NEET Physics: Gravitation and Properties of Matter · Chapter 1

Gravitation

What to remember

Gravitational force between two point masses is F = G m₁ m₂ / r². It is always attractive, acts along the line joining the masses, and falls as 1/r².
Acceleration due to gravity is g = GM/R². It is largest at the poles, smaller at the equator, decreases with height as g(1 − 2h/R) for small h and with depth as g(1 − d/R), and is zero at the centre of the Earth.
Escape speed is √(2GM/R) = √(2gR), about 11.2 km s⁻¹ for Earth. Orbital speed near the surface is √(gR), about 7.9 km s⁻¹. The total energy of a satellite is −GMm/2r, so its binding energy is +GMm/2r.
Kepler's laws: orbits are ellipses with the Sun at one focus; the radius vector sweeps equal areas in equal times; T² is proportional to a³. A geostationary satellite has a period of 24 h and a height of about 36,000 km.

Kepler's laws of planetary motion

Kepler stated three laws that describe how planets move round the Sun.

1. Law of orbits: every planet moves in an elliptical orbit with the Sun at one focus of the ellipse.
2. Law of areas: the line joining the Sun and a planet sweeps out equal areas in equal intervals of time. The areal velocity (area swept per unit time) is constant.
3. Law of periods: the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its orbit. T² ∝ a³.

The law of areas follows from conservation of angular momentum. The gravitational force on a planet is central (it acts along the line to the Sun), so its torque about the Sun is zero and angular momentum L = m v r is constant. Areal velocity = L / 2m, which is constant. A planet moves fastest at the point nearest the Sun (perihelion) and slowest at the point farthest from the Sun (aphelion).

For two planets, (T₁/T₂)² = (a₁/a₂)³. For a circular orbit the semi-major axis is the radius r.

Law	Statement	Physical basis
First (orbits)	Elliptical path, Sun at one focus	Inverse-square force
Second (areas)	Equal areas in equal times	Conservation of angular momentum
Third (periods)	T² ∝ a³	Inverse-square force and centripetal need

Universal law of gravitation and the gravitational constant

Newton's law: every particle attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

F = G m₁ m₂ / r²

Here G is the universal gravitational constant. G = 6.67 × 10⁻¹¹ N m² kg⁻². Its dimensions are [M⁻¹ L³ T⁻²]. G has the same value everywhere in the universe and does not depend on the medium between the masses.

Features of the gravitational force:

It is always attractive.
The forces on the two masses form an action–reaction pair: equal in size, opposite in direction.
It is a central force and a conservative force.
It is the weakest of the basic forces, but it acts over long range.
It obeys the superposition principle: the net force on a mass is the vector sum of the forces from all other masses.
A uniform thin spherical shell acts, for points outside it, as if all its mass were at its centre. Inside a hollow shell the net force is zero.

Acceleration due to gravity and its variation

At the surface of the Earth (mass M, radius R), the weight of a body of mass m is mg = GMm/R². So

g = GM/R²

On the surface g is about 9.8 m s⁻². The mass of the Earth is about 6 × 10²⁴ kg and its radius is about 6400 km. Since M = (4/3)πR³ρ, g = (4/3)πGρR, so g is proportional to R for a given mean density ρ.

Variation with height h above the surface

g′ = g R² / (R + h)² = g (1 + h/R)⁻²

For h much smaller than R, g′ ≈ g (1 − 2h/R). The fractional decrease is 2h/R.

Variation with depth d below the surface

Treat the Earth as a uniform sphere. Only the inner sphere of radius (R − d) attracts the body.

g′ = g (1 − d/R)

So g decreases linearly with depth and becomes zero at the centre.

Variation with latitude and rotation (qualitative)

The Earth is not a perfect sphere. Its radius is slightly smaller at the poles than at the equator. The Earth also rotates about its axis with angular speed ω. At latitude λ the effective value is

g′ = g − Rω² cos² λ

At the poles (λ = 90°) rotation has no effect and g is maximum. At the equator (λ = 0°) the reduction is Rω² and g is minimum. If the Earth stopped rotating, g would increase at the equator but stay unchanged at the poles.

Place	Value of g	Reason
Surface	g = GM/R²	Reference value
Height h (small)	g (1 − 2h/R)	Larger distance from the centre
Depth d	g (1 − d/R)	Less mass pulls inward
Centre of Earth	0	Forces cancel
Equator	g − Rω²	Rotation, flatter shape
Poles	maximum	No rotational effect, nearer the centre

Gravitational potential energy, potential and escape speed

The gravitational potential energy of two point masses m₁ and m₂ separated by r is the work done in bringing them from infinity to that separation. The zero of potential energy is taken at infinite separation.

U = − G m₁ m₂ / r

The value is negative because the force is attractive: work must be supplied to separate the masses.

Gravitational potential at a point is the potential energy per unit mass there. For the field of mass M at distance r:

V = − GM / r (unit: J kg⁻¹)

Potential energy of a mass m at that point is U = mV. At the surface of the Earth, V = −GM/R = −gR.

Potential energy near the surface. The change in potential energy in raising a body of mass m through a small height h is

ΔU = mgh

This is the familiar formula, valid when h is much smaller than R. The exact change is GMm h / [R (R + h)].

Escape speed. Escape speed is the minimum speed that must be given to a body at the surface so that it reaches infinity (with zero speed there). By energy conservation:

½ m v_e² − GMm/R = 0, so v_e = √(2GM/R) = √(2gR)

For Earth, v_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km s⁻¹. Escape speed does not depend on the mass of the body or on the direction of projection (if air resistance is ignored). For the Moon it is about 2.4 km s⁻¹. Because this is low, the Moon cannot hold an atmosphere. If a body is projected at a speed greater than v_e, it reaches infinity with some speed left.

Also v_e = √2 × v₀, where v₀ is the orbital speed near the surface.

Motion of satellites

A satellite in a circular orbit of radius r = R + h round the Earth has gravity as the centripetal force:

G M m / r² = m v² / r

Orbital speed: v = √(GM/r) = √(g R² / (R + h))

Near the surface (h much smaller than R): v₀ = √(gR) ≈ 7.9 km s⁻¹.

Time period: T = 2π r / v = 2π √(r³ / GM). So T² ∝ r³ (Kepler's third law). Near the surface, T = 2π √(R/g) ≈ 84 minutes.

Orbital speed decreases as the orbit gets larger. The period does not depend on the mass of the satellite.

Geostationary (geosynchronous) satellite. It has a period of 24 hours, the same as the Earth's rotation. It orbits in the equatorial plane in the same direction as the Earth rotates (west to east), so it appears fixed in the sky. Its orbital radius is about 42,400 km from the centre of the Earth, which is a height of about 36,000 km above the surface. It is used for communication and broadcasting.

Polar satellite. It moves in a low orbit, at a height of about 500 to 800 km, passing over the north and south poles. Its period is about 100 minutes. As the Earth rotates below it, it can view the whole surface in successive orbits. It is used for remote sensing, weather watching and mapping.

Satellite	Height above surface	Period	Plane of orbit
Near-Earth	very small	about 84 min	any
Polar	about 500–800 km	about 100 min	passes over the poles
Geostationary	about 36,000 km	24 h	equatorial

Energy of an orbiting satellite, and weightlessness

For a satellite of mass m in a circular orbit of radius r:

Kinetic energy: K = ½ m v² = GMm / 2r
Potential energy: U = − GMm / r
Total energy: E = K + U = − GMm / 2r

So U = −2K, and E = −K = U/2. The total energy is negative, which shows the satellite is bound to the Earth. The binding energy is the energy needed to take the satellite to infinity: + GMm / 2r. As the orbit becomes larger, the total energy increases (becomes less negative) and the kinetic energy decreases. If a satellite loses energy to air drag, it moves to a lower orbit and its speed increases.

Weightlessness. The apparent weight of a body is the normal reaction it feels from its support. In a freely falling frame, or in an orbiting satellite, the body and its surroundings fall with the same acceleration, so no normal reaction acts. The body feels weightless. Gravity still acts on it: the satellite's astronaut is not beyond Earth's gravity. Weightlessness also occurs in a freely falling lift. At the centre of the Earth the true gravitational force is zero.

Worked examples

Example 1 (g at height and depth). Take g = 10 m s⁻² and R = 6400 km. Find g at a height of 1600 km above the surface and at a depth of 1600 km.

Height: g′ = g R² / (R + h)² = 10 × (6400/8000)² = 10 × 0.64 = 6.4 m s⁻².

Depth: g′ = g (1 − d/R) = 10 × (1 − 0.25) = 7.5 m s⁻².

Example 2 (escape speed of another planet). A planet has the same mean density as the Earth but a radius half that of the Earth. Escape speed from the Earth is 11.2 km s⁻¹. Since g is proportional to R at constant density, v_e = √(2gR) is proportional to R. So v_e = 11.2 / 2 = 5.6 km s⁻¹.

Example 3 (satellite period by Kepler's third law). A satellite orbits at r = 4R from the centre of the Earth. A near-surface satellite has T₀ = 84 minutes. T = T₀ × (4)^(3/2) = 84 × 8 = 672 minutes.

Example 4 (energies). A satellite of mass 1000 kg orbits at a distance where its kinetic energy is 3 × 10⁹ J. Its total energy is −3 × 10⁹ J, its potential energy is −6 × 10⁹ J, and its binding energy is 3 × 10⁹ J.

Formula summary

Quantity	Formula	Symbols and unit
Gravitational force	F = G m₁ m₂ / r²	N; G = 6.67 × 10⁻¹¹ N m² kg⁻²
g at surface	g = GM/R²	m s⁻²
g at height h	g (R/(R+h))² ≈ g(1 − 2h/R)	h much smaller than R for the approximation
g at depth d	g (1 − d/R)	uniform Earth
g at latitude λ	g − Rω² cos² λ	ω in rad s⁻¹
Potential energy	U = −G m₁ m₂ / r	J
Potential	V = −GM / r	J kg⁻¹
Small height change	ΔU = mgh	J
Escape speed	√(2GM/R) = √(2gR)	about 11.2 km s⁻¹ (Earth)
Orbital speed	√(GM/r)	near surface √(gR), about 7.9 km s⁻¹
Time period	2π √(r³/GM)	s
Satellite energies	K = GMm/2r, U = −GMm/r, E = −GMm/2r	J
Kepler's third law	T² ∝ a³	for any two planets of one Sun

NEET traps

1. G and g are different: G is a universal constant; g changes with place.
2. Escape speed does not depend on the mass or direction of the projected body, but orbital speed near the surface is smaller than escape speed by the factor √2.
3. Total energy of a bound satellite is negative; kinetic energy is positive and equal to the magnitude of the total energy; potential energy is twice the total energy.
4. g falls linearly with depth but as an inverse square with height. Use g(1 − 2h/R) only for h much smaller than R.
5. Weightlessness in a satellite does not mean gravity is zero. It means the satellite and its contents are in free fall.
6. At a given height g is the same for all bodies, but the weight mg depends on the mass.
7. A geostationary satellite is above the equator only; a satellite above any other latitude cannot be geostationary.
8. Rotation lowers g at the equator, and has no effect at the poles. The value of g is largest at the poles.
9. Kepler's second law means a planet moves fastest at perihelion, not at aphelion.
10. The orbital period of a satellite does not depend on its own mass, but it does depend on the orbit radius.

One-liners

1. G = 6.67 × 10⁻¹¹ N m² kg⁻² and its dimensions are [M⁻¹ L³ T⁻²].
2. The gravitational force is central, conservative and attractive.
3. g = GM/R² is about 9.8 m s⁻² at the surface of the Earth.
4. At height h much smaller than R, g falls by the fraction 2h/R.
5. At depth d, g′ = g (1 − d/R).
6. g is zero at the centre of the Earth.
7. Gravitational potential at distance r from mass M is −GM/r.
8. The escape speed from Earth is about 11.2 km s⁻¹.
9. The orbital speed near the surface of Earth is about 7.9 km s⁻¹.
10. Escape speed = √2 × orbital speed near the surface.
11. A near-surface satellite has a period of about 84 minutes.
12. The height of a geostationary satellite is about 36,000 km.
13. A polar satellite passes over the poles at a height of about 500–800 km.
14. The total energy of a satellite in orbit is −GMm/2r.
15. Binding energy of a satellite is GMm/2r.

Practice questions

The dimensional formula of the universal gravitational constant G is
1. [M⁻¹ L³ T⁻²]
2. [M⁻¹ L² T⁻²]
3. [M L³ T⁻²]
4. [M⁻¹ L³ T⁻¹]
Answer
A. [M⁻¹ L³ T⁻²]
From F = G m₁m₂/r², G = F r²/(m₁m₂) = [M L T⁻²][L²]/[M²] = [M⁻¹ L³ T⁻²].
Kepler's law of areas (equal areas in equal times) is a direct result of conservation of
1. mass
2. mechanical energy
3. angular momentum
4. linear momentum
Answer
C. angular momentum
The gravitational force of the Sun is central, so torque about the Sun is zero and angular momentum is constant. Areal velocity is L/2m.
The mean orbital radius of a planet P round the Sun is 4 times that of the Earth. If the Earth's period is 1 year, the period of P is
1. 2 years
2. 8 years
3. 16 years
4. 4 years
Answer
B. 8 years
T² ∝ r³, so T = 1 × 4^(3/2) = 8 years. Check: T² = 64 and r³ = 64 in Earth units.
Take the value of g on the surface of the Earth as 10 m s⁻². At a height equal to the radius of the Earth above the surface, g is
1. 7.5 m s⁻²
2. 5.0 m s⁻²
3. 1.25 m s⁻²
4. 2.5 m s⁻²
Answer
D. 2.5 m s⁻²
g′ = g R²/(R + R)² = g/4 = 2.5 m s⁻². Check: distance from centre doubles, so g falls to one quarter.
Take g = 10 m s⁻² at the surface and R = 6400 km. The value of g at a depth of 1280 km below the surface (uniform Earth) is
1. 8.0 m s⁻²
2. 2.0 m s⁻²
3. 6.9 m s⁻²
4. 6.0 m s⁻²
Answer
A. 8.0 m s⁻²
g′ = g (1 − d/R) = 10 × (1 − 0.2) = 8.0 m s⁻². Check: only 80% of the radius remains, and g is proportional to the radius inside.
The escape speed of a body from the surface of the Earth does not depend on
1. the acceleration due to gravity at the surface
2. the mass of the body
3. the radius of the Earth
4. the mass of the Earth
Answer
B. the mass of the body
v_e = √(2GM/R) = √(2gR). It contains M, R and g of the Earth but not the mass of the projected body.
A planet has 9 times the mass and 4 times the radius of the Earth. If the Earth's escape speed is 11.2 km s⁻¹, the escape speed from this planet is
1. 33.6 km s⁻¹
2. 7.5 km s⁻¹
3. 16.8 km s⁻¹
4. 25.2 km s⁻¹
Answer
C. 16.8 km s⁻¹
v_e ∝ √(M/R) = √(9/4) = 1.5, so v = 11.2 × 1.5 = 16.8 km s⁻¹.
The ratio of the escape speed to the orbital speed of a satellite revolving very close to the surface of a planet is
1. 2
2. 1/√2
3. 1
4. √2
Answer
D. √2
v_e = √(2gR) and v₀ = √(gR), so v_e/v₀ = √2.
Take g = 10 m s⁻² and R = 6400 km. The orbital speed of a satellite revolving very close to the Earth's surface is
1. 5.7 km s⁻¹
2. 11.3 km s⁻¹
3. 8.0 km s⁻¹
4. 16 km s⁻¹
Answer
C. 8.0 km s⁻¹
v₀ = √(gR) = √(10 × 6.4 × 10⁶) = √(6.4 × 10⁷) = 8000 m s⁻¹. Check: √2 × 8 = 11.3 km s⁻¹ is the escape speed.
The time period of a satellite in a circular orbit round the Earth depends on
1. the mass of the satellite and the radius of the orbit
2. the mass of the Earth and the radius of the orbit
3. the mass of the satellite only
4. the mass of the satellite and the mass of the Earth only
Answer
B. the mass of the Earth and the radius of the orbit
T = 2π √(r³/GM). It depends on r and M of the Earth, not on the satellite's own mass.
The total energy of a satellite of mass m in a circular orbit of radius r round the Earth (mass M) is
1. −GMm/2r
2. +GMm/r
3. +GMm/2r
4. −GMm/r
Answer
A. −GMm/2r
K = GMm/2r and U = −GMm/r, so E = K + U = −GMm/2r.
Take g = 10 m s⁻² and R = 6400 km. The binding energy of an 800 kg satellite in a circular orbit of radius 2R (from the centre of the Earth) is
1. 2.56 × 10¹⁰ J
2. 5.12 × 10¹⁰ J
3. 6.4 × 10⁹ J
4. 1.28 × 10¹⁰ J
Answer
D. 1.28 × 10¹⁰ J
Binding energy = GMm/2r = (gR²)m/(2 × 2R) = mgR/4 = 800 × 10 × 6.4 × 10⁶/4 = 1.28 × 10¹⁰ J.
The height of a geostationary satellite above the surface of the Earth is about
1. 36,000 km
2. 6,400 km
3. 3,600 km
4. 84,000 km
Answer
A. 36,000 km
A period of 24 h gives an orbital radius of about 42,400 km, which is about 36,000 km above the surface.
Take g = 10 m s⁻² and R = 6400 km. The increase in potential energy of a 2 kg mass when it is raised from the surface of the Earth to a height equal to R is
1. 6.4 × 10⁶ J
2. 6.4 × 10⁷ J
3. 1.28 × 10⁸ J
4. 3.2 × 10⁷ J
Answer
B. 6.4 × 10⁷ J
ΔU = GMm(1/R − 1/2R) = mgR/2 = 2 × 10 × 6.4 × 10⁶/2 = 6.4 × 10⁷ J. Using mgh = mgR would wrongly give 1.28 × 10⁸ J.
Take R = 6400 km. The height above the surface at which g falls by 1% of its surface value is about
1. 320 km
2. 16 km
3. 64 km
4. 32 km
Answer
D. 32 km
Fractional fall = 2h/R = 0.01, so h = R/200 = 6400/200 = 32 km. Check: g(R/(R+h))² = 0.9901 g.
Take g = 10 m s⁻² and R = 6400 km. The gravitational potential at the surface of the Earth (zero at infinity) is
1. +6.4 × 10⁷ J kg⁻¹
2. −6.4 × 10⁶ J kg⁻¹
3. −6.4 × 10⁷ J kg⁻¹
4. −3.2 × 10⁷ J kg⁻¹
Answer
C. −6.4 × 10⁷ J kg⁻¹
V = −GM/R = −gR = −10 × 6.4 × 10⁶ = −6.4 × 10⁷ J kg⁻¹.
Given below are two statements. Statement I: The value of g at the equator is less than its value at the poles. Statement II: Rotation of the Earth reduces the effective g at the equator by Rω². Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are correct. Effective g at latitude λ is g − Rω² cos²λ, which is least at the equator and unaffected at the poles.
Given below are two statements. Statement I: Inside the Earth, g decreases linearly with depth. Statement II: The value of g at the centre of the Earth is maximum. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
g′ = g(1 − d/R) falls linearly and is zero at the centre, so Statement II is wrong.
Given below are two statements. Statement I: A planet moves slowest when it is nearest to the Sun. Statement II: The angular momentum of a planet about the Sun is constant. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
m v r is constant, so speed is greatest where r is least (perihelion). Statement I is wrong, Statement II is correct.
Given below are two statements. Statement I: The escape speed of a body depends on its mass. Statement II: The escape speed of a body depends on the direction in which it is projected. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
v_e = √(2GM/R) depends only on the planet's M and R, not on the body's mass or the direction of projection.
Which of the following statements are correct? 1. The orbital speed of a satellite decreases as the orbit becomes larger. 2. The period of a satellite does not depend on its mass. 3. The total energy of a satellite in a circular orbit is positive.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
v = √(GM/r) and T = 2π√(r³/GM) support 1 and 2. The total energy −GMm/2r is negative, so 3 is wrong.
Which of the following statements are correct? 1. A geostationary satellite revolves from west to east. 2. A geostationary satellite can be placed above any latitude. 3. The period of a geostationary satellite is 24 hours.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
It moves with the Earth's rotation (west to east) and has T = 24 h, but must be in the equatorial plane, so 2 is wrong.
Given below are two statements. Statement I: The kinetic energy of a satellite in a circular orbit equals the magnitude of its total energy. Statement II: The kinetic energy of the satellite equals the magnitude of its potential energy. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
K = GMm/2r and |E| = GMm/2r, so I is correct. |U| = GMm/r = 2K, so II is wrong.
Given below are two statements. Statement I: An astronaut in an orbiting satellite feels weightless because gravity is zero at that height. Statement II: A person in a freely falling lift also feels weightless. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Gravity still acts at satellite height; weightlessness is due to free fall. Statement II is correct.
Which of the following statements are correct? 1. The escape speed from the Moon is greater than that from the Earth. 2. The Moon cannot hold an atmosphere because its escape speed is low. 3. The escape speed from the Earth is about 11.2 km s⁻¹.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
The Moon's escape speed (about 2.4 km s⁻¹) is much smaller than the Earth's, so 1 is wrong; 2 and 3 are correct.
Given below are two statements. Statement I: When a satellite moves to a lower orbit because of air drag, its speed increases. Statement II: Its total energy becomes more negative. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
v = √(GM/r) rises as r falls, and E = −GMm/2r becomes more negative. Both are correct.
Assertion (A): A satellite revolving very close to the Earth has an orbital speed of about 7.9 km s⁻¹. Reason (R): The escape speed from the surface of the Earth is about 11.2 km s⁻¹.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both facts are true, but the escape speed does not explain the orbital speed, which comes from √(gR).
Assertion (A): The weight of a body is zero at the centre of the Earth. Reason (R): The value of g is maximum at the centre of the Earth.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
g is zero at the centre, so A is true and R is false.
Assertion (A): A geostationary satellite can be placed above the north pole. Reason (R): The orbit of a geostationary satellite must lie in the equatorial plane.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A is false because the orbit must be equatorial; R is true.
Assertion (A): A planet moves faster when it is nearer the Sun. Reason (R): The angular momentum of the planet about the Sun is conserved.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
m v r = constant, so v is larger at smaller r. R explains A.
Assertion (A): The gravitational force between two masses does not change when a third mass is brought near them. Reason (R): Gravitational force follows an inverse square law.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
The pair force depends only on the two masses and their distance, so A is true. The inverse square law is true but does not explain independence from a third body.
Match List I with List II. List I: a. Law of orbits b. Law of areas c. Law of periods d. Gravitational constant G. List II: i. T² ∝ a³ ii. Elliptical path with the Sun at a focus iii. Conservation of angular momentum iv. Dimensions [M⁻¹ L³ T⁻²]
1. a-ii, b-i, c-iii, d-iv
2. a-i, b-ii, c-iii, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-ii, b-iii, c-i, d-iv
Answer
D. a-ii, b-iii, c-i, d-iv
Pairs: orbits - ellipse (ii); areas - angular momentum (iii); periods - T² ∝ a³ (i); G - dimensions (iv).
Match List I with List II. List I: a. Escape speed from the Earth b. Orbital speed near the Earth's surface c. Height of a geostationary satellite d. Period of a near-surface satellite. List II: i. About 84 minutes ii. About 36,000 km iii. About 7.9 km s⁻¹ iv. About 11.2 km s⁻¹
1. a-iii, b-iv, c-ii, d-i
2. a-iv, b-iii, c-ii, d-i
3. a-ii, b-iii, c-iv, d-i
4. a-iv, b-iii, c-i, d-ii
Answer
B. a-iv, b-iii, c-ii, d-i
Pairs: 11.2 km/s (iv), 7.9 km/s (iii), 36,000 km (ii), 84 min (i).
Match List I with List II for a satellite of mass m in a circular orbit of radius r. List I: a. Kinetic energy b. Potential energy c. Total energy d. Ratio of potential energy to kinetic energy. List II: i. −GMm/2r ii. −GMm/r iii. −2 iv. GMm/2r
1. a-ii, b-iv, c-i, d-iii
2. a-iv, b-ii, c-iii, d-i
3. a-iv, b-ii, c-i, d-iii
4. a-iv, b-i, c-ii, d-iii
Answer
C. a-iv, b-ii, c-i, d-iii
K = GMm/2r (iv), U = −GMm/r (ii), E = −GMm/2r (i), U/K = −2 (iii).
Match List I with List II. List I: a. Centre of the Earth b. Equator c. At depth d d. At small height h. List II (value of g): i. g − Rω² ii. g(1 − 2h/R) iii. g(1 − d/R) iv. Zero
1. a-iv, b-i, c-ii, d-iii
2. a-i, b-iv, c-iii, d-ii
3. a-iii, b-i, c-iv, d-ii
4. a-iv, b-i, c-iii, d-ii
Answer
D. a-iv, b-i, c-iii, d-ii
Centre: 0 (iv); equator: g − Rω² (i); depth: g(1 − d/R) (iii); height: g(1 − 2h/R) (ii).
Match List I with List II. List I: a. Geostationary satellite b. Polar satellite c. Satellite very close to the Earth d. Satellite at a distance 4R from the centre of the Earth. List II: i. Period about 84 minutes ii. Period 24 hours iii. Height about 500 to 800 km iv. Period about 672 minutes (taking 84 minutes for the near-surface satellite)
1. a-ii, b-iii, c-i, d-iv
2. a-i, b-iii, c-ii, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-ii, b-iii, c-iv, d-i
Answer
A. a-ii, b-iii, c-i, d-iv
Geostationary: 24 h (ii); polar: 500-800 km (iii); near surface: 84 min (i); at 4R: 84 × 4^(3/2) = 672 min (iv).
Match List I with List II. List I: a. Escape speed b. Orbital speed near the surface c. Gravitational potential at the surface d. Period near the surface. List II: i. √(gR) ii. −gR iii. 2π√(R/g) iv. √(2gR)
1. a-i, b-iv, c-ii, d-iii
2. a-iv, b-i, c-iii, d-ii
3. a-iv, b-i, c-ii, d-iii
4. a-iii, b-i, c-ii, d-iv
Answer
C. a-iv, b-i, c-ii, d-iii
v_e = √(2gR) (iv); v₀ = √(gR) (i); V = −GM/R = −gR (ii); T = 2πR/v₀ = 2π√(R/g) (iii).
A planet moves round the Sun in an elliptical orbit. Its closest distance from the Sun is one quarter of its farthest distance. The ratio of its speed at the closest point to its speed at the farthest point is
1. 2
2. 16
3. 4
4. 1/4
Answer
C. 4
Angular momentum is conserved: v₁r₁ = v₂r₂, so v₁/v₂ = r₂/r₁ = 4.
A body weighs 900 N on the surface of the Earth. Its weight at a height of twice the radius of the Earth above the surface is
1. 300 N
2. 100 N
3. 450 N
4. 225 N
Answer
B. 100 N
Distance from the centre is 3R, so weight = 900 × (1/3)² = 100 N.
Take R = 6400 km. A body is projected vertically upward from the Earth's surface with a speed equal to the orbital speed near the surface (√(gR)). Neglecting air resistance and rotation, the maximum height it reaches above the surface is
1. 12800 km
2. 19200 km
3. 3200 km
4. 6400 km
Answer
D. 6400 km
½v² − gR = −gR²/(R+h) gives −gR/2 = −gR²/(R+h), so R + h = 2R and h = R = 6400 km. Using v²/2g would wrongly give 3200 km.
Take g = 10 m s⁻² and R = 6400 km. A satellite orbits at a height of 3R above the surface of the Earth. Its orbital speed is
1. 4.0 km s⁻¹
2. 5.7 km s⁻¹
3. 2.0 km s⁻¹
4. 8.0 km s⁻¹
Answer
A. 4.0 km s⁻¹
r = 4R, so v = √(gR²/4R) = ½√(gR) = ½ × 8 = 4 km s⁻¹.
Take g = 10 m s⁻² and R = 6400 km. The extra energy needed to shift a 1000 kg satellite from a circular orbit of radius 2R to a circular orbit of radius 4R (radii from the centre of the Earth) is
1. 1.6 × 10¹⁰ J
2. 8.0 × 10⁹ J
3. 4.0 × 10⁹ J
4. 3.2 × 10¹⁰ J
Answer
B. 8.0 × 10⁹ J
ΔE = (GMm/2)(1/2R − 1/4R) = GMm/8R = mgR/8 = 1000 × 10 × 6.4 × 10⁶/8 = 8 × 10⁹ J. Using the potential energy change alone gives 1.6 × 10¹⁰ J.
A geostationary satellite stays above the same point on the equator. Its angular speed about the Earth's axis, in rad s⁻¹ (taking π = 3.14), is about
1. 4.4 × 10⁻³
2. 7.3 × 10⁻⁴
3. 1.2 × 10⁻³
4. 7.3 × 10⁻⁵
Answer
D. 7.3 × 10⁻⁵
ω = 2π/T = 2 × 3.14/86400 = 7.3 × 10⁻⁵ rad s⁻¹. Check: 360° in 24 h is 15° per hour.
Take R = 6400 km and ω = 7.5 × 10⁻⁵ rad s⁻¹ for the Earth. The decrease in the value of g because of rotation at a place of latitude 60° is
1. 0.018 m s⁻²
2. 0.009 m s⁻²
3. 0.036 m s⁻²
4. 0.0045 m s⁻²
Answer
B. 0.009 m s⁻²
Δg = Rω² cos²λ = 6.4 × 10⁶ × 5.625 × 10⁻⁹ × 0.25 = 0.009 m s⁻². Without cos² the answer would be 0.036 m s⁻².

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