Chapter 9: Gravitation
Exercise & In-Text Questions & Answers

Answer & Reference:
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects. (Page 101)

Answer & Reference:
The formula is: F = G × (M × m) / R²
Where:
- F is the gravitational force
- G is the universal gravitational constant
- M is the mass of the earth
- m is the mass of the object
- R is the radius of the earth (distance between the object and the earth's center)
(Page 103, Eq. 9.8)

Answer & Reference:
Whenever objects fall towards the earth under the gravitational force alone, we say that the objects are in free fall. (Page 102)

Answer & Reference:
The acceleration experienced by an object during free fall due to the earth's gravitational force is called acceleration due to gravity. It is denoted by 'g' and its value is approximately 9.8 m s⁻². (Page 103)

Answer & Reference:

Mass	Weight
Mass is the measure of inertia of an object.	Weight is the force with which an object is attracted towards the earth.
Mass is constant and does not change from place to place.	Weight depends on location because it depends on 'g' which varies with location.
SI unit is kilogram (kg).	SI unit is newton (N).
It is a scalar quantity.	It is a vector quantity (has direction towards the center of the earth).

(Page 105)

Answer & Reference:
The weight of an object on the moon is 1/6th its weight on the earth because the mass of the moon is less than that of the earth, and the radius of the moon is smaller. Using the universal law of gravitation:
W_m/W_e = (M_m/M_e) × (R_e/R_m)²
Substituting values from Table 9.1 (Page 105):
M_m = 7.36 × 10²² kg, M_e = 5.98 × 10²⁴ kg
R_m = 1.74 × 10⁶ m, R_e = 6.37 × 10⁶ m
W_m/W_e = (7.36 × 10²² / 5.98 × 10²⁴) × (6.37 × 10⁶ / 1.74 × 10⁶)² ≈ 1/6
(Page 105, Eq. 9.19)

Answer & Reference:
It is difficult to hold a school bag with a thin strap because the pressure exerted on the shoulders is large. Pressure is inversely proportional to the area on which the force acts. A thin strap has a smaller area, resulting in higher pressure on the shoulders. (Page 107, Eq. 9.20)

Answer & Reference:
Buoyancy is the upward force exerted by a fluid on an object immersed in it. This force is also known as upthrust. (Page 108)

Answer & Reference:
An object floats if its density is less than that of water, because the upthrust of water on the object is greater than its weight. An object sinks if its density is more than that of water, because the upthrust is less than its weight. (Page 109)

Answer & Reference:
Your actual mass is more than 42 kg. The weighing machine shows the apparent weight, which is less than the actual weight due to the buoyant force exerted by air. Since mass = weight/g, the measured mass is slightly less than the actual mass. (Page 110)

Answer & Reference:
The bag of cotton is actually heavier than the iron bar. Although both show the same mass on the weighing machine, the cotton has a larger volume and thus experiences a greater buoyant force from air. Therefore, its actual weight (and mass) is greater than what is measured. (Page 110)

Answer & Reference:
According to the universal law of gravitation, F ∝ 1/d². If the distance is reduced to half (d/2), the new force F' ∝ 1/(d/2)² = 4/d². Therefore, the force becomes four times the original force. (Page 101, Eq. 9.2)

Answer & Reference:
All objects fall with the same acceleration due to gravity (g) regardless of their mass. From F = mg and F = GMm/R², we get g = GM/R², which is independent of the mass of the falling object. Hence, heavy and light objects fall at the same rate. (Page 104)

Answer & Reference:
Using F = G × (M × m) / R²
G = 6.7 × 10⁻¹¹ N m² kg⁻², M = 6 × 10²⁴ kg, m = 1 kg, R = 6.4 × 10⁶ m
F = (6.7 × 10⁻¹¹ × 6 × 10²⁴ × 1) / (6.4 × 10⁶)²
F = (40.2 × 10¹³) / (40.96 × 10¹²) = 9.8 N
(Page 103, Eq. 9.9)

Answer & Reference:
The earth attracts the moon with the same force as the moon attracts the earth. According to Newton's third law of motion, forces of action and reaction are equal and opposite. (Page 101)

Answer & Reference:
The earth does not move towards the moon because the mass of the earth is much larger than that of the moon. According to Newton's second law (F = ma), for the same force, acceleration is inversely proportional to mass. Hence, the earth's acceleration towards the moon is negligible compared to the moon's acceleration towards the earth. (Page 101)

Answer & Reference:
(i) If mass of one object is doubled, force doubles (F ∝ m₁).
(ii) If distance is doubled, force becomes 1/4th (F ∝ 1/d²). If distance is tripled, force becomes 1/9th.
(iii) If masses of both objects are doubled, force becomes four times (F ∝ m₁m₂).
(Page 101, Eq. 9.3)

Answer & Reference:
The universal law of gravitation successfully explains:
(i) The force that binds us to the earth
(ii) The motion of the moon around the earth
(iii) The motion of planets around the Sun
(iv) The tides due to the moon and the Sun
(Page 102)

Answer & Reference:
The acceleration of free fall is the acceleration experienced by an object during free fall due to the earth's gravitational force. Its value is 9.8 m s⁻². (Page 103)

Answer & Reference:
The gravitational force between the earth and an object is called the weight of the object. (Page 105)

Answer & Reference:
No, the friend will not agree with the weight of gold bought. The value of 'g' is greater at the poles than at the equator. Since weight = mg, the gold weighs more at the poles than at the equator. When weighed at the equator, it will show less weight. (Page 103)

Answer & Reference:
A sheet of paper falls slower than a crumpled ball because it experiences more air resistance due to its larger surface area. The crumpled ball has less surface area exposed to air resistance, so it falls faster. (Page 103, Activity 9.3)

Answer & Reference:
Mass of object, m = 10 kg
On earth: W_e = m × g = 10 kg × 9.8 m s⁻² = 98 N
On moon: W_m = (1/6) × W_e = (1/6) × 98 N = 16.33 N
(Page 106, Examples 9.4 and 9.5)

Answer & Reference:
Initial velocity, u = 49 m/s
Final velocity at max height, v = 0 m/s
Acceleration, a = -g = -9.8 m/s²
(i) Using v² = u² + 2as
0 = (49)² + 2(-9.8)s
s = 2401 / 19.6 = 122.5 m
(ii) Time to reach max height: v = u + at
0 = 49 - 9.8t → t = 5 s
Total time = 2 × 5 = 10 s
(Page 104, Example 9.3)

Answer & Reference:
Height, s = 19.6 m
Initial velocity, u = 0 m/s
Acceleration, a = g = 9.8 m/s²
Using v² = u² + 2as
v² = 0 + 2(9.8)(19.6) = 384.16
v = √384.16 = 19.6 m/s
(Page 104, Eq. 9.12)

Answer & Reference:
Initial velocity, u = 40 m/s
Final velocity at max height, v = 0 m/s
Acceleration, a = -g = -10 m/s²
Using v² = u² + 2as
0 = (40)² + 2(-10)s
s = 1600 / 20 = 80 m (maximum height)
Total distance covered = 80 m (up) + 80 m (down) = 160 m
Net displacement = 0 (since it returns to starting point)
(Page 104, Eq. 9.12)

Answer & Reference:
M_e = 6 × 10²⁴ kg, M_s = 2 × 10³⁰ kg, d = 1.5 × 10¹¹ m
G = 6.7 × 10⁻¹¹ N m² kg⁻²
F = G × (M_e × M_s) / d²
F = (6.7 × 10⁻¹¹ × 6 × 10²⁴ × 2 × 10³⁰) / (1.5 × 10¹¹)²
F = (80.4 × 10⁴³) / (2.25 × 10²²) = 3.57 × 10²² N
(Page 102, Example 9.1)

Answer & Reference:
Let the stones meet after time 't' at height 'h' from ground.
For falling stone: Distance covered = 100 - h = (1/2)gt²
For rising stone: h = 25t - (1/2)gt²
Adding both equations: 100 = 25t → t = 4 s
Then h = 25(4) - (1/2)(9.8)(4)² = 100 - 78.4 = 21.6 m
So they meet after 4 seconds at 21.6 m above ground.
(Page 104, Eqs. 9.11 and 9.12)

Answer & Reference:
Total time = 6 s → Time to reach max height = 3 s
(a) Using v = u + at → 0 = u - 9.8(3) → u = 29.4 m/s
(b) Using s = ut + (1/2)at² → s = 29.4(3) - (1/2)(9.8)(3)² = 44.1 m
(c) After 4 s: It has spent 3 s going up and 1 s coming down.
Distance fallen in 1 s = (1/2)(9.8)(1)² = 4.9 m
Position = 44.1 - 4.9 = 39.2 m above ground
(Page 104, Eqs. 9.10 and 9.11)

Answer & Reference:
The buoyant force on an object immersed in a liquid acts in the vertically upward direction. (Page 108)

Answer & Reference:
A block of plastic comes up to the surface because its density is less than that of water. The buoyant force acting on it is greater than its weight, resulting in a net upward force. (Page 109)

Answer & Reference:
Density of substance = mass/volume = 50 g / 20 cm³ = 2.5 g/cm³
Since density of substance (2.5 g/cm³) > density of water (1 g/cm³), the substance will sink. (Page 109)

Answer & Reference:
Density of packet = 500 g / 350 cm³ ≈ 1.43 g/cm³
Since density of packet > density of water, it will sink.
Mass of water displaced = volume of packet × density of water
= 350 cm³ × 1 g/cm³ = 350 g
(Page 109)