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Chapter 1 - Gravitation

Chapter 1 : Gravitation

Textbook Exercise Solutions

Q1) Study the entries in the following table and rewrite them putting the connected items together:
Answer :

The correctly matched columns showing physical quantities, their SI units, and defining formulas are as follows:

Column I (Quantity) Column II (Formula / Law) Column III (SI Unit / Property)
1. Mass Measure of inertia kg
2. Weight W = mg N (Newton)
3. Acceleration due to gravity (g) g = GM / R² m/s²
4. Gravitational potential energy -GMm / (R + h) J (Joule)
Q2) Answer the following questions:
1. What is the difference between mass and weight of an object? Will the mass and weight of an object on the earth be the same as their values on Mars? Why?
Answer :

(A) Differences between Mass and Weight:

Mass Weight
Mass is the total amount of matter contained in a body. Weight is the gravitational force by which the Earth (or celestial body) attracts the object towards its center.
It is a scalar quantity (has magnitude only). It is a vector quantity (directed towards the center of gravity).
Mass remains constant everywhere in the universe. Weight changes from place to place depending on the local value of acceleration due to gravity (g).
SI unit: kilogram (kg). SI unit: newton (N).
Measured using a beam balance. Measured using a spring balance.
Mass can never be zero for any physical object. Weight becomes zero in free fall or in deep space where g = 0.

(B) Comparison on Earth and Mars:

  • Mass: The mass of an object will be exactly the same on both Earth and Mars because mass is the actual amount of matter in a body and does not depend on gravity.
  • Weight: The weight on Mars will not be the same as on Earth. Since weight is calculated as W = mg, and the acceleration due to gravity on Mars (gMars ≈ 3.73 m/s²) is much less than on Earth (gEarth ≈ 9.8 m/s²), the object will weigh much less on Mars (about 38% of its Earth weight).
2. What are (i) Free fall, (ii) Acceleration due to gravity, (iii) Escape velocity, and (iv) Centripetal force?
Answer :
  1. Free Fall: When an object moves towards the Earth under the influence of only the gravitational force of the Earth (without air resistance), its motion is called free fall. True free fall is possible only in a vacuum.
  2. Acceleration due to Gravity (g): The uniform acceleration produced in a freely falling body due to the gravitational pull of the Earth is called acceleration due to gravity. Its standard average value on Earth's surface is 9.8 m/s² (directed downwards towards the center of Earth).
  3. Escape Velocity (vesc): The minimum initial velocity with which an object must be projected upward from the surface of a celestial body so that it overcomes the gravitational pull and escapes into space without falling back is called escape velocity. For Earth, vesc = √(2GM / R) ≈ 11.2 km/s.
  4. Centripetal Force: The center-seeking force that acts on an object moving along a circular path, directing it continuously towards the center of the circle to maintain circular motion, is called centripetal force (F = mv² / r).
3. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?
Answer :

(A) Kepler's Three Laws of Planetary Motion:

  1. Kepler's First Law (Law of Orbits): The orbit of a planet is an ellipse with the Sun situated at one of the two foci.
  2. Kepler's Second Law (Law of Equal Areas): The line joining the planet and the Sun sweeps out equal areas in equal intervals of time. (Planets travel faster near perihelion and slower near aphelion).
  3. Kepler's Third Law (Law of Periods): The square of the orbital period of revolution (T) of a planet around the Sun is directly proportional to the cube of the mean distance (r) of the planet from the Sun:
    T² ∝ r³  ⇒  T² / r³ = constant (K)
Sun (Focus F₁) Area A Area B Equal Areas Swept in Equal Times: Area A = Area B
Fig 1.1: Kepler's Laws of Planetary Motion

(B) Deduction of Newton's Inverse Square Law:

For a planet of mass m orbiting at radius r with speed v, the required centripetal force is:

F = mv² / r
Speed of the planet is distance divided by period: v = 2πr / T
Substitute v: F = m(2πr / T)² / r = 4π²mr / T²
Multiply and divide by r²: F = (4π²m / r²) × (r³ / T²)
According to Kepler’s third law: T² / r³ = K  ⇒  r³ / T² = 1/K (constant)
Therefore: F = (4π²m / K) × (1 / r²)
Since (4π²m / K) is constant, F ∝ 1 / r². Newton thus concluded that the gravitational force is inversely proportional to the square of the distance.
4. A stone thrown vertically upwards with initial velocity u reaches a height 'h' before coming down. Show that the time taken to go up is same as the time taken to come down.
Answer :

Let an object be thrown upwards with initial velocity u reaching maximum height h. At the highest point, final velocity v = 0.

Case 1: Upward Motion (to maximum height h):

Initial velocity = u, Final velocity = 0, Acceleration a = -g, Time taken = t₁.
Using the first kinematic equation: v = u + at
0 = u - gt₁  ⇒  t₁ = u / g  ...(Equation 1)
Using the third kinematic equation: v² = u² + 2as
0 = u² - 2gh  ⇒  u² = 2gh  ⇒  u = √(2gh)
Substitute u in Equation 1:
t₁ = √(2gh) / g = √(2h / g)  ...(Equation 2)

Case 2: Downward Motion (from height h to ground):

Initial velocity at highest point u = 0, Distance fallen s = h, Acceleration a = +g, Time taken = t₂.
Using the second kinematic equation: s = ut + ½ at²
h = (0 × t₂) + ½ gt₂² = ½ gt₂²
t₂² = 2h / g
t₂ = √(2h / g)  ...(Equation 3)

Conclusion:

Comparing Equation 2 and Equation 3:

t₁ = t₂

Hence, the time taken by the object to ascend to its maximum height is exactly equal to the time taken to descend back to the initial level.

5. If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?
Answer :
  • The weight of a heavy object resting on the floor is given by W = mg. If g doubles to 2g, the weight also doubles to 2W.
  • By Newton's third law, the normal reaction force N exerted by the floor on the object equals its weight: N = W = mg.
  • The frictional force opposing the horizontal pulling motion is directly proportional to the normal reaction: f = μ × N = μ × mg (where μ is the coefficient of friction).
  • When g becomes twice, the normal reaction doubles, causing the frictional force to double: f' = 2f.
  • Therefore, pulling the heavy object along the floor requires double the force to overcome friction, making it twice as difficult.
Q3) Explain why the value of g is zero at the centre of the earth.
Answer :
  • The acceleration due to gravity at depth d below the Earth's surface depends solely on the mass of the smaller concentric sphere of radius r = (R - d) beneath the object.
  • As depth increases towards the center, the effective mass M' of the Earth attracting the object decreases proportionally with r³, while distance decreases with r²: g = GM' / r² ∝ r.
  • At the exact center of the Earth, the radial distance becomes zero (r = 0).
  • Moreover, at the center, the matter of the Earth surrounds the object uniformly in all directions. The gravitational pull exerted by opposite sectors cancels out completely symmetrically.
  • Therefore, the net gravitational force acting on the object is zero, and acceleration due to gravity is zero (g = 0) at the center of the Earth.
Q4) Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be √8 T.
Answer :
Given Data:
Initial distance from star: R₁ = R, Initial orbital period: T₁ = T
New distance from star: R₂ = 2R, New orbital period: T₂ = ?
According to Kepler's Third Law:
The square of the orbital period is directly proportional to the cube of the orbital radius:
T² ∝ R³  ⇒  (T₂ / T₁)² = (R₂ / R₁)³
Calculation:
(T₂ / T)² = (2R / R)³ = 2³ = 8
Taking square root on both sides:
T₂ / T = √8 = 2√2
T₂ = √8 T   (or   2√2 T). Hence proved.
Q5) Solve the following problems:
1. An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?
Answer :
Given: Initial velocity u = 0, Height s = 5 m, Time t = 5 s
Formula: s = ut + ½ gt²
5 = (0 × 5) + ½ × g × (5)²
5 = ½ × g × 25
g = (5 × 2) / 25 = 10 / 25 = 0.4 m/s²
The value of g on the planet is 0.4 m/s².
2. The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?
Answer :
Given Data:
• Radius of planet A: RA = RB / 2  ⇒  RB = 2 RA
• Acceleration due to gravity on planet B: gB = gA / 2 = ½ gA
• Mass of planet A = MA,   Mass of planet B = MB = ?
Formula:
The acceleration due to gravity on the surface of a planet is given by: g = GM / R²
Step-by-step Calculation:
• For planet A: gA = G MA / RA²
• For planet B: gB = G MB / RB² = G MB / (2 RA)² = G MB / (4 RA²)
According to the given condition: gB = ½ gA
G MB / (4 RA²) = ½ × (G MA / RA²)
Cancelling G / RA² from both sides:
MB / 4 = MA / 2
MB = (4 / 2) × MA = 2 MA
The mass of planet B must be 2 MA (i.e., twice the mass of planet A).
3. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.
Answer :
Given: Mass on Earth m = 5 kg, Weight on Earth WEarth = 49 N, gMoon = (1/6) gEarth
1. Mass on Moon:
Mass is a constant property and does not change: Mass on Moon = 5 kg.
2. Weight on Moon:
WMoon = m × gMoon = m × (gEarth / 6) = WEarth / 6
WMoon = 49 / 6 ≈ 8.17 N
Mass on the moon = 5 kg; Weight on the moon = 8.17 N.
4. An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s².
Answer :
Given: Maximum height s = 500 m, Final velocity at highest point v = 0, g = 10 m/s² (upward: a = -10 m/s²).
1. Initial Velocity (u):
v² = u² + 2as  ⇒  0 = u² + 2(-10)(500)
0 = u² - 10000  ⇒  u² = 10000  ⇒  u = 100 m/s
2. Time to Reach Maximum Height (t₁):
v = u + at  ⇒  0 = 100 - 10t₁  ⇒  10t₁ = 100  ⇒  t₁ = 10 s
3. Total Time to Return to Ground:
Total time = Time of ascent (t₁) + Time of descent (t₂) = 10 s + 10 s = 20 s.
Initial velocity = 100 m/s; Total time to return = 20 seconds.
5. A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s², calculate its speed on reaching the ground and the height of the table.
Answer :
Given: Initial velocity u = 0, Time t = 1 s, g = 10 m/s²
1. Speed on reaching ground (v):
v = u + gt = 0 + (10 × 1) = 10 m/s
2. Height of the table (h):
h = ut + ½ gt² = (0 × 1) + ½ × 10 × (1)² = 5 m
Speed on reaching ground = 10 m/s; Height of table = 5 meters.
6. The masses of the earth and the moon are 6 × 10²⁴ kg and 7.4 × 10²² kg respectively. The distance between them is 3.84 × 10⁵ km. Calculate the gravitational force of attraction between them. Use G = 6.7 × 10⁻¹¹ N·m²/kg².
Answer :
Given: M₁ = 6 × 10²⁴ kg, M₂ = 7.4 × 10²² kg
Distance r = 3.84 × 10⁵ km = 3.84 × 10⁸ m
G = 6.7 × 10⁻¹¹ N·m²/kg²
Formula: F = G(M₁M₂) / r²
F = [6.7 × 10⁻¹¹ × 6 × 10²⁴ × 7.4 × 10²²] / (3.84 × 10⁸)²
F = [297.48 × 10³⁵] / [14.7456 × 10¹⁶]
F = 20.174 × 10¹⁹ N = 2.02 × 10²⁰ N
The gravitational force of attraction between the Earth and Moon is 2.02 × 10²⁰ N.
7. The mass of the earth is 6 × 10²⁴ kg. The distance between the earth and the sun is 1.5 × 10¹¹ m. If the gravitational force between the two is 3.5 × 10²² N, what is the mass of the sun? Use G = 6.7 × 10⁻¹¹ N·m²/kg².
Answer :
Given: Mass of Earth MEarth = 6 × 10²⁴ kg, Distance r = 1.5 × 10¹¹ m, Force F = 3.5 × 10²² N
G = 6.7 × 10⁻¹¹ N·m²/kg²
Formula: F = G(MSun × MEarth) / r²  ⇒  MSun = (F × r²) / (G × MEarth)
MSun = [3.5 × 10²² × (1.5 × 10¹¹)²] / [6.7 × 10⁻¹¹ × 6 × 10²⁴]
MSun = [3.5 × 10²² × 2.25 × 10²²] / [40.2 × 10¹³]
MSun = [7.875 × 10⁴⁴] / [40.2 × 10¹³] = 0.1959 × 10³¹ kg ≈ 1.96 × 10³⁰ kg
The mass of the Sun is approximately 1.96 × 10³⁰ kg.
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Chapter Overview: Gravitation

Class 10 Maharashtra State Board Science 1 Chapter 1 introduces students to the universal law of gravitation, Kepler's three laws of planetary motion, circular motion and centripetal force, variations in the value of 'g' (with altitude, depth, and latitude), mass versus weight, gravitational potential energy, free fall, and escape velocity.

Key Formulas Summary

Frequently Asked Questions (FAQs)

Q1. Why does the value of 'g' vary at different points on Earth's surface?

The Earth is not a perfect sphere; it bulges at the equator and is flattened at the poles. Since g ∝ 1/R², radius R is greatest at the equator (making g minimum ≈ 9.78 m/s²) and least at the poles (making g maximum ≈ 9.83 m/s²).

Q2. What is the weight of an object during free fall?

During free fall, both the object and its supporting surface accelerate downwards at the same rate 'g'. As the normal reaction force becomes zero, the object experiences weightlessness (apparent weight = 0).

Q3. What is the difference between G and g?

'G' is the Universal Gravitational Constant (6.67 × 10⁻¹¹ N·m²/kg²), a scalar quantity that never changes anywhere in the universe. 'g' is the acceleration due to gravity (9.8 m/s²), a vector quantity that varies with height, depth, and celestial bodies.

Q4. Why is escape velocity independent of the mass of the object?

From the derivation vesc = √(2GM / R), the formula depends only on the mass (M) and radius (R) of the celestial planet, not on the mass of the escaping object. Hence, a feather and a rocket require the exact same escape velocity.

Related Study Topics

Class 10 Gravitation Solutions Kepler's Three Laws Newton's Inverse Square Law Mass vs Weight Escape Velocity Derivation Free Fall Kinematics Gravitation Numericals Solved SSC Board Science 1
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