Exercises
Q7.1. Answer the following
(a) Can you shield a body from gravitational influence by putting it inside a hollow sphere?
(b) Can an astronaut in a large space station detect gravity?
(c) Why is the tidal effect of the moon greater than that of the Sun, even though the Sun's gravitational pull on Earth is greater?
Ans.
(a) No. Unlike electrical forces, there is no gravitational equivalent of a conductor that can shield a body. Gravitational force cannot be shielded by any material. Every mass attracts every other mass, and no barrier blocks gravity.
(b) Yes, if the station is large enough. Inside a freely falling/orbiting small space station, everything is in free fall and the astronaut feels weightless. But in a large space station, the gravitational force varies across the station (tidal effects). The astronaut can detect the differential gravitational force (tidal force) — part of the body nearer to Earth feels stronger gravity than the part farther away. This difference becomes detectable over a large station.
(c) Tidal effects depend on the gradient (rate of change) of gravitational force with distance, not on the absolute gravitational force. The tidal force ∝ M/r³. Moon: M_moon/r_moon³ is larger than M_sun/r_sun³ because the moon is much closer. Although the Sun is more massive, it is ~390 times farther than the Moon. The Moon's proximity dominates → Moon's tidal effect is greater.
2. (Answer missing from excerpt — standard exercise on Kepler's law)
Answer
Choose the correct alternative: The gravitational force on a point mass due to two other point masses is found by:
(a) vector addition (b) scalar addition (c) can't be added
Ans. (a) Vector addition — Gravitational forces are vectors. The net gravitational force on a mass due to multiple sources is found by vector addition of individual forces (principle of superposition).
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 Earth?
Answer
Using Kepler's third law: T² ∝ R³
If the planet's orbital period T' = T_Earth/2 (twice as fast means half the time period):
(T'/T_E)² = (R'/R_E)³
(1/2)² = (R'/R_E)³
1/4 = (R'/R_E)³
R'/R_E = (1/4)^(1/3) = (0.25)^(0.333) ≈ 0.63
The orbital radius would be about 0.63 times that of Earth's orbit ≈ 0.63 AU.
4. Io, one of the moons of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the Sun.
Answer
From Kepler's 3rd law for a satellite orbiting a planet: T² = 4π²r³/(GM)
M_Jupiter = 4π²r³/(GT²)
T = 1.769 days = 1.769 × 86400 = 1.528 × 10⁵ s
r = 4.22 × 10⁸ m; G = 6.67 × 10⁻¹¹ N m² kg⁻²
M_J = 4π² × (4.22 × 10⁸)³ / (6.67 × 10⁻¹¹ × (1.528 × 10⁵)²)
= 4π² × 7.52 × 10²⁵ / (6.67 × 10⁻¹¹ × 2.335 × 10¹⁰)
= 4 × 9.87 × 7.52 × 10²⁵ / (1.557)
= 4π² × 7.52 × 10²⁵ / 1.557
Numerator: 4 × 9.87 × 7.52 × 10²⁵ = 297.0 × 10²⁵ = 2.97 × 10²⁷
M_J = 2.97 × 10²⁷ / 1.557 ≈ 1.91 × 10²⁷ kg
M_Sun = 2.0 × 10³⁰ kg
M_J/M_Sun = 1.91 × 10²⁷ / 2.0 × 10³⁰ ≈ 1/1000 ✓
5. Let us assume that our galaxy consists of 2.5 × 10¹¹ stars, each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution?
Answer
Total mass M = 2.5 × 10¹¹ × M_sun = 2.5 × 10¹¹ × 2.0 × 10³⁰ = 5.0 × 10⁴¹ kg
r = 50,000 ly = 5 × 10⁴ × 9.46 × 10¹⁵ m = 4.73 × 10²⁰ m
T = 2πr^(3/2) / √(GM) = 2π × (4.73 × 10²⁰)^(3/2) / √(6.67 × 10⁻¹¹ × 5.0 × 10⁴¹)
√(GM) = √(6.67 × 10⁻¹¹ × 5 × 10⁴¹) = √(3.335 × 10³¹) = 5.776 × 10¹⁵ m^(3/2) s⁻¹... (using Kepler's form)
r^(3/2) = (4.73 × 10²⁰)^(3/2) = (4.73)^(1.5) × 10³⁰ = 10.28 × 10³⁰ = 1.028 × 10³¹
T = 2π × 1.028 × 10³¹ / 5.776 × 10¹⁵ = 6.283 × 1.028 × 10³¹ / 5.776 × 10¹⁵
= 6.46 × 10³¹ / 5.776 × 10¹⁵ = 1.118 × 10¹⁶ s
In years: 1.118 × 10¹⁶ / 3.156 × 10⁷ ≈ 3.54 × 10⁸ years ≈ 3.5 × 10⁸ years
6. Choose the correct alternative: The escape velocity of a body projected vertically upwards from the surface of Earth is independent of... (many options). Justify.
Answer
Escape velocity v_e = √(2GM/R) = √(2gR)
It depends only on the mass of Earth (M) and radius of Earth (R) — or equivalently, on g and R. It is independent of the mass of the body projected, the direction of projection (for vertical upward), and the location from where it is projected (assuming it's from the surface).
7. Does the escape speed of a body from the Earth depend on: (a) mass of body (b) location from which it is projected (c) direction of projection (d) height of location?
Answer
v_e = √(2GM_E/r), where r is the distance from Earth's centre.
(a) No — mass of body cancels out in the energy equation.
(b) Yes — escape speed depends on the location because r changes. From different points on Earth's surface, r ≈ R_E but slight variations due to Earth's non-spherical shape exist. From a higher location (r > R_E), v_e is smaller.
(c) No — direction doesn't matter. Escape speed is the minimum speed needed; direction affects the path but not the minimum speed (for conservative forces).
(d) Yes — at height h: v_e = √(2GM_E/(R_E + h)), which decreases with increasing height.
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?
Answer
(a) Linear speed: No — varies (fastest at perihelion, slowest at aphelion).
(b) Angular speed: No — varies with distance.
(c) Angular momentum: Yes — conserved (no torque; gravity is central force).
(d) Kinetic energy: No — varies (KE largest at perihelion).
(e) Potential energy: No — varies with distance from Sun.
(f) Total energy: Yes — conserved throughout orbit (gravity is conservative).
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?
Answer
In space (weightlessness), body fluids shift upward (toward the head), since there's no gravity to keep fluid in the lower body.
(a) Swollen feet: No — without gravity, blood doesn't pool in feet.
(b) Swollen face: Yes — fluids accumulate in the upper body and face due to absence of gravity drainage.
(c) Headache: Yes — fluid accumulation causes increased pressure in the head and sinuses.
(d) Orientational problem: Yes — in weightlessness, there is no "up" or "down"; inner ear and body orientation sense are confused.
Most likely: (b), (c), and (d)
10. In the following two statements, which one is more accurate and why?
(a) "g is zero at the centre of the Earth"
(b) "g is zero at the centre of the Earth"... (both state same — from textbook: distinguish between g and field)
Answer
At the centre of the Earth, the gravitational field g = 0 (since equal and opposite pulls from all directions cancel). Therefore both the acceleration due to gravity and the gravitational field intensity are zero at the centre of the Earth. The weight of a body (mg) is also zero there.
11. As you go from the equator to the poles, g increases. Why?
Answer
There are two reasons:
1. Earth's shape: Earth is not a perfect sphere but an oblate spheroid (flattened at poles). The polar radius is smaller than equatorial radius (R_pole ≈ 6356 km < R_equator ≈ 6378 km). Since g = GM/R², smaller R at poles means larger g.
2. Rotation: Due to Earth's rotation, effective g = g₀ − ω²R cosλ (where λ is latitude). At the equator, the centrifugal reduction is maximum. At the poles, no centrifugal effect. So effective g is smaller at equator.
12. The gravitational intensity at the surface of Earth is g. What is it at a height h above the surface (h << R)?
Answer
g(h) = GM/(R + h)² = g × R²/(R + h)²
= g × 1/(1 + h/R)² ≈ g(1 − 2h/R) for h << R
Fractional decrease: Δg/g = 2h/R
13. A body weighs 63 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?
Answer
Weight on surface W = 63 N; height h = R/2
g(h) = GM/(R+h)² = GM/(R + R/2)² = GM/(3R/2)² = GM/(9R²/4) = 4GM/(9R²) = (4/9)g
W(h) = m × g(h) = W × (4/9) = 63 × 4/9 = 252/9 = 28 N
14. Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the Earth if it weighed 250 N on the surface?
Answer
g at depth d = g(1 − d/R)
At d = R/2: g' = g(1 − 1/2) = g/2
Weight W' = mg' = W/2 = 250/2 = 125 N
15. A rocket is fired vertically with speed 5 km s⁻¹. How far from Earth does it go? (M_E = 6.0 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻²)
Answer
Using energy conservation: ½mv² = GMm(1/R − 1/r_max)
v = 5 km/s = 5000 m/s
r_max = 1/(1/R − v²/2GM)
v² = 25 × 10⁶ m² s⁻²
2GM = 2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ = 8.004 × 10¹⁴
v²/(2GM) = 25 × 10⁶ / 8.004 × 10¹⁴ = 3.124 × 10⁻⁸ m⁻¹
1/R = 1/(6.4 × 10⁶) = 1.5625 × 10⁻⁷ m⁻¹
1/r_max = 1.5625 × 10⁻⁷ − 3.124 × 10⁻⁸ = 1.2501 × 10⁻⁷ m⁻¹
r_max = 1/1.2501 × 10⁻⁷ = 8.0 × 10⁶ m
Height above surface = r_max − R = 8.0 × 10⁶ − 6.4 × 10⁶ = 1.6 × 10⁶ m = 1600 km
16. A body weighs 250 N on surface. How much halfway down to the centre?
Answer
(Same as Q7.14 — see above) 125 N
17. Rocket fired vertically at 5 km s⁻¹. How far does it go? (Same as Q7.15 — standard exercise)
Answer
1.6 × 10⁶ m = 1600 km above Earth's surface (see Q7.15 solution above)
18. Escape speed from Earth = 11.2 km s⁻¹. A body is projected at 3 times escape speed. Find speed far away from Earth.
Answer
Initial speed u = 3 × 11.2 = 33.6 km s⁻¹
Using energy conservation: ½mu² − GMm/R = ½mv_∞²
½v_∞² = ½u² − GM/R
Escape speed v_e = √(2GM/R)
⇒ GM/R = v_e²/2
½v_∞² = ½u² − v_e²/2 = ½(u² − v_e²)
v_∞ = √(u² − v_e²) = √((33.6)² − (11.2)²) = √(1128.96 − 125.44) = √1003.52 ≈ 31.7 km s⁻¹
19. A satellite orbits at 400 km height. Energy to rocket it out of Earth's gravity? (m_satellite = 200 kg, M_E = 6.0 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻²)
Answer
Total energy of satellite in orbit at radius r = R_E + h:
r = 6.4 × 10⁶ + 4 × 10⁵ = 6.8 × 10⁶ m
E_orbit = −GMm/(2r) = −6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200 / (2 × 6.8 × 10⁶)
= −6.67 × 10⁻¹¹ × 1.2 × 10²⁷ / 1.36 × 10⁷
= −8.004 × 10¹⁶ / 1.36 × 10⁷
= −5.885 × 10⁹ J
Energy to escape = −E_orbit = 5.885 × 10⁹ J ≈ 5.9 × 10⁹ J
20. Two stars each of solar mass (2 × 10³⁰ kg) at separation 10⁹ km approach each other. Each star has radius 10⁴ km. Find collision speed.
Answer
Initial separation r₀ = 10⁹ km = 10¹² m; final separation r_f = 2 × 10⁴ km = 2 × 10⁷ m (touching)
Initial speeds ≈ 0; by conservation of energy:
0 + (−GM²/r₀) = ½mv₁² + ½mv₂² − GM²/r_f
By symmetry v₁ = v₂ = v (equal masses, equal speeds toward each other)
−GM²/r₀ = mv² − GM²/r_f
mv² = GM²(1/r_f − 1/r₀)
v² = GM(1/r_f − 1/r₀) = 6.67 × 10⁻¹¹ × 2 × 10³⁰ × (1/(2 × 10⁷) − 1/10¹²)
= 1.334 × 10²⁰ × (5 × 10⁻⁸ − 10⁻¹²)
≈ 1.334 × 10²⁰ × 5 × 10⁻⁸ = 6.67 × 10¹²
v = √(6.67 × 10¹²) = 2.58 × 10⁶ m s⁻¹ ≈ 2.6 × 10⁶ m s⁻¹
21. Two heavy spheres (100 kg, radius 0.10 m) are placed 1.0 m apart. Find gravitational force and potential at the midpoint. Is an object placed there in equilibrium? Stable or unstable?
Answer
At midpoint (0.5 m from each sphere):
Gravitational force: Force from sphere 1 on a test mass m at midpoint = GMm/(0.5)² toward sphere 1
Force from sphere 2 = GMm/(0.5)² toward sphere 2 (equal magnitude, opposite direction)
Net force = zero
Gravitational potential: V = −GM/r₁ − GM/r₂ = −2GM/0.5
= −2 × 6.67 × 10⁻¹¹ × 100 / 0.5 = −2 × 6.67 × 10⁻⁹ / 0.5 = −2.67 × 10⁻⁸ J kg⁻¹
Equilibrium: Yes, the object is in equilibrium (net force = 0).
Stability: The equilibrium is unstable. If the object is displaced slightly toward either sphere, the gravitational pull from that sphere (now closer) becomes stronger, pulling it away from the midpoint. The object does not return to equilibrium.
Important Extra Questions
Q. State Newton's Law of Universal Gravitation and give its mathematical form.
Ans. Every particle in the universe attracts every other particle with a force that is:
(i) directly proportional to the product of their masses, and
(ii) inversely proportional to the square of the distance between them.
F = Gm₁m₂/r²
where G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant.
The force is directed along the line joining the two particles and is attractive.
Q. State Kepler's three laws of planetary motion.
Ans.
1. Law of Orbits: All planets move in elliptical orbits with the Sun at one focus.
2. Law of Areas (Law of Equal Areas): The line joining a planet to the Sun sweeps equal areas in equal intervals of time. This means planets move faster when nearer the Sun (perihelion) and slower when farther (aphelion).
3. Law of Periods: The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit: T² ∝ a³, or T² = (4π²/GM) × a³.
Q. Define escape velocity. Derive an expression for it.
Ans. Escape velocity is the minimum speed needed to project a body from the Earth's surface such that it escapes Earth's gravitational field and never returns.
Derivation: At surface, KE = ½mv_e²; PE = −GMm/R
For escape, total energy ≥ 0: ½mv_e² − GMm/R ≥ 0
v_e = √(2GM/R) = √(2gR)
v_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km s⁻¹
Q. What is a geostationary satellite? Find the height of a geostationary orbit. (G = 6.67 × 10⁻¹¹, M_E = 6 × 10²⁴ kg, T = 24 h)
Answer
A geostationary satellite is one that orbits Earth in the equatorial plane with the same period as Earth's rotation (24 hours), making it appear stationary above a fixed point on the equator.
T² = 4π²r³/(GM) → r³ = GMT²/(4π²)
r³ = 6.67 × 10⁻¹¹ × 6 × 10²⁴ × (24 × 3600)² / (4π²)
= 6.67 × 10⁻¹¹ × 6 × 10²⁴ × 7.464 × 10⁹ / 39.48
= 4 × 10¹⁴ × 7.464 × 10⁹ / 39.48
= 2.99 × 10²⁴ / 39.48 = 7.57 × 10²² m³
r = (7.57 × 10²²)^(1/3) ≈ 4.22 × 10⁷ m = 42,200 km
Height above surface = 42,200 − 6,400 = 35,800 km ≈ 3.58 × 10⁷ m
Q. A satellite of mass 500 kg orbits Earth at a height of 2000 km. Find its orbital velocity, period, and total energy. (R_E = 6.4 × 10⁶ m, g = 9.8 m s⁻², M_E = 6 × 10²⁴ kg)
Ans.
r = R_E + h = 6.4 × 10⁶ + 2 × 10⁶ = 8.4 × 10⁶ m
Orbital velocity: v = √(GM/r) = √(6.67 × 10⁻¹¹ × 6 × 10²⁴ / 8.4 × 10⁶)
= √(4.002 × 10¹⁴ / 8.4 × 10⁶) = √(4.764 × 10⁷) ≈ 6899 m s⁻¹ ≈ 6.9 km s⁻¹
Period: T = 2πr/v = 2π × 8.4 × 10⁶ / 6899 = 7.65 × 10³ s ≈ 127.5 min
Total energy = −GMm/(2r) = −6.67 × 10⁻¹¹ × 6 × 10²⁴ × 500 / (2 × 8.4 × 10⁶)
= −2.001 × 10¹⁷ / 1.68 × 10⁷ ≈ −1.19 × 10¹⁰ J