NCERT Solutions for Class 11 Physics Chapter 7: Gravitation

Concept. These three questions probe the special properties of the gravitational force: it is always attractive (there is no negative mass), it cannot be screened, it is the same for all masses (equivalence), and its variation with distance produces tidal effects. Answer each with the underlying principle.

(a) Can gravity be shielded?

No. Electric shielding works because the free charges of a conductor rearrange themselves on the surface (induced charges of both signs) to cancel the field inside. In gravity there is only one kind of ‘charge’ (mass), which is always positive, and gravity is always attractive, so no distribution of matter can produce an opposite (repulsive) field to cancel the external gravitational field at the point inside. A hollow sphere shields the body from the gravity of the sphere itself (the force due to a uniform shell on a particle inside is zero, Newton’s shell theorem), but it does not shield it from the gravitational influence of the outside matter. Gravitational forces cannot be screened because the force on a mass is determined by the sum of all attractions.

(b) Large space station

Yes. In a small spacecraft in free fall the astronaut and everything around him have the same acceleration (that of the gravity at the location), so he feels weightless and cannot detect gravity. But gravity varies from point to point (it is not uniform): in a large station the gravitational field at different places differs in magnitude (1/r²) and direction (towards the Earth’s centre). Objects at different places then accelerate by slightly different amounts relative to the station, i.e. tidal effects show up: for example two balls placed at different heights, or at different horizontal positions, will move relative to each other (the tidal force). So with sensitive measurements over a large size one can detect gravity.

(c) Tides

The gravitational force of a body on the Earth is F = GMm/d², so the Sun (larger mass, much larger distance) pulls harder than the Moon: F_Sun/F_Moon ≈ 180. But the tidal effect depends on the difference of the force between the near and far sides of the Earth, i.e. on the gradient dF/dd, which varies as 1/d³: tidal effect ∝ M/d³. Since the Moon is much closer (3.84 × 10⁸ m against 1.5 × 10¹¹ m), M/d³ for the Moon exceeds that of the Sun: (7.3 × 10²²/(3.84 × 10⁸)³) versus (2 × 10³⁰/(1.5 × 10¹¹)³) = 1.3 × 10⁻³ versus 5.9 × 10⁻⁴ (kg m⁻³) ⇒ the Moon’s tidal effect is about 2.2 times larger. Hence the Moon’s tidal effect is greater.

Extra depth. Tidal bulges form on the side nearest the Moon and on the far side because the near side is pulled more than the Earth’s centre, while the centre is pulled more than the far side. The Sun’s tidal effect is about 45% of the Moon’s; at new moon and full moon (syzygy) the two add up to give spring tides, and at the quarter phases they partly cancel, giving neap tides. The same differential gravity (tidal force) explains why a body falling into a black hole is stretched (‘spaghettification’).

More to know — equivalence and tides. The tidal acceleration across the Earth due to a body of mass M at distance d is Δg = 2GMR_E/d³. For the Moon Δg = 1.1 × 10⁻⁶ m s⁻²; for the Sun 0.5 × 10⁻⁶ m s⁻² — so the Moon’s tidal effect is 2.2 times that of the Sun, and the highest spring tides occur when the Sun, Moon and Earth are in line. The principle that all bodies fall with the same acceleration (the equivalence of gravitational and inertial mass) is why an astronaut in a small spaceship cannot detect gravity, but in a large one the differences of g across the ship (tides) become measurable.

Conclusion: (a) No; (b) yes, through tidal effects; (c) tidal effect varies as 1/d³, so the nearer Moon has the greater tidal effect.

Concept. Recall the formulae: g = GM/R² at the surface; g(h) = g(1 − 2h/R) above, g(d) = g(1 − d/R) below the surface; potential energy difference exact: ΔU = −GMm(1/r₂ − 1/r₁).

1. At the surface g = 9.8 m s⁻²; at a height h = R_E (6400 km) g falls to g/4 = 2.45 m s⁻²; at depth d = R_E/2, g = g/2 = 4.9 m s⁻².
2. (c) In the formula for the Earth’s surface g depends on M_E and R_E: it is independent of the body’s mass.
3. (d) Check: −GMm(1/r₂ − 1/r₁) = GMm(r₂ − r₁)/(r_1r2) ≈ GMm(r₂ − r₁)/r² = mg(r₂ − r₁) when r₁ ≈ r₂ ≈ R and g = GM/R².

Choices: (a) decreases; (b) decreases; (c) mass of the body; (d) more.

Extra depth. For (b), remember the assumption in the question (uniform density). In reality g first increases slightly with depth (the Earth’s core is denser) reaching a maximum near the core–mantle boundary at about 10.7 m s⁻² and then decreases to zero at the centre. Parts (a) and (b) can be remembered with one picture: g is maximum at the surface (for the uniform Earth) and falls as 1/r² above and linearly below. Part (d) shows the proper use of gravitational potential energy in Ex 7.17–7.19.

More to know — g at different places. g falls from 9.83 m s⁻² at the poles to 9.78 m s⁻² at the equator (the Earth’s rotation, centrifugal effect mω²R cos²λ, and the oblate shape); at the top of Mt Everest (8.8 km) g = 9.77 m s⁻², a decrease of 0.28% (= 2h/R_E). In a mine 3 km deep g(d) = g(1 − d/R) would be 0.05% smaller for a uniform Earth. These small differences are used in geophysics (gravimetry) to find ore bodies and oil.

Conclusion: (a) decreases; (b) decreases; (c) mass of the body; (d) more.

Concept. Use Kepler’s third law: T² ∝ r³ (semi-major axis or radius). ‘Twice as fast’ means that the planet completes its revolution in half the time: T_p = T_E/2 (the period is halved, not the speed in m/s doubled; for the same orbit, twice the angular speed). Compare the orbits with the ratio of periods.

1. Kepler’s law: T² = k r³ with the same constant k = 4π²/GM_S for all planets.
2. (T_p/T_E)² = (r_p/r_E)³ ⇒ (1/2)² = (r_p/r_E)³ ⇒ r_p/r_E = (1/4)^1/3 = 0.63.
3. r_p = 0.63 r_E (about 0.63 AU, i.e. 9.4 × 10¹⁰ m).

Remark: the planet is smaller than the Earth’s orbit (inside it), between Mercury (0.39 AU) and Venus (0.72 AU). If ‘twice as fast’ were taken as twice the orbital speed then v ∝ 1/√r gives r = r_E/4 and T = T_E/8; the NCERT solution interprets ‘twice as fast’ as an orbital period half that of the Earth: r = 0.63 r_E.

Extra depth. In terms of the law: planets closer to the Sun move faster and have shorter periods (Mercury 88 days, Earth 365 days, Neptune 165 years). The relation r ∝ T^2/3 is used all the time in astronomy: a planet with period 8 years would orbit at r = 4 r_E (the cube root of 64 is 4). Kepler’s third law holds for all bodies bound to the same central mass (e.g., all satellites of Jupiter).

More to know — Kepler’s third law in its general form. For a body of mass m orbiting a central mass M in a circular orbit: T² = 4π²r³/GM; the constant depends only on M, so planets of the Sun share one constant (2.97 × 10⁻¹⁹ s² m⁻³) and satellites of Jupiter another. Corollary used above: if the period is halved the radius is multiplied by (1/2)^2/3 = 0.63; if the radius is doubled the period becomes 2^3/2 = 2.83 times as long. Exact Newtonian form with both masses: T² = 4π² a³/G(M + m).

Conclusion: The planet’s orbital radius is (1/4)^1/3 = 0.63 times that of the Earth.

Concept. Io moves around Jupiter under Jupiter’s gravity just as a planet moves around the Sun: the centripetal force is supplied by gravity, giving T² = 4π²r³/(GM). So M = 4π²r³/(GT²). Compute M_J from the data of Io; compute (or recall) the mass of the Sun (2 × 10³⁰ kg, found from the Earth’s orbit in Ex 7.13) and compare.

1. T = 1.769 d = 1.769 × 86400 s = 1.528 × 10⁵ s; r = 4.22 × 10⁸ m.
2. M_J = 4π²r³/(GT²) = 4 × (3.14)² × (4.22 × 10⁸)³/[6.67 × 10⁻¹¹ × (1.528 × 10⁵)²] = (39.5 × 7.52 × 10²⁵)/(6.67 × 10⁻¹¹ × 2.335 × 10¹⁰) = 2.97 × 10²⁷/1.557 ≈ 1.9 × 10²⁷ kg.
3. Mass of the Sun M_S = 2.0 × 10³⁰ kg (from the Earth’s orbit: M_S = 4π²r_E³/(GT_E²), Ex 7.13).
4. Ratio M_J/M_S = 1.9 × 10²⁷/2.0 × 10³⁰ = 0.95 × 10⁻³ ≈ 1/1000. Hence the mass of Jupiter is about one-thousandth that of the Sun. Shown.

Comparison: Jupiter is 318 times as massive as the Earth (6.0 × 10²⁴ kg) and the Sun is 1050 times as massive as Jupiter (the exact ratio is 1/1047).

Extra depth. This is the standard method to weigh any celestial body that has a satellite: measure the period and the radius of the orbit of the satellite. The Galilean moons (Io, Europa, Ganymede, Callisto) all give the same M_J, confirming Kepler’s third law with a constant 4π²/GM_J. Note that the mass of the satellite does not matter (it cancels), provided it is much smaller than that of the planet.

More to know — the Galilean moons. Jupiter’s four large moons obey the same law T² ∝ r³ with the constant 4π²/GM_J = 3.1 × 10⁻¹⁶ s² m⁻³: Io (1.77 d, 4.22 × 10⁸ m), Europa (3.55 d, 6.71 × 10⁸ m), Ganymede (7.15 d, 1.07 × 10⁹ m), Callisto (16.7 d, 1.88 × 10⁹ m). All give M_J = 1.9 × 10²⁷ kg. Thus the satellites of a planet serve as natural balances: Mars’s Phobos gives the mass of Mars (6.5 × 10²³ kg, Example 7.5), the Moon gives the mass of the Earth (6.0 × 10²⁴ kg, Example 7.6) and the Earth’s orbit gives the mass of the Sun (Ex 7.13).

Conclusion: M_Jupiter = 1.9 × 10²⁷ kg, about 10⁻³ times the mass of the Sun (2 × 10³⁰ kg).

Concept. Assume that the whole mass of the galaxy is concentrated near the galactic centre (the star orbits as if a planet around a central mass) and apply the formula for a circular orbit under gravity: T = 2π√(r³/GM) with M = 2.5 × 10¹¹ solar masses and r = 5 × 10⁴ ly (the star is at the edge: radius = half of the diameter 10⁵ ly).

1. M = 2.5 × 10¹¹ × 2.0 × 10³⁰ kg = 5.0 × 10⁴¹ kg.
2. r = 5 × 10⁴ ly = 5 × 10⁴ × 9.46 × 10¹⁵ m = 4.73 × 10²⁰ m (1 ly = 9.46 × 10¹⁵ m).
3. T = 2π√(r³/GM) = 2 × 3.14 × √[(4.73 × 10²⁰)³/(6.67 × 10⁻¹¹ × 5.0 × 10⁴¹)] = 6.28 × √[1.058 × 10⁶²/3.335 × 10³¹] = 6.28 × √(3.17 × 10³⁰) = 6.28 × 1.78 × 10¹⁵ = 1.12 × 10¹⁶ s.
4. In years: 1.12 × 10¹⁶/(3.156 × 10⁷) = 3.55 × 10⁸ years (about 355 million years).

Sense of scale. The star’s orbital speed is v = 2πr/T = 2.65 × 10⁵ m s⁻¹ = 265 km s⁻¹ (the actual Sun, at 26,000 ly, takes about 2.3 × 10⁸ years for one galactic revolution).

Extra depth. The method shows how astronomers estimate the mass of a galaxy from the rotation periods of stars: if the observed period is shorter than predicted from the visible matter, more mass (dark matter) must be present. Treating the entire mass as concentrated at the centre is a good approximation only for a spherically symmetric distribution (by Newton’s shell theorem, only the mass within the orbit attracts the star); the real galaxy is a flattened disc, so the result is approximate (it agrees in order of magnitude with the Sun’s galactic year of about 2 × 10⁸ years).

Conclusion: The star takes about 1.12 × 10¹⁶ s (3.5 × 10⁸ years) to complete one revolution.

Concept. For a satellite in a circular orbit of radius r: K = GMm/2r, U = −GMm/r (zero at infinity) and E = K + U = −GMm/2r. These relations decide both parts.

(a)

E = −GMm/2r and K = +GMm/2r, so E = −K: the total energy is the negative of the kinetic energy (also E = U/2). Hence the answer is ‘kinetic’. (The total energy is negative because the satellite is bound to the Earth; E < 0 means a closed orbit.)

(b)

1. To escape from the Earth’s influence the final energy must be zero (at infinity with zero speed).
2. Orbiting satellite at height h (radius r): initial energy E₁ = −GMm/2r; energy to be supplied = 0 − E₁ = +GMm/2r.
3. Stationary object at the same height: its kinetic energy is zero and U = −GMm/r, so E₂ = −GMm/r; energy to be supplied = +GMm/r.
4. Comparison: GMm/2r < GMm/r. The orbiting satellite already possesses kinetic energy GMm/2r, which reduces the energy needed. So the energy required is less for the orbiting satellite (half of that for the stationary object).

Check with numbers: for a 200 kg satellite at 400 km (r = 6.8 × 10⁶ m) the escape energy is GMm/2r = 5.9 × 10⁹ J (Ex 7.19), while a stationary 200 kg object at the same height would need 1.18 × 10¹⁰ J.

Extra depth. The relations K = −E and U = 2E are valid for all circular orbits and also for elliptical orbits (with r replaced by the semi-major axis a for E). They give rise to the apparent paradox of Ex 7.5c in Chapter 5: air drag decreases E, so K = −E increases and the satellite speeds up as it falls to a lower orbit. In mission planning this means that it takes less energy to push a satellite from orbit to escape than to raise a body from the ground to the same height: the satellite’s orbital kinetic energy is ‘free’.

Conclusion: (a) kinetic; (b) less.

Concept. The escape speed is found by requiring that the total energy of the body at launch is zero: ½mv_e² − GMm/r = 0, which gives v_e = √(2GM/r) where r is the distance of the launch point from the centre of the Earth. This formula has no m and no direction in it, and contains the distance r only.

1. Escape speed from the surface v_e = √(2gR_E) = √(2 × 9.8 × 6.4 × 10⁶) = 11.2 km s⁻¹.
2. At a height h = R_E, r = 2R_E and v_e = 11.2/√2 = 7.9 km s⁻¹.
3. Therefore the escape speed depends on the location (distance from the centre) and height, but not on the mass of the body or the direction of projection.

Note. In the NCERT key the answers are: (a) No; (b) Yes; (c) No; (d) Yes. Strictly ‘location’ matters only via r (and, very slightly, through the Earth’s rotation, which assists a launch eastwards at the equator); direction matters only if one counts the rotation (a body projected eastwards has the Earth’s surface speed added).

Extra depth. That the escape speed is independent of the mass means that a tiny gas molecule and a rocket need the same v_e = 11.2 km s⁻¹ to leave the Earth; that is why lighter gases (H₂, He) have more chance to escape since their rms speeds are larger at the same temperature (v_rms ∝ 1/√M), which explains why the Earth’s atmosphere has almost no hydrogen or helium, while the Moon (v_e = 2.4 km s⁻¹) has no atmosphere at all.

Conclusion: (a) No; (b) Yes (through the distance from the centre); (c) No; (d) Yes.

Concept. The comet moves under the gravitational attraction of the Sun, a central, conservative force. For such a motion the angular momentum about the Sun and the total energy are conserved, while the speed, the kinetic energy and the potential energy vary with the distance from the Sun (Kepler’s law of areas: the comet moves faster near the Sun and slower far from it).

1. Perihelion: smallest r, largest speed and K, most negative U.
2. Aphelion: largest r, smallest speed and K, least negative U.
3. The decrease of U on approach equals the increase of K, so the total E stays constant. The areal speed dA/dt = L/2m is constant.

Answer: constant: (c) angular momentum and (f) total energy; not constant: (a), (b), (d), (e).

Extra depth. Halley’s comet (period 76 years) has a perihelion distance of 0.59 AU and an aphelion of 35 AU, so its speed varies from about 54 km s⁻¹ at perihelion to 0.9 km s⁻¹ at aphelion (a ratio 35/0.59 = 59). A comet loses mass (the tail) near the Sun, which the question asks us to neglect. In a circular orbit all six quantities are constant; the elliptical orbit makes (a), (b), (d), (e) vary.

More to know — Kepler’s laws and comets. Perihelion and aphelion speeds of a comet are in the inverse ratio of their distances (v_p/v_A = r_A/r_p). The areal velocity dA/dt = L/2m = ½ r v sinθ is constant, so a comet near the Sun sweeps a long thin sector quickly, and the orbital period follows T² ∝ a³ where a = (r_p + r_A)/2 (for Halley’s comet a = 17.8 AU, T = 76 y). Energy: E = −GMm/2a is the same at every point of the orbit, K + U = const.

Conclusion: The comet’s angular momentum and total energy are constant; its linear speed, angular speed, kinetic energy and potential energy are not.

Concept. In an orbiting spacecraft the astronaut is in free fall (apparent weightlessness): the downward pull of gravity no longer pulls body fluids towards the feet, there is no ‘up’ or ‘down’, and the inner ear (organ of balance) gets no gravitational cue.

1. Body fluids shift from the legs to the upper body (about 1–2 litres) within the first hours in orbit.
2. The vestibular organs of the inner ear cannot sense the direction of gravity; the brain receives conflicting signals from eyes and the inner ear.
3. Long stays also cause loss of bone and muscle mass because the body bears no weight.

Answer: (b), (c) and (d) are likely; (a) is not.

Extra depth. The cause is the continuous free fall of the spacecraft, which makes the effective gravity zero inside it: not the absence of gravity itself (at 400 km height g is still 8.7 m s⁻²). To protect their health astronauts exercise for about two hours a day, wear compression garments and take medicines for space sickness. Tests with the ‘vomit comet’ (aircraft in parabolic free fall) are used to train astronauts.

More to know — effects of weightlessness on the body. In orbit the heart no longer works against gravity, so about 1–2 litres of fluid shift from the legs to the chest and head within hours; astronauts show thin ‘bird legs’ and a puffy face. The vestibular system of the inner ear, which senses linear acceleration and the direction of gravity, gives conflicting signals and causes space adaptation sickness in about half of astronauts during the first 2–3 days. Over months bone mineral is lost at about 1–1.5% per month and muscles waste unless the crew exercises. The ISS has gravity g = 8.7 m s⁻², so the ‘weightlessness’ is due to free fall only.

Conclusion: Swollen face, headache and orientational problems are likely; swollen feet are not.

Concept. For a complete uniform spherical shell the gravitational intensity inside is zero. A hemispherical shell is half of such a shell. The intensity at the centre due to the two halves must add to zero, so the field due to the lower (given) half is equal and opposite to that of the missing upper half. Both are not zero, so the answer is not (iv).

1. The shell lies in the lower half; the centre C is at the middle of the rim plane.
2. Every mass element of the shell attracts the unit mass at C towards itself. By symmetry the horizontal components of the attractions of elements on opposite sides cancel; the components along the axis (vertical) all point downward, towards the shell, and add up.
3. Hence the net intensity at C is directed along the axis of symmetry towards the shell, i.e. in the direction c.
4. The correct alternative is (iii) c.

Check with a limiting case. If the missing upper hemisphere were added, its field at C (upward) would cancel the field of the lower hemisphere (downward): total zero, as required for a full shell; so the field of the lower hemisphere alone is non-zero and downward.

Extra depth. This result can be generalised: the field at the centre of a hemispherical shell of mass M and radius R is g = GM/2R² directed along the axis towards the shell (half of what a point mass M at the centre of the hollow would produce, and the same magnitude as the uniform field of a half shell). The direction is the axis of symmetry in all such problems; use symmetry to cancel perpendicular components before computing.

Conclusion: The intensity at the centre of the hemispherical shell is directed along c, (iii).

Concept. This question refers to the same hemispherical shell as Ex 7.10, with P an arbitrary point inside the rim plane. For a complete spherical shell the intensity is zero everywhere inside, so the field of the lower hemisphere at P equals the negative of the field of the missing upper hemisphere. The two halves have a symmetry property that fixes the direction.

1. At any point P inside the hemispherical shell, in the plane of its rim (or inside the bowl), the intensity due to the shell is the same in direction as at the centre: along the axis of symmetry, i.e. perpendicular to the plane of the rim, directed towards the shell (downward).
2. Reason (superposition and mirror symmetry). A complete shell has zero field at P. So the field of the lower half at P is the negative of the field of the upper half. The upper half is the mirror image of the lower half in the plane of the rim, and P lies in that plane: the in-plane (horizontal) components of the two fields are equal, while their vertical components are opposite. If the lower half had a horizontal component the two halves would add it up instead of cancelling it, contradicting the zero total; hence the horizontal component is zero and only a vertical component remains.
3. Hence at every point of the plane of the rim inside the shell the field has only a vertical component, pointing downward: arrow e.

Answer: (ii) e: the intensity at P is vertical (parallel to the axis), pointing towards the shell.

Extra depth. This is a classic result: the gravitational (or electric) field at any point on the plane of the rim of a uniformly charged/massive hemispherical shell is normal to the plane. For points off the plane, inside the bowl, the field is still approximately along the axis but not exactly. Using the symmetry argument saves an integration. Compare with the field at the centre (Ex 7.10), which has the same direction (c) and the same sense.

Conclusion: The intensity at P is directed along e (vertically downward, towards the shell).

Concept. Between the Earth and the Sun there is a point where the Earth’s pull on the rocket is exactly cancelled by the Sun’s pull (a neutral point, like the point N of Example 7.4). At such a point GM_Em/x² = GM_Sm/(r − x)², where x is the distance from the Earth’s centre and r the Earth–Sun distance.

1. Set the magnitudes equal: M_E/x² = M_S/(r − x)² (G and m cancel).
2. Take the square root: (r − x)/x = √(M_S/M_E) = √(2 × 10³⁰/6 × 10²⁴) = √(3.33 × 10⁵) = 577.
3. r − x = 577x ⇒ r = 578x ⇒ x = r/578 = 1.5 × 10¹¹/578 = 2.6 × 10⁸ m from the centre of the Earth.

Remark: the point is only about 2.6 × 10⁸ m (260,000 km) from the Earth, a distance comparable to the Earth–Moon separation (3.84 × 10⁸ m). Beyond it the Sun’s attraction dominates the rocket. (The true ‘sphere of influence’ of the Earth is bigger, about 9.2 × 10⁸ m, since the Sun also accelerates the Earth.)

Extra depth. The root x = r/(1 − 577)… also exists on the far side of the Earth? The equation (r − x)/x = ±577 has another solution with the negative sign, x = r/(1 − 577) = −2.6 × 10⁸ m, which lies on the opposite side of the Earth from the Sun: there the forces of the two bodies are in the same direction, so they cannot cancel; therefore it is not a point of zero force. Only the point between the two bodies counts.

Conclusion: The gravitational force on the rocket is zero at 2.6 × 10⁸ m from the centre of the Earth, towards the Sun.

Concept. To ‘weigh’ the Sun we use the motion of a planet (the Earth) around it. The gravitational pull of the Sun on the Earth provides the centripetal force of the Earth’s (nearly circular) orbit: GM_SME/r² = M_E(4π²/T²)r. The mass of the Earth cancels, and M_S = 4π²r³/(GT²).

1. Orbital radius r = 1.5 × 10⁸ km = 1.5 × 10¹¹ m; period T = 1 year = 365.25 × 24 × 3600 s = 3.156 × 10⁷ s; G = 6.67 × 10⁻¹¹ N m² kg⁻².
2. Equate gravity to the centripetal force: GM_SME/r² = M_Eω²r with ω = 2π/T ⇒ M_S = 4π²r³/(GT²).
3. r³ = (1.5 × 10¹¹)³ = 3.375 × 10³³ m³; 4π² = 39.48; numerator = 1.332 × 10³⁵; T² = 9.96 × 10¹⁴ s²; GT² = 6.64 × 10⁴.
4. M_S = 1.332 × 10³⁵/6.64 × 10⁴ = 2.0 × 10³⁰ kg.

Cross-check by Kepler’s law: T²/r³ = 4π²/GM_S = 2.96 × 10⁻¹⁹ s² m⁻³, in agreement with Kepler’s constant (2.96 × 10⁻³⁴ y² m⁻³ in Table 7.1 when T is in years: dividing by (3.156 × 10⁷)² gives 2.97 × 10⁻³⁴).

Extra depth. The Sun contains 99.86% of the mass of the solar system. The same procedure works for any planet: Jupiter’s year of 11.9 y and orbital radius 7.78 × 10¹¹ m give the same M_S. It does not require the knowledge of the Earth’s mass; all planetary data must agree on one value of M_S, confirming the law. Likewise the Earth’s mass is found from the Moon’s orbit (Example 7.6) and Jupiter’s from Io (Ex 7.4).

Conclusion: M_S = 4π²r³/(GT²) ≈ 2 × 10³⁰ kg.

Concept. Both Saturn and the Earth orbit the Sun, so Kepler’s third law applies: T² ∝ r³, i.e. (T_S/T_E)² = (r_S/r_E)³. Hence r_S = r_E (T_S/T_E)^2/3.

1. T_S/T_E = 29.5; r_E = 1.50 × 10⁸ km.
2. (r_S/r_E)³ = (29.5)² = 870.25 ⇒ r_S/r_E = (870.25)^1/3 = 9.55.
3. r_S = 9.55 × 1.50 × 10⁸ km = 1.43 × 10⁹ km (1.43 × 10¹² m).

In astronomical units: r_S = 9.55 AU (actual value 9.54 AU).

Extra depth. Kepler’s third law gives only the ratio of orbital radii if the central mass is the same; to get an absolute radius you need one reference value (here the Earth’s orbital radius). The shortcut is r ∝ T^2/3: for T = 29.5 the factor is 9.55. The speed of Saturn in its orbit is smaller than the Earth’s by the factor √(1/9.55) = 0.32, i.e. 9.6 km s⁻¹ against 29.8 km s⁻¹.

More to know — periods and speeds of all planets. From T ∝ r^3/2: Mercury (r = 0.39 AU) T = 0.24 y; Venus 0.72 AU, 0.615 y; Earth 1 AU, 1 y; Mars 1.52 AU, 1.88 y; Jupiter 5.2 AU, 11.9 y; Saturn 9.55 AU, 29.5 y; Uranus 19.2 AU, 84 y; Neptune 30.1 AU, 165 y. Orbital speeds fall as 1/√r: Mercury 47.9, Earth 29.8, Saturn 9.7, Neptune 5.4 km s⁻¹. Since T²/r³ = 4π²/GM_S is the same for all, the mass of the Sun can be computed from any planet; e.g. using Saturn: M_S = 4π² (1.43 × 10¹²)³/(6.67 × 10⁻¹¹ × (9.31 × 10⁸)²) = 2.0 × 10³⁰ kg.

Conclusion: Saturn is 1.43 × 10⁹ km (9.55 AU) from the Sun.

Concept. The weight at the surface is W = mg = GMm/R² = 63 N. At a distance r from the Earth’s centre the gravitational force is F = GMm/r². At a height h = R/2, r = R + R/2 = 3R/2, so F = GMm/(1.5R)² = W/(1.5)².

1. Weight at the surface W = GMm/R² = 63 N.
2. At height h = R/2: r = R + h = 1.5R.
3. F = GMm/r² = GMm/(1.5R)² = (1/2.25) GMm/R² = W/2.25.
4. F = 63/2.25 = 28 N.

Check with the approximate formula g(h) ≈ g(1 − 2h/R): it gives 63 × (1 − 1) = 0 for h = R/2, which is wrong because the expansion is valid only for h ≪ R; for large heights we must use the exact inverse-square formula.

Extra depth. The mass of the body is m = 63/9.8 = 6.4 kg; it stays the same at height, only its weight changes. At the height R/2 (3200 km), the force is 44% of the surface value. Do not use g(h) = g(1 − 2h/R) when h is not small compared with R (the NCERT solution uses the exact formula).

More to know — how weight varies. Weight W(h) = W(0)(R/(R + h))². At the height of the International Space Station (400 km) the weight is still 88% of the surface weight (the astronauts are weightless only because they are in free fall, not because gravity is absent). At the geostationary height (3.6 × 10⁷ m = 5.6 R) the weight is only 2.3% and at the Moon’s distance (60 R) 0.028%. Mass is constant; weight depends on the location, which is why a spring balance reads differently at the equator and the poles (by about 0.5%).

More to know — check with formula. W′ = W[R/(R + h)]² = 63 × (2/3)² = 63 × 0.444 = 28 N. The first-order formula g(h) = g(1 − 2h/R) would give zero; the exact inverse-square relation must be used for large h. Compare with altitude of the ISS (h = 0.0625R): exact factor 0.884 versus first-order 0.875, which differ by only 1%. This shows the range of validity of the binomial approximation (h ≤ R/16 for 1% accuracy).

Conclusion: The gravitational force at a height R/2 is 28 N.

Concept. For a uniform Earth, the gravitational force on a body at depth d is due only to the mass inside the sphere of radius (R − d) (the outer shell exerts no net force): g(d) = g(0)(1 − d/R), so the weight at depth d is W(d) = W(0)(1 − d/R). Half way down to the centre means d = R/2.

1. Surface weight W(0) = 250 N.
2. At d = R/2: W(d) = 250 × (1 − 1/2) = 125 N.
3. Derivation: g(d) = GM(r)/r² with M(r) = M (r/R)³ and r = R − d ⇒ g = GM r/R³ = g(0)(r/R); at r = R/2, g = g(0)/2.

Contrast with Ex 7.15: at a height R/2 above the surface the weight falls to 28 N out of 63 N (44%), but at a depth R/2 below the surface it falls to 50%: the weight falls off faster above (inverse square) than below (linear), for the same magnitude of displacement.

Extra depth. The linear decrease of g inside a uniform Earth is also the reason for the simple harmonic motion of a body dropped in a tunnel through the centre of the Earth: F = −(mg/R)x, with period T = 2π√(R/g) = 84.4 min (the same as the period of a satellite orbiting just above the surface). For the real Earth the density increases towards the core, so g does not fall linearly.

More to know — tunnel through the Earth. Inside a uniform Earth the restoring force on a body at distance r from the centre is F = −(mg/R) r: simple harmonic motion with angular frequency ω = √(g/R) and period T = 2π√(R/g) = 84.4 min. A body dropped into a tunnel through the centre would pass the centre with speed v = √(gR) = 7.9 km s⁻¹ and emerge on the other side 42 minutes later: the same time as for any straight tunnel (chord) between two points on the surface. The weight half way down is 125 N; at the centre it is zero (all attractions cancel).

More to know — the real Earth. Seismology shows that the Earth’s density rises from 2.7 g cm⁻³ near the crust to about 13 g cm⁻³ at the centre; g(r) therefore stays almost constant at 9.8–10.7 m s⁻² through the mantle and only falls to zero in the inner core (the PREM model). The uniform-density model of the question gives half the weight at half depth; for the real Earth the weight half way down is nearly as large as at the surface (about 125–230 N for a 250 N body, depending on the radius chosen for ‘half way’).

Conclusion: The body would weigh 125 N half way down to the centre.

Concept. The rocket moves under gravity alone (neglect air resistance and the Earth’s rotation), so the total mechanical energy is conserved. It rises until its speed becomes zero at the maximum height h above the surface, then falls back. Use the exact potential energy U = −GMm/r (not mgh) because the height is large.

1. Initial energy at the surface (r = R): E_i = ½mv² − GMm/R, with v = 5 × 10³ m s⁻¹.
2. At the highest point (r = R + h) the speed is zero: E_f = −GMm/(R + h).
3. Conservation: ½mv² − GMm/R = −GMm/(R + h).
4. Divide by m: ½v² = GM[1/R − 1/(R + h)]. GM = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ = 4.002 × 10¹⁴; GM/R = 4.002 × 10¹⁴/6.4 × 10⁶ = 6.253 × 10⁷ J kg⁻¹; ½v² = ½(25 × 10⁶) = 1.25 × 10⁷ J kg⁻¹.
5. So GM/(R + h) = 6.253 × 10⁷ − 1.25 × 10⁷ = 5.003 × 10⁷ ⇒ R + h = 4.002 × 10¹⁴/5.003 × 10⁷ = 8.0 × 10⁶ m.
6. h = 8.0 × 10⁶ − 6.4 × 10⁶ = 1.6 × 10⁶ m above the surface (distance from the centre 8.0 × 10⁶ m).

Comparison with the mgh formula: h = v²/2g = (5000)²/19.6 = 1.28 × 10⁶ m (1280 km), which underestimates the true height by 20% because g decreases with height; the exact method gives 1600 km.

Extra depth. The rocket speed 5 km s⁻¹ is less than the escape speed (11.2 km s⁻¹), so the rocket returns. In terms of the escape speed: h = R v²/(v_e² − v²) = 6.4 × 10⁶ × 25/(125.4 − 25) = 1.6 × 10⁶ m. The same energy-conservation approach is used in all questions on orbits and escape.

Conclusion: The rocket rises to 1.6 × 10⁶ m above the Earth’s surface (8.0 × 10⁶ m from the centre).

Concept. A body projected with speed v = 3v_e has total energy E = ½mv² − GMm/R. At infinity (far away) the potential energy is zero, so the total energy is purely kinetic: ½mv_∞² = E. With ½mv_e² = GMm/R this gives v_∞² = v² − v_e².

1. v = 3v_e = 3 × 11.2 = 33.6 km s⁻¹.
2. Energy conservation: ½mv² − GMm/R = ½mv_∞².
3. Since GMm/R = ½mv_e²: v_∞² = v² − v_e² = (3v_e)² − v_e² = 8v_e².
4. v_∞ = √8 v_e = 2.83 × 11.2 = 31.7 km s⁻¹ (3.17 × 10⁴ m s⁻¹).

Interpretation. The speed far away is a bit less than the launch speed because the body gives up an amount of kinetic energy equal to the escape energy (½mv_e² per unit mass = 6.27 × 10⁷ J kg⁻¹) to climb out of the Earth’s potential well; the remaining kinetic energy is 8/9 of the initial value.

Extra depth. This is the principle of hyperbolic escape: for E > 0 the orbit is a hyperbola and the body retains a residual speed v_∞ at infinity (called the hyperbolic excess speed); for E = 0 it is a parabola (v_∞ = 0, projection at exactly v_e); for E < 0 it is an ellipse or circle (bound). Interplanetary probes are launched with a speed above v_e to reach the other planets with a useful v_∞.

More to know — types of orbit by total energy. E < 0: bound orbit (circle or ellipse); E = 0: parabolic escape, v_∞ = 0; E > 0: hyperbolic escape, v_∞ = √(2E/m). The launch speeds corresponding to the surface of the Earth: 7.9 km s⁻¹ (circular orbit), 11.2 km s⁻¹ (parabolic, escape from the Earth), 16.7 km s⁻¹ (escape from the solar system when launched along the Earth’s orbital motion). With launch speed 33.6 km s⁻¹ the probe is far above even the solar-system escape speed.

Conclusion: The speed far away is √8 v_e ≈ 31.7 km s⁻¹.

Concept. The energy needed to make the satellite escape is the amount that raises its total energy to zero. The total energy of a satellite in a circular orbit is E = −GMm/2r (negative: bound). The energy to be supplied is therefore +GMm/2r (the binding energy), where r = R + h = 6.4 × 10⁶ + 4.0 × 10⁵ = 6.8 × 10⁶ m.

1. r = R_E + h = 6.4 × 10⁶ + 0.4 × 10⁶ = 6.8 × 10⁶ m.
2. Total energy E = −GMm/2r = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200)/(2 × 6.8 × 10⁶).
3. GMm = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200 = 8.0 × 10¹⁶ J m; 2r = 1.36 × 10⁷ m; E = −8.0 × 10¹⁶/1.36 × 10⁷ = −5.9 × 10⁹ J.
4. Energy required to take it to infinity at rest = 0 − E = 5.9 × 10⁹ J.

Orbital speed: v = √(GM/r) = √(4.0 × 10¹⁴/6.8 × 10⁶) = 7.67 km s⁻¹; the satellite already has K = ½ × 200 × (7670)² = 5.9 × 10⁹ J, exactly the energy needed: it is enough to double its speed’s square (v → √2 v = escape speed at that height).

Extra depth. The energy needed to escape from the surface itself would be GMm/R = 1.25 × 10¹⁰ J for the 200 kg satellite, about twice the figure above, because the satellite in orbit already has kinetic energy and sits at a higher potential (Ex 7.6b). It is also the energy released as heat if the satellite re-enters from orbit: about 5.9 × 10⁹ J of kinetic energy plus the potential-energy change.

Conclusion: 5.9 × 10⁹ J must be supplied to take the satellite out of the Earth’s gravitational influence.

Concept. Starting from rest at a large separation, the two stars fall towards each other under their mutual attraction. The total mechanical energy of the system (kinetic + gravitational potential energy) is conserved. When the stars touch, the distance between their centres is 2R_star = 2 × 10⁴ km = 2 × 10⁷ m, and each star has the same speed v (equal masses, by momentum conservation).

1. Masses M = 2 × 10³⁰ kg each; initial separation r_i = 10⁹ km = 10¹² m; final separation r_f = 2 × 10⁷ m; initial speeds ~ 0.
2. Initial energy E_i = 0 + U_i = −GM²/r_i. Final energy E_f = 2 × ½Mv² − GM²/r_f = Mv² − GM²/r_f.
3. Conservation: Mv² = GM²(1/r_f − 1/r_i) ⇒ v² = GM(1/r_f − 1/r_i).
4. 1/r_f = 1/(2 × 10⁷) = 5 × 10⁻⁸ m⁻¹; 1/r_i = 10⁻¹² m⁻¹ (negligible); GM = 6.67 × 10⁻¹¹ × 2 × 10³⁰ = 1.334 × 10²⁰.
5. v² = 1.334 × 10²⁰ × (5 × 10⁻⁸ − 10⁻¹²) = 6.67 × 10¹² m² s⁻² ⇒ v = 2.6 × 10⁶ m s⁻¹.

Relative speed of approach = 2v = 5.2 × 10⁶ m s⁻¹. The kinetic energy released as the stars collide is 2 × ½ × 2 × 10³⁰ × (2.58 × 10⁶)² = 1.3 × 10⁴³ J.

Extra depth. The initial separation of 10⁹ km (about 7 AU) is much larger than the final separation, so the initial potential energy is almost zero; the result depends mainly on the distance at collision. Newtonian mechanics is adequate since v is only about 1% of c. Each star acquires the same speed because the two have equal masses: by momentum conservation the speeds are equal and opposite. For unequal masses M_1v1 = M_2v2.

Conclusion: The stars collide with speed about 2.6 × 10⁶ m s⁻¹ each.

Concept. At the mid-point of the line joining two identical spheres the gravitational attractions of the two spheres on any object are equal in magnitude and opposite in direction, so the net force is zero. The potential, a scalar, is the algebraic sum of the potentials due to each sphere (superposition); each sphere acts as a point mass at its centre. Then examine the stability by displacing the object.

1. Force: F₁ = Gm_1m/r² towards sphere 1; F₂ = Gm_2m/r² towards sphere 2; with m₁ = m₂ = 100 kg and r = 0.5 m, F₁ = F₂ and they are opposite: net force = 0.
2. Potential: V = −GM/r₁ − GM/r₂ = −2GM/r with M = 100 kg, r = 0.50 m: V = −2 × 6.67 × 10⁻¹¹ × 100/0.5 = −2.67 × 10⁻⁸ J kg⁻¹.
3. Equilibrium: since the net force is zero the object is in equilibrium at that point.
4. Stability: displace the object slightly along the line joining the centres towards one sphere: the attraction of the nearer sphere increases (inverse square law) and that of the farther one decreases, so the net force points further towards the nearer sphere: the displacement grows — unstable. (A displacement perpendicular to the line would be restored, but a single unstable direction is enough to make the equilibrium unstable.)

Potential energy view: along the line the potential V(x) = −GM[1/x + 1/(1 − x)] has a maximum at x = 0.5 m (the mid-point): so the equilibrium is unstable along the line (the mass would roll off towards either sphere), although the potential has a minimum with respect to perpendicular displacements, so the point is a saddle point in the potential energy surface.

Extra depth. The ‘force is zero but the potential is not’ is typical: the force is the negative gradient of the potential, and the potential is a scalar with no direction, so potentials add arithmetically while fields cancel vectorially. The potential here is negative everywhere, as for any attractive masses. Note that the radius of the spheres (0.10 m) does not matter because the point is outside both of them: the spheres act as point masses at their centres.

Conclusion: Net force = 0; potential = −2.67 × 10⁻⁸ J kg⁻¹; the equilibrium is unstable.