of the Earth, and $$T_p$$, $$R_p$$ the same for the hypothetical planet. "Twice as fast" means the planet completes one orbit in half the time:

$$T_p = \dfrac{T_E}{2}$$

Kepler's third law for planets orbiting the same Sun gives

$$\dfrac{T_p^2}{T_E^2} = \dfrac{R_p^3}{R_E^3}$$

$$\dfrac{R_p^3}{R_E^3} = \left(\dfrac{T_p}{T_E}\right)^2 = \dfrac{1}{4}$$

$$\dfrac{R_p}{R_E} = \left(\dfrac{1}{4}\right)^{1/3} = 4^{-1/3} \approx 0.63$$

So the planet's orbital radius is about $$0.63\,R_E \approx 0.63$$ AU — roughly 63% of the Earth-Sun distance.

Convert Io's orbital period to SI units:

$$T = 1.769\,\mathrm{day} \times 86400\,\mathrm{s/day} \approx 1.528 \times 10^5\,\mathrm{s}$$

Apply Kepler's third law to the Io-Jupiter system (Io orbits Jupiter):

$$T^2 = \dfrac{4\pi^2 r^3}{G M_J} \;\Rightarrow\; M_J = \dfrac{4\pi^2 r^3}{G T^2}$$

$$M_J = \dfrac{4\pi^2 (4.22 \times 10^8)^3}{(6.67 \times 10^{-11})(1.528 \times 10^5)^2}$$

$$M_J = \dfrac{4\pi^2 \times 7.51 \times 10^{25}}{(6.67 \times 10^{-11})(2.34 \times 10^{10})}$$

$$M_J = \dfrac{2.965 \times 10^{27}}{1.559} \approx 1.90 \times 10^{27}\,\mathrm{kg}$$

The Sun's mass is $$M_\odot \approx 2 \times 10^{30}$$ kg, so

$$\dfrac{M_J}{M_\odot} = \dfrac{1.90 \times 10^{27}}{2 \times 10^{30}} \approx 9.5 \times 10^{-4} \approx \dfrac{1}{1050}$$

Hence Jupiter is about one-thousandth the mass of the Sun, as required.

The star is at the rim of the disk ($$50000\,\mathrm{ly} = \tfrac{1}{2} \times 10^5\,\mathrm{ly}$$), so the entire galactic mass interior to its orbit is effectively the total stellar mass.

Total mass of the galaxy:

$$M = (2.5 \times 10^{11})(2 \times 10^{30}\,\mathrm{kg}) = 5 \times 10^{41}\,\mathrm{kg}$$

Orbital radius in SI units (1 ly $$= 9.46 \times 10^{15}$$ m):

$$r = (5 \times 10^4)(9.46 \times 10^{15}) \approx 4.73 \times 10^{20}\,\mathrm{m}$$

Kepler's third law (treating the rest of the galaxy as a point mass at the centre):

$$T^2 = \dfrac{4\pi^2 r^3}{G M}$$

$$r^3 = (4.73 \times 10^{20})^3 \approx 1.058 \times 10^{62}\,\mathrm{m^3}$$

$$T^2 = \dfrac{4\pi^2 \times 1.058 \times 10^{62}}{(6.67 \times 10^{-11})(5 \times 10^{41})} = \dfrac{4.18 \times 10^{63}}{3.34 \times 10^{31}} \approx 1.25 \times 10^{32}\,\mathrm{s^2}$$

$$T \approx 1.12 \times 10^{16}\,\mathrm{s}$$

Converting to years using $$1\,\mathrm{yr} = 3.156 \times 10^7$$ s:

$$T \approx \dfrac{1.12 \times 10^{16}}{3.156 \times 10^7} \approx 3.55 \times 10^{8}\,\mathrm{yr}$$

The escape speed from a point at distance $$r$$ from the centre of a spherical body of mass $$M$$ is obtained from energy conservation (set $$KE_\infty = 0$$ and $$PE_\infty = 0$$):

$$\tfrac12 m v_e^2 - \dfrac{GMm}{r} = 0 \;\Rightarrow\; v_e = \sqrt{\dfrac{2GM}{r}}$$

(a) Does not depend on the mass of the body: $$m$$ cancels in the expression for $$v_e$$.

(b) Does not depend on the location on the Earth's surface (assuming a spherical Earth): all surface points have the same radial distance $$R_E$$ from the centre, so the formula gives the same $$v_e$$ everywhere on the surface.

(c) Does not depend on the direction of projection: the escape condition is purely energetic. (One must, of course, ensure the trajectory does not re-intersect the Earth — but the threshold speed itself is direction-independent.)

(d) Does depend on the height $$h$$ of the launching point, because then $$r = R_E + h$$ and

$$v_e = \sqrt{\dfrac{2GM}{R_E + h}}$$

which decreases as $$h$$ increases.

The Sun's gravitational pull on the comet is a central, conservative force, so it exerts zero torque about the Sun and does no non-conservative work.

(a) Linear speed — No. By conservation of angular momentum, $$r v_\perp = \text{const}$$, so $$v$$ is largest at perihelion and smallest at aphelion.

(b) Angular speed — No. $$\omega = v_\perp/r$$ varies for the same reason as (a).

(c) Angular momentum — Yes. Gravity is a central force; its torque about the Sun is zero, so $$\vec L$$ is conserved.

(d) Kinetic energy — No. Speed varies along the orbit, so KE varies.

(e) Potential energy — No. $$U = -GM_\odot m / r$$ depends on $$r$$, which changes.

(f) Total mechanical energy — Yes. Gravity is conservative and we are told to neglect mass loss; with no dissipation, $$E = KE + PE$$ is conserved.

On Earth, gravity normally pulls bodily fluids (especially blood) downward; the cardiovascular system compensates by pumping harder against this hydrostatic pressure. In micro-gravity (free fall) the downward bias is lost, and the fluids redistribute, accumulating in the upper body.

• (a) Swollen feet — No. Fluids actually move away from the legs ("chicken-leg syndrome"), so the feet do not swell.
• (b) Swollen face — Yes. Excess fluid in the head and upper body produces a noticeable puffiness of the face.
• (c) Headache — Yes. Higher intracranial pressure from the fluid shift commonly produces headaches.
• (d) Orientational problem — Yes. The vestibular (balance) system relies on gravity to sense "up". Without it, astronauts experience disorientation and space sickness.

Think of a complete uniform spherical shell — by Newton's shell theorem, the gravitational field inside is exactly zero everywhere. Decompose this complete shell into two hemispherical shells (upper and lower). If the field due to the complete shell vanishes, the fields due to the two hemispheres at the common centre must be equal in magnitude and opposite in direction.

By symmetry, the field at the centre of the flat face of each hemispherical shell must lie along the axis of symmetry of the hemisphere (the components perpendicular to that axis cancel pair-wise). The direction is along the axis, into the body of the hemisphere — towards the curved cap where the mass lies. That is the direction shown by the arrow c in Fig. 7.11.

Apply the same decomposition argument used in 7.10. Inside a complete uniform spherical shell, the gravitational field is zero at every interior point P. So the field at P due to the upper hemispherical shell plus the field at P due to the lower hemispherical shell must vanish:

$$\vec g_{\text{upper}}(P) + \vec g_{\text{lower}}(P) = \vec 0$$

Thus the field due to the upper hemisphere alone is equal in magnitude and opposite in direction to that of the lower hemisphere. The field due to either hemisphere at the arbitrary internal point P is parallel to the axis of symmetry of the shell (since the perpendicular components cancel as discussed in 7.10). The direction is along the axis, in the sense shown by arrow e in Fig. 7.11.

Let $$x$$ be the distance from the Earth's centre at which the gravitational forces of the Sun and the Earth on the rocket are equal and opposite. The Sun is at $$r = 1.5 \times 10^{11}$$ m from the Earth, so the rocket is $$r - x$$ from the Sun.

Set the magnitudes equal:

$$\dfrac{G M_E m}{x^2} = \dfrac{G M_S m}{(r - x)^2}$$

$$\left(\dfrac{r - x}{x}\right)^2 = \dfrac{M_S}{M_E} = \dfrac{2 \times 10^{30}}{6 \times 10^{24}} = \dfrac{10^6}{3}$$

$$\dfrac{r - x}{x} = \dfrac{10^3}{\sqrt 3} \approx 577.35$$

$$r = x\,(1 + 577.35) \approx 578.35\,x$$

$$x = \dfrac{r}{578.35} = \dfrac{1.5 \times 10^{11}}{578.35} \approx 2.59 \times 10^{8}\,\mathrm{m}$$

So the null point lies about $$2.6 \times 10^{8}$$ m from the Earth's centre (towards the Sun) — well inside the Earth's orbital radius.

Treat the Earth as a planet moving in a (nearly) circular orbit of radius $$r$$ about the Sun with period $$T$$. The Sun's gravitational pull provides the centripetal force:

$$\dfrac{G M_S m_E}{r^2} = m_E\,\omega^2 r = m_E\left(\dfrac{2\pi}{T}\right)^2 r$$

$$M_S = \dfrac{4\pi^2 r^3}{G T^2}$$

So one only needs the radius of the Earth's orbit and its period — the Earth's own mass cancels.

Plug in $$r = 1.5 \times 10^{8}$$ km $$= 1.5 \times 10^{11}$$ m and $$T = 1$$ year $$= 365.25 \times 86400 \approx 3.156 \times 10^{7}$$ s:

$$r^3 = (1.5 \times 10^{11})^3 = 3.375 \times 10^{33}\,\mathrm{m^3}$$

$$T^2 = (3.156 \times 10^{7})^2 \approx 9.96 \times 10^{14}\,\mathrm{s^2}$$

$$M_S = \dfrac{4\pi^2 \times 3.375 \times 10^{33}}{(6.67 \times 10^{-11})(9.96 \times 10^{14})} = \dfrac{1.332 \times 10^{35}}{6.643 \times 10^{4}}$$

$$M_S \approx 2.0 \times 10^{30}\,\mathrm{kg}$$

Apply Kepler's third law to two planets orbiting the same Sun:

$$\dfrac{T_S^2}{T_E^2} = \dfrac{r_S^3}{r_E^3} \;\Rightarrow\; \dfrac{r_S}{r_E} = \left(\dfrac{T_S}{T_E}\right)^{2/3}$$

With $$T_S/T_E = 29.5$$:

$$\dfrac{r_S}{r_E} = (29.5)^{2/3} = \left[(29.5)^2\right]^{1/3} = (870.25)^{1/3} \approx 9.55$$

$$r_S \approx 9.55 \times (1.50 \times 10^{8}\,\mathrm{km}) \approx 1.43 \times 10^{9}\,\mathrm{km}$$

Surface weight: $$W = mg = 63\,\mathrm{N}$$.

At height $$h = R_E/2$$, the radial distance from the Earth's centre is $$R_E + h = \tfrac{3}{2}R_E$$, so

$$g' = \dfrac{G M_E}{(R_E + h)^2} = g\left(\dfrac{R_E}{R_E + h}\right)^2 = g\left(\dfrac{R_E}{(3/2)R_E}\right)^2 = g\left(\dfrac{2}{3}\right)^2 = \dfrac{4g}{9}$$

Hence the gravitational force on the body at that height is

$$W' = m g' = \dfrac{4}{9}(mg) = \dfrac{4}{9}(63\,\mathrm{N}) = 28\,\mathrm{N}$$

For a uniform-density Earth, by Newton's shell theorem only the mass within radius $$R_E - d$$ contributes to gravity at depth $$d$$. This gives

$$g_d = g\left(1 - \dfrac{d}{R_E}\right)$$

"Half way down to the centre" means $$d = R_E/2$$, so

$$g_d = g\left(1 - \dfrac{1}{2}\right) = \dfrac{g}{2}$$

Hence the weight there is

$$W_d = m g_d = \dfrac{1}{2}(mg) = \dfrac{1}{2}(250\,\mathrm{N}) = 125\,\mathrm{N}$$

Let $$h$$ be the maximum height above the Earth's surface (so the apex is at distance $$R_E + h$$ from the centre). Apply energy conservation between the launch point and the apex (where KE = 0):

$$\tfrac12 m v^2 - \dfrac{G M_E m}{R_E} = 0 - \dfrac{G M_E m}{R_E + h}$$

$$\tfrac12 v^2 = G M_E\left(\dfrac{1}{R_E} - \dfrac{1}{R_E + h}\right) = \dfrac{G M_E\,h}{R_E (R_E + h)}$$

Solving for $$h$$:

$$v^2 R_E (R_E + h) = 2 G M_E h \;\Rightarrow\; h = \dfrac{v^2 R_E^2}{2 G M_E - v^2 R_E}$$

Plug in $$v = 5 \times 10^3$$ m/s, $$R_E = 6.4 \times 10^6$$ m, $$G M_E = (6.67 \times 10^{-11})(6.0 \times 10^{24}) = 4.00 \times 10^{14}$$ m³/s²:

$$v^2 R_E^2 = (2.5 \times 10^{7})(4.096 \times 10^{13}) = 1.024 \times 10^{21}\,\mathrm{m^4\,s^{-2}}$$

$$2 G M_E = 8.00 \times 10^{14},\quad v^2 R_E = 1.60 \times 10^{14}$$

$$h = \dfrac{1.024 \times 10^{21}}{(8.00 - 1.60) \times 10^{14}} = \dfrac{1.024 \times 10^{21}}{6.40 \times 10^{14}} = 1.6 \times 10^{6}\,\mathrm{m}$$

So the rocket rises about 1600 km above the Earth's surface; its maximum distance from the Earth's centre is $$R_E + h = (6.4 + 1.6) \times 10^{6} = 8.0 \times 10^{6}$$ m.

Let $$v_e = 11.2$$ km/s be the escape speed, and $$v_i = 3 v_e$$ the initial speed. By definition of escape speed,

$$\tfrac12 m v_e^2 = \dfrac{G M_E m}{R_E}$$

Apply energy conservation between the surface and infinity (where PE = 0):

$$\tfrac12 m v_i^2 - \dfrac{G M_E m}{R_E} = \tfrac12 m v_\infty^2$$

Substitute $$G M_E m / R_E = \tfrac12 m v_e^2$$:

$$\tfrac12 m v_i^2 - \tfrac12 m v_e^2 = \tfrac12 m v_\infty^2$$

$$v_\infty^2 = v_i^2 - v_e^2 = (3 v_e)^2 - v_e^2 = 8\, v_e^2$$

$$v_\infty = v_e\sqrt{8} = 2\sqrt{2}\,v_e = 2\sqrt{2}\,(11.2\,\mathrm{km/s}) \approx 31.7\,\mathrm{km\,s^{-1}}$$

Orbital radius:

$$r = R_E + h = (6.4 \times 10^{6}) + (4 \times 10^{5}) = 6.8 \times 10^{6}\,\mathrm{m}$$

For a satellite in a circular orbit, the total mechanical energy is

$$E_{\text{orb}} = -\dfrac{G M_E m}{2 r}$$

To "rocket it out of the Earth's gravitational influence" means to give it enough additional energy that its total energy becomes zero (just barely escaping):

$$\Delta E = 0 - E_{\text{orb}} = +\dfrac{G M_E m}{2 r}$$

Plug in $$G = 6.67 \times 10^{-11}$$, $$M_E = 6.0 \times 10^{24}$$ kg, $$m = 200$$ kg, $$r = 6.8 \times 10^{6}$$ m:

$$\Delta E = \dfrac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(200)}{2 \times 6.8 \times 10^{6}}$$

$$\Delta E = \dfrac{8.004 \times 10^{16}}{1.36 \times 10^{7}} \approx 5.89 \times 10^{9}\,\mathrm{J}$$

Each star has mass $$M = 2 \times 10^{30}$$ kg and radius $$R = 10^4$$ km $$= 10^{7}$$ m. Initial separation $$d = 10^9$$ km $$= 10^{12}$$ m. Initial speeds are negligible.

By symmetry (equal masses), in the centre-of-mass frame each star always has the same speed $$v$$. Just before collision, the centres are separated by $$2R$$.

Conservation of mechanical energy (KE initial $$\approx 0$$):

$$0 + \left(-\dfrac{G M^2}{d}\right) = 2 \times \tfrac12 M v^2 + \left(-\dfrac{G M^2}{2R}\right)$$

$$M v^2 = \dfrac{G M^2}{2R} - \dfrac{G M^2}{d}$$

$$v^2 = G M\left(\dfrac{1}{2R} - \dfrac{1}{d}\right)$$

Numerical values:

$$G M = (6.67 \times 10^{-11})(2 \times 10^{30}) = 1.334 \times 10^{20}\,\mathrm{m^3\,s^{-2}}$$

$$\dfrac{1}{2R} = \dfrac{1}{2 \times 10^{7}} = 5 \times 10^{-8}\,\mathrm{m^{-1}}, \quad \dfrac{1}{d} = 10^{-12}\,\mathrm{m^{-1}}\;(\text{negligible})$$

$$v^2 \approx (1.334 \times 10^{20})(5 \times 10^{-8}) = 6.67 \times 10^{12}\,\mathrm{m^2\,s^{-2}}$$

$$v \approx 2.58 \times 10^{6}\,\mathrm{m\,s^{-1}}$$

This is the speed of each star. Their relative speed of approach at the moment of collision is

$$v_{\text{rel}} = 2 v \approx 5.2 \times 10^{6}\,\mathrm{m\,s^{-1}}$$

Put the origin at the midpoint M of the line joining the centres. Each sphere (mass $$m_0 = 100$$ kg) is at distance $$r = 0.50$$ m from M, and M lies outside both spheres (since $$r > 0.10$$ m), so we can treat each sphere as a point mass at its centre.

Gravitational field (force per unit mass) at M. The two pulls have equal magnitudes

$$\dfrac{G m_0}{r^2} = \dfrac{(6.67 \times 10^{-11})(100)}{(0.5)^2} = 2.67 \times 10^{-8}\,\mathrm{m\,s^{-2}}$$

but they point in opposite directions along the line of centres, so their vector sum is zero. Net gravitational force on a test mass placed at M is $$\boxed{0}$$.

Gravitational potential at M. Potential is a scalar; contributions add:

$$V = -\dfrac{G m_0}{r} - \dfrac{G m_0}{r} = -\dfrac{2 G m_0}{r}$$

$$V = -\dfrac{2 \times (6.67 \times 10^{-11}) \times 100}{0.5} = -2.67 \times 10^{-8}\,\mathrm{J\,kg^{-1}}$$

Equilibrium and its stability. The net force on a test object placed at M is zero, so the object is in equilibrium.

To test stability, displace the object slightly along the line of centres towards (say) the left-hand sphere by $$\epsilon$$. Its distance from the left sphere becomes $$r - \epsilon$$ and from the right sphere $$r + \epsilon$$. The net leftward pull is

$$F_{\text{net}} = \dfrac{G m_0 m}{(r-\epsilon)^2} - \dfrac{G m_0 m}{(r+\epsilon)^2}$$

which is positive (to the left) and grows as $$\epsilon$$ increases. So the small displacement is amplified — the test object accelerates further from the equilibrium point along the line. Hence the equilibrium is unstable.