NCERT Solutions Class 11 Physics Chapter 7 — Gravitation

NCERT Class 11 Physics — Gravitation, Chapter 7 Exercises. All 21 questions solved.

The exercises at the end of Chapter 7 fall into three groups. Questions 7.1, 7.2 and 7.6 to 7.11 are mostly reasoning: shielding, tides, how $g$ changes above and below the surface, what stays constant along an orbit, and the pull of a hemispherical shell. Questions 7.3 to 7.5 and 7.12 to 7.16 use the law of gravitation and Kepler’s third law to find masses, orbit sizes and weights. Questions 7.17 to 7.21 are energy calculations. Almost all the arithmetic runs on three results:

• Newton’s law of gravitation — $F = \dfrac{Gm_1m_2}{r^2}$, with $G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$. Outside a spherical body, $r$ is measured from its centre, as though all its mass were there.
• Kepler’s third law for a circular orbit — setting the pull equal to the centripetal force, $\dfrac{GMm}{r^2} = m\left(\dfrac{2\pi}{T}\right)^2 r$, gives $T^2 = \dfrac{4\pi^2}{GM}\,r^3$.
• Gravitational potential energy — $U = -\dfrac{GMm}{r}$, zero at infinity. From it come the escape speed $v_e = \sqrt{2GM/R}$ and the total energy of a satellite in a circular orbit, $E = -\dfrac{GMm}{2r}$.

Two consequences are used repeatedly: $g(h) = g\,\dfrac{R^2}{(R+h)^2}$ at a height $h$ above the surface, and $g(d) = g\left(1 – \dfrac{d}{R}\right)$ at a depth $d$ below it.

Key insight. Two ideas carry almost the whole exercise. The first is the shell theorem: a spherical body pulls on anything outside it as though all its mass sat at its centre, and a shell pulls nothing inside it at all. That makes 7.5, 7.15, 7.16 and 7.21 one-line calculations, and completing a hemisphere into a full shell settles 7.10 and 7.11 without any integration. The second is energy: gravity is conservative, so $\tfrac12mv^2 – GMm/r$ stays the same all along a path, and 7.17 to 7.20 are that one equation written at two points. Kepler’s third law, $T^2\propto r^3$, is the same force law applied to circular orbits, and it turns 7.3, 7.4, 7.13 and 7.14 into ratios.

Question 7.1

Answer the following: (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means? (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity? (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises.) However, the tidal effect of the moon’s pull is greater than the tidal effect of the sun. Why?

Question 7.1 (a)

Solution. Electrical shielding works for two reasons. A conductor has free charges that move until the field inside is zero, and charge comes in two signs, so the charges gathered on the conductor can cancel an outside field. Mass has only one sign — gravity always attracts — so nothing can collect on a shell to cancel an outside pull. A hollow sphere exerts no net force of its own on a body inside it, but the pull of matter outside passes straight through it, and no other material or arrangement is known that blocks gravity.

Question 7.1 (b)

Solution. The astronaut and the ship are both in free fall towards the earth with the same acceleration, so in a small ship nothing moves relative to anything else and gravity seems to vanish. But $g$ is not quite the same everywhere in the ship. It is slightly larger on the side nearer the earth and slightly smaller on the far side, and it points in slightly different directions at the two ends. In a large enough station, objects released at different places drift slowly apart — a tidal effect — and with sensitive enough instruments the astronaut can detect it.

Question 7.1 (c)

Solution. The tides are not raised by the pull itself but by the difference between the pull on the side of the earth nearest the body and the pull on the far side. The pull per unit mass is $GM/r^2$; over the earth’s diameter it changes by about $\dfrac{d}{dr}\!\left(\dfrac{GM}{r^2}\right)\times 2R_E = -\dfrac{4GMR_E}{r^3}$. So the tidal effect goes as $M/r^3$, while the pull goes as $M/r^2$. The extra power of $r$ favours the much nearer moon.

With the sun’s mass $2\times10^{30}\ \text{kg}$ at $1.5\times10^{11}\ \text{m}$ (the values in Question 7.12) and the moon’s $7.36\times10^{22}\ \text{kg}$ at $3.84\times10^8\ \text{m}$:

$$\frac{\text{sun’s pull}}{\text{moon’s pull}} = \frac{M_S/r_S^2}{M_M/r_M^2} \approx 178, \qquad \frac{\text{moon’s tidal effect}}{\text{sun’s tidal effect}} = \frac{M_M/r_M^3}{M_S/r_S^3} \approx 2.2$$

Question 7.2

Choose the correct alternative: (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula $-GMm(1/r_2 – 1/r_1)$ is more/less accurate than the formula $mg(r_2 – r_1)$ for the difference of potential energy between two points $r_2$ and $r_1$ distance away from the centre of the earth.

Solution. (a) Above the surface, $g(h) = GM/(R+h)^2$, which falls as $h$ grows. (b) Below the surface only the mass nearer the centre than the body pulls on it; the shell outside contributes nothing. That inner mass shrinks as $r^3$ while the distance shrinks as $r$, so $g(d) = g(1 – d/R)$, which falls to zero at the centre. (c) $g = GM/R^2$ contains the earth’s mass $M$ but not the body’s mass, which cancels between the force $GMm/R^2$ and $F = ma$. (d) $-GMm(1/r_2 – 1/r_1)$ comes straight from $U = -GMm/r$ and is exact. Rewriting it as $\dfrac{GMm(r_2-r_1)}{r_1r_2}$ shows that it becomes $mg(r_2-r_1)$ only when $r_1 \approx r_2 \approx R$, that is, when $g$ can be treated as constant between the two points.

Question 7.3

Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Solution. “Twice as fast” means it completes an orbit in half the time, $T_p = T_E/2$. Both planets orbit the same sun, so Kepler’s third law, $T^2\propto R^3$, compares them without needing the sun’s mass:

$$\frac{R_p}{R_E} = \left(\frac{T_p}{T_E}\right)^{2/3} = \left(\frac12\right)^{2/3} = 0.63$$

Question 7.4

Io, one of the satellites of Jupiter, has an orbital period of $1.769$ days and the radius of the orbit is $4.22\times10^8\ \text{m}$. Show that the mass of Jupiter is about one-thousandth that of the sun.

Solution. For any body in a circular orbit, gravity supplies the centripetal force, so the central mass is $M = \dfrac{4\pi^2R^3}{GT^2}$. Writing this once for Io about Jupiter and once for the earth about the sun, and dividing, removes $G$ and $4\pi^2$:

$$\frac{M_J}{M_S} = \left(\frac{R_{\text{Io}}}{R_E}\right)^3\left(\frac{T_E}{T_{\text{Io}}}\right)^2$$

With the earth’s orbit $R_E = 1.496\times10^{11}\ \text{m}$ and $T_E = 365.25$ days:

$$\frac{M_J}{M_S} = \left(\frac{4.22\times10^8}{1.496\times10^{11}}\right)^3\left(\frac{365.25}{1.769}\right)^2 = (2.245\times10^{-8})(4.26\times10^4) = 9.6\times10^{-4} \approx \frac{1}{1045}$$

As a check, putting Io’s figures directly into $M = 4\pi^2R^3/GT^2$, with $T = 1.769\times86400 = 1.528\times10^5\ \text{s}$, gives $M_J = 1.9\times10^{27}\ \text{kg}$, which is about a thousandth of the sun’s $2\times10^{30}\ \text{kg}$.

Question 7.5

Let us assume that our galaxy consists of $2.5\times10^{11}$ stars each of one solar mass. How long will a star at a distance of $50{,}000\ \text{ly}$ from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be $10^5\ \text{ly}$.

Solution. A diameter of $10^5\ \text{ly}$ means a radius of $5\times10^4\ \text{ly}$, so this star is at the edge of the galaxy and practically all of the galaxy’s mass lies inside its orbit. Treating that mass as though it were concentrated at the centre, the star moves like a planet round a sun of mass

$$M = (2.5\times10^{11})(2\times10^{30}\ \text{kg}) = 5\times10^{41}\ \text{kg}$$

The radius, with $1\ \text{ly} = 9.46\times10^{15}\ \text{m}$, is $r = 5\times10^4\times9.46\times10^{15} = 4.73\times10^{20}\ \text{m}$. Kepler’s third law then gives the period:

$$T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(4.73\times10^{20})^3}{(6.67\times10^{-11})(5\times10^{41})}} = 2\pi\sqrt{3.17\times10^{30}} = 1.12\times10^{16}\ \text{s}$$

Dividing by $3.156\times10^7\ \text{s}$ per year gives $3.55\times10^8$ years. The key’s $3.54\times10^8$ years differs only in the rounding of the light year and the year.

Question 7.6

Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

Solution. (a) In a circular orbit of radius $r$, gravity provides the centripetal force, $\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$, so the kinetic energy is $K = \dfrac{GMm}{2r}$. The potential energy is $U = -\dfrac{GMm}{r}$, so the total is

$$E = K + U = -\frac{GMm}{2r} = -K$$

It is also $U/2$, not $-U$.

(b) To escape, a body must be brought up to a total energy of zero. The stationary object at that height has only potential energy, $-GMm/r$, and needs $GMm/r$. The satellite has $-GMm/2r$, because it already carries kinetic energy, and needs only $GMm/2r$ — half as much.

Question 7.7

Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?

Solution. A body escapes if its kinetic energy is enough to raise its total energy to zero: $\tfrac12mv_e^2 = \dfrac{GMm}{r}$, so $v_e = \sqrt{2GM/r}$, where $r$ is the distance of the launch point from the earth’s centre.

(a) The mass $m$ cancels, so the escape speed does not depend on it. (b) For the uniform spherical earth of the chapter, every point on the surface is at the same distance $R$ from the centre and so has the same potential, which means the location makes no difference. For the real earth, which is not a perfect sphere, the potential varies slightly with latitude, as the answer key’s note points out, so there is a very slight effect. (c) Energy is a scalar, so any direction will do, provided the path does not run into the earth (air resistance and the earth’s rotation are ignored). (d) From a height $h$, $r = R + h$ and $v_e = \sqrt{2GM/(R+h)}$, which is smaller, so the height does matter.

Question 7.8

A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Solution. The sun’s pull is always directed towards the sun, so it exerts no torque about the sun, and the comet’s angular momentum $L = mvr_\perp$ is conserved. That is Kepler’s second law, and it forces the comet to move fastest at its closest approach, where $r$ is smallest. So its linear speed and angular speed both change, and so does its kinetic energy, which depends on speed. Its potential energy, $-GMm/r$, changes with $r$. Gravity is a conservative force, so the total energy, $K + U$, stays constant: the comet trades potential energy for kinetic energy as it falls in and back again as it moves out.

Question 7.9

Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.

Solution. On the earth, gravity pulls the body’s blood and other fluids towards the legs, and the body pumps against this. In free fall there is no such pull, so the fluids shift upwards into the chest and head. The face swells and the extra pressure in the head causes headaches, while the legs, far from swelling, become thinner. Our sense of up and down comes largely from the balance organs of the inner ear, which rely on gravity; without it they give no reference, and astronauts become disoriented.

Question 7.10

In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig. 7.11).

Solution. Adding up the pull of every piece of the bowl directly would need an integral. Instead, imagine putting an identical upper hemisphere on top to complete a full spherical shell. Inside a uniform spherical shell the gravitational intensity is zero at every point, so at $C$ the pull of the bowl and the pull of the added upper half must be equal and opposite.

The bowl is symmetric about the vertical line through $C$, so the sideways pulls of pieces on opposite sides of the bowl cancel, and its net pull at $C$ can only point along that line. All of the bowl’s mass lies below the rim, so every piece pulls $C$ downwards and nothing cancels the downward pull. The intensity at $C$ therefore points straight down, along $c$. It is not zero, because nothing above the rim balances it.

Question 7.11

For the above problem, the direction of the gravitational intensity at an arbitrary point $P$ is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Solution. At $P$ the symmetry argument of Question 7.10 no longer works, since $P$ is off the axis, so use the gravitational potential instead. Complete the shell again, as in the figure above. Inside a uniform shell the intensity is zero, which means the potential is the same, $V_0$ say, at every point inside. The bowl and the added upper half are mirror images of each other in the plane of the rim, and $P$ lies in that plane, so the two halves contribute equal potentials there: each gives $V_0/2$. This holds at every point of the flat circular face, so the bowl’s own potential is constant all over that face.

A constant potential along a surface means no intensity along the surface: the component of the field in any direction is the rate at which the potential falls in that direction, and the potential does not change along the face. So the bowl’s intensity at $P$ has no component in the plane of the rim, and must point perpendicular to it. Since all the bowl’s mass lies below the rim, it points down, along $e$. The same argument gives $c$ at $C$, which is simply the special case of $P$ at the centre.

Question 7.12

A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun $= 2\times10^{30}\ \text{kg}$, mass of the earth $= 6\times10^{24}\ \text{kg}$. Neglect the effect of other planets etc. (orbital radius $= 1.5\times10^{11}\ \text{m}$).

Solution. Between the earth and the sun, the earth pulls the rocket back towards the earth and the sun pulls it on towards the sun. The net force is zero at the point $N$ where the two pulls are equal. If $N$ is a distance $x$ from the earth’s centre, it is $r – x$ from the sun’s, and

$$\frac{GM_Sm}{(r-x)^2} = \frac{GM_Em}{x^2} \quad\Longrightarrow\quad \frac{r-x}{x} = \sqrt{\frac{M_S}{M_E}} = \sqrt{\frac{2\times10^{30}}{6\times10^{24}}} = \sqrt{3.33\times10^5} = 577$$

So $r = 578x$ and

$$x = \frac{1.5\times10^{11}}{578} = 2.6\times10^8\ \text{m}$$

This is less than the moon’s distance: the neutral point lies very close to the earth because the sun, although far heavier, is so much farther away.

Question 7.13

How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5\times10^8\ \text{km}$.

Solution. The earth moves in a (nearly) circular orbit, and the only force keeping it there is the sun’s pull. Setting that pull equal to the centripetal force needed,

$$\frac{GM_SM_E}{r^2} = M_E\left(\frac{2\pi}{T}\right)^2 r \quad\Longrightarrow\quad M_S = \frac{4\pi^2r^3}{GT^2}$$

The earth’s own mass cancels, which is what lets us weigh the sun from the earth’s orbit alone. With $r = 1.5\times10^{11}\ \text{m}$ and $T = 365$ days $= 3.15\times10^7\ \text{s}$:

$$M_S = \frac{4\pi^2(1.5\times10^{11})^3}{(6.67\times10^{-11})(3.15\times10^7)^2} = \frac{1.33\times10^{35}}{6.63\times10^4} = 2.0\times10^{30}\ \text{kg}$$

Question 7.14

A saturn year is $29.5$ times the earth year. How far is the saturn from the sun if the earth is $1.50\times10^8\ \text{km}$ away from the sun?

Solution. Saturn and the earth orbit the same sun, so Kepler’s third law compares them directly:

$$\frac{r_{\text{Sat}}}{r_E} = \left(\frac{T_{\text{Sat}}}{T_E}\right)^{2/3} = (29.5)^{2/3} = 9.55$$

$$r_{\text{Sat}} = 9.55\times1.50\times10^8\ \text{km} = 1.43\times10^9\ \text{km} = 1.43\times10^{12}\ \text{m}$$

Question 7.15

A body weighs $63\ \text{N}$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Solution. The earth pulls as though all its mass were at its centre, so the force falls as the inverse square of the distance from the centre. At a height $R/2$ that distance is $3R/2$:

$$F = 63\times\left(\frac{R}{3R/2}\right)^2 = 63\times\frac49 = 28\ \text{N}$$

The approximation $g(h)\approx g(1 – 2h/R)$ cannot be used here: it holds only for $h \ll R$, and with $h = R/2$ it would give zero.

Question 7.16

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed $250\ \text{N}$ on the surface?

Solution. At a depth $d$, the shell of the earth outside the body exerts no net force on it, so only the sphere of radius $R – d$ below it pulls. For uniform density that inner mass is $M(R-d)^3/R^3$, and it acts from a distance $R – d$, so

$$g(d) = \frac{G M (R-d)^3/R^3}{(R-d)^2} = g\left(1 – \frac dR\right)$$

Half way down, $d = R/2$, and $g$ is halved:

$$W = 250\times\left(1 – \frac12\right) = 125\ \text{N}$$

Question 7.17

A rocket is fired vertically with a speed of $5\ \text{km s}^{-1}$ from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $= 6.0\times10^{24}\ \text{kg}$; mean radius of the earth $= 6.4\times10^6\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$.

Solution. The rocket rises far enough for $g$ to weaken appreciably, so $h = v^2/2g$ does not apply. Use energy conservation with the full potential energy $-GMm/r$ instead. At its highest point, a distance $r$ from the centre, the rocket is momentarily at rest:

$$\frac12mv^2 – \frac{GMm}{R} = -\frac{GMm}{r}$$

The rocket’s mass cancels. With $GM = (6.67\times10^{-11})(6.0\times10^{24}) = 4.00\times10^{14}\ \text{m}^3\,\text{s}^{-2}$:

$$\frac{GM}{r} = \frac{GM}{R} – \frac{v^2}{2} = 6.25\times10^7 – 1.25\times10^7 = 5.00\times10^7\ \text{J kg}^{-1}$$

$$r = \frac{4.00\times10^{14}}{5.00\times10^7} = 8.0\times10^6\ \text{m}$$

That is $8.0\times10^6 – 6.4\times10^6 = 1.6\times10^6\ \text{m}$ above the surface. Assuming constant $g$ would have given only $1.28\times10^6\ \text{m}$.

Question 7.18

The escape speed of a projectile on the earth’s surface is $11.2\ \text{km s}^{-1}$. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Solution. Far away the potential energy is zero, so energy conservation gives $\tfrac12mv^2 – \dfrac{GMm}{R} = \tfrac12mv_f^2$. The escape speed is defined by $\tfrac12mv_e^2 = \dfrac{GMm}{R}$, so this becomes

$$v_f^2 = v^2 – v_e^2 = (3v_e)^2 – v_e^2 = 8v_e^2$$

$$v_f = 2\sqrt2\times11.2 = 31.7\ \text{km s}^{-1}$$

It is not $3v_e – v_e$: energies subtract, not speeds.

Question 7.19

A satellite orbits the earth at a height of $400\ \text{km}$ above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite $= 200\ \text{kg}$; mass of the earth $= 6.0\times10^{24}\ \text{kg}$; radius of the earth $= 6.4\times10^6\ \text{m}$; $G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$.

Solution. The satellite orbits at $r = R + h = 6.4\times10^6 + 0.4\times10^6 = 6.8\times10^6\ \text{m}$ from the centre, and in a circular orbit its total energy is $-GMm/2r$. To leave the earth’s influence it must be brought to zero total energy, so the energy needed is

$$\Delta E = \frac{GMm}{2r} = \frac{(6.67\times10^{-11})(6.0\times10^{24})(200)}{2\times6.8\times10^6} = 5.9\times10^9\ \text{J}$$

Only half of $GMm/r$ is needed, because the satellite already has kinetic energy equal to the other half (Question 7.6).

Question 7.20

Two stars each of one solar mass ($= 2\times10^{30}\ \text{kg}$) are approaching each other for a head on collision. When they are a distance $10^9\ \text{km}$, their speeds are negligible. What is the speed with which they collide? The radius of each star is $10^4\ \text{km}$. Assume the stars to remain undistorted until they collide. (Use the known value of $G$.)

Solution. The stars start from rest, so their total momentum is zero and stays zero; being of equal mass, they must always move with equal speeds $v$ in opposite directions. They touch when their centres are two radii apart, $2R = 2\times10^7\ \text{m}$, having started $d = 10^{12}\ \text{m}$ apart. Energy is conserved between the two positions:

$$-\frac{GM^2}{d} = 2\times\frac12Mv^2 – \frac{GM^2}{2R}$$

$$v^2 = GM\left(\frac{1}{2R} – \frac1d\right) = (6.67\times10^{-11})(2\times10^{30})\left(5\times10^{-8} – 10^{-12}\right) = 6.67\times10^{12}\ \text{m}^2\,\text{s}^{-2}$$

$$v = 2.6\times10^6\ \text{m s}^{-1}$$

The $1/d$ term hardly matters: almost all the kinetic energy is gained in the last stages of the fall. Each star hits at $2.6\times10^6\ \text{m s}^{-1}$, so they close on each other at $5.2\times10^6\ \text{m s}^{-1}$.

Question 7.21

Two heavy spheres each of mass $100\ \text{kg}$ and radius $0.10\ \text{m}$ are placed $1.0\ \text{m}$ apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Solution. Each sphere acts as a point mass at its centre, and the midpoint is $0.5\ \text{m}$ from both. Their pulls there are equal in size and opposite in direction, so the net gravitational force (and intensity) is zero.

Potential is a scalar, so the two contributions simply add, and each is negative:

$$V = -\frac{GM}{r} – \frac{GM}{r} = -\frac{2(6.67\times10^{-11})(100)}{0.5} = -2.7\times10^{-8}\ \text{J kg}^{-1}$$

An object at the midpoint feels no net force, so it is in equilibrium. Displace it slightly towards one sphere, though, and it is now nearer that sphere, whose pull grows, and farther from the other, whose pull weakens; the net force carries it further from the midpoint. The equilibrium is therefore unstable. (A small displacement at right angles to the line is pulled back, but an equilibrium that fails in any one direction is unstable.)

Note on the book’s answer. The key prints the potential as $2.7\times10^{-8}\ \text{J/kg}$, without a sign. With the zero of potential at infinity, the gravitational potential of any mass is negative everywhere, so the value is $-2.7\times10^{-8}\ \text{J/kg}$.

Common mistakes

• Question 7.3: taking the orbit size as proportional to the period. “Twice as fast” halves the period, but by Kepler’s third law $R\propto T^{2/3}$, so the orbit shrinks to $0.63$ of the earth’s, not to a half or a quarter.
• Question 7.5: using the diameter as the orbital radius, or leaving it in light years. The star is at $5\times10^4\ \text{ly}$ from the centre, and that must become $4.73\times10^{20}\ \text{m}$ before it goes into $T = 2\pi\sqrt{r^3/GM}$ — which is why the diameter is given only to show that the star is at the edge.
• Question 7.12: forgetting the square root. Equating the two pulls gives $(r-x)^2/x^2 = M_S/M_E$, so the distances are in the ratio $\sqrt{M_S/M_E} = 577$, not $3.3\times10^5$. Students also measure $x$ from the sun when the question asks for the distance from the earth’s centre.
• Question 7.15: using $g(1 – 2h/R)$. That approximation needs $h \ll R$. At $h = R/2$ it gives a weight of zero, while the exact inverse-square law gives $28\ \text{N}$.
• Question 7.16: using the height formula for depth. Below the surface $g$ falls linearly, $g(1 – d/R)$, because the outer shell stops pulling. Using $R^2/(R+d)^2$ gives $111\ \text{N}$ instead of $125\ \text{N}$.
• Question 7.17: using $v^2 = 2gh$. With $g$ treated as constant the rocket seems to reach only $1280\ \text{km}$. The pull weakens as it climbs, so it actually reaches $1600\ \text{km}$ above the surface, and the question’s “how far from the earth” is often answered with the height when the key gives the distance from the centre, or the other way round.
• Question 7.19: using $GMm/r$ instead of $GMm/2r$. That is the energy needed to free a stationary object at that height, $1.2\times10^{10}\ \text{J}$; the orbiting satellite already has half of it as kinetic energy.
• Question 7.21: reporting the potential as positive. Gravitational potential with its zero at infinity is always negative; the book’s own key drops the sign here.

Practise next

• Chapter 8, Mechanical Properties of Solids — the next chapter, which turns from the forces between bodies to how a solid body deforms under forces.
• Chapter 5, Work, Energy and Power — Questions 7.17 to 7.20 are conservation of mechanical energy with a force that changes with distance; revise the work–energy theorem there.
• Chapter 3, Motion in a Plane — the centripetal acceleration $v^2/r = \omega^2r$ behind every orbit calculation from 7.3 to 7.14.