Gravitation

Given: Distance between two objects is reduced to half, i.e., new distance $r' = \dfrac{r}{2}$.

Formula used (Newton's Law of Gravitation):
$$F = G\frac{m_1 m_2}{r^2}$$

Working:

Original force:
$$F = G\frac{m_1 m_2}{r^2}$$

New force when distance is halved:
$$F' = G\frac{m_1 m_2}{\left(\dfrac{r}{2}\right)^2} = G\frac{m_1 m_2}{\dfrac{r^2}{4}} = 4 \times G\frac{m_1 m_2}{r^2} = 4F$$

Conclusion: When the distance between two objects is reduced to half, the gravitational force between them becomes four times the original force.

Concept: Although the gravitational force on a heavier object is greater (since $F = mg$), the acceleration produced depends on the mass of the object.

Explanation:

For a freely falling object, applying Newton's second law:
$$F = ma \implies mg = ma \implies a = g$$

The acceleration $g$ is independent of the mass $m$ of the object. Although a heavier object experiences a greater gravitational force, it also has greater inertia (resistance to motion). These two effects exactly cancel each other.

Conclusion: All objects, regardless of their mass, fall with the same acceleration $g \approx 9.8\, \text{m/s}^2$ (in the absence of air resistance). Hence, a heavy object does not fall faster than a light object.

Given:
- Mass of object, $m = 1\,\text{kg}$
- Mass of Earth, $M = 6 \times 10^{24}\,\text{kg}$
- Radius of Earth, $R = 6.4 \times 10^{6}\,\text{m}$
- Universal gravitational constant, $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$

Formula:
$$F = G\frac{Mm}{R^2}$$

Calculation:
$$F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^{6})^2}$$

$$F = \frac{6.67 \times 6 \times 10^{-11+24}}{40.96 \times 10^{12}}$$

$$F = \frac{40.02 \times 10^{13}}{40.96 \times 10^{12}}$$

$$F = \frac{400.2 \times 10^{12}}{40.96 \times 10^{12}} \approx 9.77\,\text{N}$$

Answer: The gravitational force between the Earth and the 1 kg object is approximately $\mathbf{9.8\,N}$.

Answer: The Earth attracts the Moon with a force that is equal to the force with which the Moon attracts the Earth.

Reason: According to Newton's Third Law of Motion, every action has an equal and opposite reaction. The gravitational force is a mutual force between two bodies. By Newton's Law of Gravitation:
$$F = G\frac{M_{\text{Earth}} \times M_{\text{Moon}}}{d^2}$$
This formula gives a single value of force for the pair. The force exerted by the Earth on the Moon and the force exerted by the Moon on the Earth are equal in magnitude but opposite in direction. Hence, both forces are the same.

Explanation:

The Moon does attract the Earth with a gravitational force, and in fact the Earth does accelerate slightly towards the Moon. However, the Earth also has a very large mass compared to the Moon.

Using Newton's Second Law: $a = \dfrac{F}{m}$

Since the mass of the Earth ($M_E = 6 \times 10^{24}\,\text{kg}$) is very large, the acceleration produced in the Earth by the Moon's gravitational pull is extremely small (negligible).

Additionally, both the Earth and the Moon are in a state of continuous revolution around their common centre of mass (barycentre). The gravitational force between them provides the centripetal force needed for this orbital motion.

Conclusion: The Earth does experience a force due to the Moon, but because of its enormous mass, the acceleration is negligibly small. The system is in orbital equilibrium, so the Earth does not visibly move towards the Moon.

Formula: $F = G\dfrac{m_1 m_2}{r^2}$

(i) Mass of one object is doubled ($m_1' = 2m_1$):
$$F' = G\frac{2m_1 \cdot m_2}{r^2} = 2 \times G\frac{m_1 m_2}{r^2} = 2F$$
The force becomes twice (doubles).

(ii) Distance is doubled ($r' = 2r$):
$$F' = G\frac{m_1 m_2}{(2r)^2} = G\frac{m_1 m_2}{4r^2} = \frac{F}{4}$$
The force becomes one-fourth.

Distance is tripled ($r'' = 3r$):
$$F'' = G\frac{m_1 m_2}{(3r)^2} = G\frac{m_1 m_2}{9r^2} = \frac{F}{9}$$
The force becomes one-ninth.

(iii) Masses of both objects are doubled ($m_1' = 2m_1$, $m_2' = 2m_2$):
$$F' = G\frac{2m_1 \cdot 2m_2}{r^2} = 4 \times G\frac{m_1 m_2}{r^2} = 4F$$
The force becomes four times the original force.

The Universal Law of Gravitation is important because it:

1. Explains planetary motion: It explains why planets revolve around the Sun and moons revolve around planets.

2. Explains tides: The gravitational force of the Moon (and Sun) on the Earth's oceans causes tides.

3. Explains free fall: It explains why all objects fall towards the Earth when released.

4. Predicts motion of celestial bodies: It helps in predicting the positions and motions of planets, comets, and other celestial objects.

5. Explains the binding of the atmosphere: The gravitational force of the Earth holds the atmosphere around it.

6. Space exploration: It is used to calculate the trajectories of spacecraft and satellites.

In summary, the universal law of gravitation unified terrestrial and celestial mechanics under one single law, making it one of the most significant achievements in science.

Definition: The acceleration with which an object falls freely towards the Earth (under the influence of Earth's gravitational force alone, with no air resistance) is called the acceleration due to gravity or acceleration of free fall.

Symbol: $g$

Value:
$$g = 9.8\,\text{m/s}^2 \approx 10\,\text{m/s}^2$$

This means that the velocity of a freely falling object increases by $9.8\,\text{m/s}$ every second.

Formula:
$$g = \frac{GM}{R^2}$$
where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth.

The gravitational force between the Earth and an object is called the weight of the object.

Formula:
$$W = mg$$
where:
- $W$ = weight of the object (in Newtons, N)
- $m$ = mass of the object (in kg)
- $g$ = acceleration due to gravity ($9.8\,\text{m/s}^2$)

Weight is a vector quantity directed towards the centre of the Earth. It varies from place to place (e.g., it is slightly less at the equator than at the poles), unlike mass which remains constant.

Answer: No, the friend will not agree with the weight of the gold.

Reason:

Weight is given by $W = mg$.

The value of $g$ is greater at the poles than at the equator because:
- The Earth is slightly flattened at the poles and bulges at the equator.
- The distance from the centre of the Earth to the surface is less at the poles than at the equator.
- Since $g = \dfrac{GM}{R^2}$, a smaller $R$ at the poles gives a larger $g$.

Since g_{\text{poles}} > g_{\text{equator}}, the weight of the gold at the poles will be more than its weight at the equator.

When Amit buys gold at the poles (where $g$ is higher), the balance shows a certain weight. When the same gold is weighed at the equator (where $g$ is lower), it will show a lesser weight.

Conclusion: The friend will find the gold to weigh less at the equator than what Amit paid for at the poles. The mass of the gold remains the same, but the weight changes.

Reason: A flat sheet of paper has a larger surface area compared to a crumpled ball of the same paper.

When an object falls through air, it experiences air resistance (drag force) which acts upward, opposing the motion.

- A flat sheet of paper has a large surface area, so it experiences greater air resistance, which slows it down significantly.
- A crumpled ball has a smaller surface area, so it experiences less air resistance and falls faster.

In the absence of air, both the flat sheet and the crumpled ball would fall at the same rate (same acceleration $g$), as demonstrated by Galileo's experiment.

Conclusion: The flat sheet of paper falls slower because of greater air resistance due to its larger surface area.

Given:
- Mass of object, $m = 10\,\text{kg}$
- $g_{\text{Earth}} = 9.8\,\text{m/s}^2$
- $g_{\text{Moon}} = \dfrac{1}{6} \times g_{\text{Earth}} = \dfrac{9.8}{6} \approx 1.633\,\text{m/s}^2$

Formula: $W = mg$

Weight on Earth:
$$W_{\text{Earth}} = m \times g_{\text{Earth}} = 10 \times 9.8 = \mathbf{98\,N}$$

Weight on Moon:
$$W_{\text{Moon}} = m \times g_{\text{Moon}} = 10 \times \frac{9.8}{6} = \frac{98}{6} \approx \mathbf{16.3\,N}$$

Answer:
- Weight of the object on Earth $= 98\,\text{N}$
- Weight of the object on Moon $\approx 16.3\,\text{N}$

Given:
- Initial velocity, $u = 49\,\text{m/s}$ (upward)
- At maximum height, final velocity, $v = 0$
- Acceleration due to gravity, $g = 9.8\,\text{m/s}^2$ (acting downward, so $a = -9.8\,\text{m/s}^2$)

(i) Maximum height:

Using the equation: $v^2 = u^2 + 2as$
$$0 = (49)^2 + 2 \times (-9.8) \times h$$
$$0 = 2401 - 19.6h$$
$$h = \frac{2401}{19.6} = \mathbf{122.5\,m}$$

(ii) Total time to return to the surface:

Time to reach maximum height using $v = u + at$:
$$0 = 49 + (-9.8) \times t$$
$$t = \frac{49}{9.8} = 5\,\text{s}$$

By symmetry, the time to fall back from maximum height to the ground is also $5\,\text{s}$.

$$\text{Total time} = 5 + 5 = \mathbf{10\,s}$$

Answers:
- (i) Maximum height $= 122.5\,\text{m}$
- (ii) Total time $= 10\,\text{s}$

Given:
- Initial velocity, $u = 0$ (released from rest)
- Height of tower, $s = 19.6\,\text{m}$
- Acceleration due to gravity, $g = 9.8\,\text{m/s}^2$

Formula: $v^2 = u^2 + 2gs$

Calculation:
$$v^2 = 0 + 2 \times 9.8 \times 19.6$$
$$v^2 = 2 \times 9.8 \times 19.6 = 384.16$$
$$v = \sqrt{384.16} = 19.6\,\text{m/s}$$

Answer: The final velocity of the stone just before touching the ground is $\mathbf{19.6\,m/s}$.

Given:
- Initial velocity, $u = 40\,\text{m/s}$ (upward)
- $g = 10\,\text{m/s}^2$
- At maximum height, $v = 0$

Maximum Height:

Using $v^2 = u^2 - 2gh$:
$$0 = (40)^2 - 2 \times 10 \times h$$
$$h = \frac{1600}{20} = \mathbf{80\,m}$$

Net Displacement:

The stone is thrown upward and returns to the same point (the thrower's hand/ground level).
$$\text{Net displacement} = \mathbf{0\,m}$$

Total Distance Covered:

The stone travels $80\,\text{m}$ upward and then $80\,\text{m}$ downward.
$$\text{Total distance} = 80 + 80 = \mathbf{160\,m}$$

Summary:
- Maximum height $= 80\,\text{m}$
- Net displacement $= 0\,\text{m}$
- Total distance covered $= 160\,\text{m}$

Given:
- Mass of Earth, $M_E = 6 \times 10^{24}\,\text{kg}$
- Mass of Sun, $M_S = 2 \times 10^{30}\,\text{kg}$
- Distance, $d = 1.5 \times 10^{11}\,\text{m}$
- $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$

Formula:
$$F = G\frac{M_E \times M_S}{d^2}$$

Calculation:
$$F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^2}$$

Numerator:
$$6.67 \times 6 \times 2 \times 10^{-11+24+30} = 80.04 \times 10^{43}$$

Denominator:
$$(1.5)^2 \times 10^{22} = 2.25 \times 10^{22}$$

$$F = \frac{80.04 \times 10^{43}}{2.25 \times 10^{22}} = \frac{80.04}{2.25} \times 10^{21} = 35.57 \times 10^{21}$$

$$F \approx 3.56 \times 10^{22}\,\text{N}$$

Answer: The gravitational force between the Earth and the Sun is approximately $\mathbf{3.56 \times 10^{22}\,N}$.

Given:
- Height of tower, $H = 100\,\text{m}$
- Stone 1: falls from top, initial velocity $u_1 = 0$
- Stone 2: projected upward from ground, initial velocity $u_2 = 25\,\text{m/s}$
- $g = 10\,\text{m/s}^2$

Let the two stones meet after time $t$ seconds.

Let the meeting point be at height $x$ from the ground.

For Stone 1 (falling from top):

Distance fallen from top $= H - x = 100 - x$
$$100 - x = \frac{1}{2}g t^2 = \frac{1}{2} \times 10 \times t^2 = 5t^2 \quad \cdots (1)$$

For Stone 2 (projected upward from ground):
$$x = u_2 t - \frac{1}{2}g t^2 = 25t - 5t^2 \quad \cdots (2)$$

Adding equations (1) and (2):
$$(100 - x) + x = 5t^2 + 25t - 5t^2$$
$$100 = 25t$$
$$t = 4\,\text{s}$$

Finding the height (from equation 2):
$$x = 25 \times 4 - 5 \times (4)^2 = 100 - 80 = 20\,\text{m}$$

Answer: The two stones will meet after $\mathbf{t = 4\,s}$ at a height of $\mathbf{20\,m}$ from the ground.

Given:
- Total time of flight $= 6\,\text{s}$
- $g = 9.8\,\text{m/s}^2$

Since the motion is symmetric, time to reach maximum height $= \dfrac{6}{2} = 3\,\text{s}$.

(a) Initial velocity:

At maximum height, $v = 0$. Using $v = u - gt$:
$$0 = u - 9.8 \times 3$$
$$u = 29.4\,\text{m/s}$$

The ball was thrown up with a velocity of $\mathbf{29.4\,m/s}$.

(b) Maximum height:

Using $s = ut - \dfrac{1}{2}gt^2$ for upward journey ($t = 3\,\text{s}$):
$$h = 29.4 \times 3 - \frac{1}{2} \times 9.8 \times (3)^2$$
$$h = 88.2 - \frac{1}{2} \times 9.8 \times 9$$
$$h = 88.2 - 44.1 = 44.1\,\text{m}$$

Maximum height reached $= \mathbf{44.1\,m}$.

(c) Position after 4 s:

The ball reaches maximum height at $t = 3\,\text{s}$ and then starts falling. At $t = 4\,\text{s}$, it has been falling for $4 - 3 = 1\,\text{s}$ from the maximum height.

Distance fallen in 1 s from maximum height:
$$d = \frac{1}{2}g(1)^2 = \frac{1}{2} \times 9.8 \times 1 = 4.9\,\text{m}$$

Position from ground $= 44.1 - 4.9 = 39.2\,\text{m}$

After 4 s, the ball is at a height of $\mathbf{39.2\,m}$ from the ground (on its way down).

The buoyant force (upthrust) on an object immersed in a liquid acts in the vertically upward direction.

Explanation: When an object is immersed in a liquid, the liquid exerts pressure on all surfaces of the object. The pressure on the lower surface of the object is greater than the pressure on the upper surface (since pressure increases with depth). This difference in pressure results in a net upward force called the buoyant force or upthrust, which always acts vertically upward — opposite to the direction of gravity.

Reason:

When a block of plastic is released under water, two forces act on it:
1. Weight ($W = mg$) — acting vertically downward.
2. Buoyant force (upthrust) — acting vertically upward.

The density of plastic is less than the density of water. According to Archimedes' Principle, the buoyant force equals the weight of water displaced by the plastic block.

Since the plastic has lower density than water:
\text{Buoyant force} > \text{Weight of plastic block}

The net force on the plastic block is directed upward, so it accelerates upward and rises to the surface of the water.

Conclusion: A block of plastic comes up to the surface because the buoyant force acting on it is greater than its weight.

Given:
- Mass of substance, $m = 50\,\text{g}$
- Volume of substance, $V = 20\,\text{cm}^3$
- Density of water, $\rho_{\text{water}} = 1\,\text{g cm}^{-3}$

Density of substance:
$$\rho_{\text{substance}} = \frac{m}{V} = \frac{50}{20} = 2.5\,\text{g cm}^{-3}$$

Comparison:
\rho_{\text{substance}} = 2.5\,\text{g cm}^{-3} > \rho_{\text{water}} = 1\,\text{g cm}^{-3}

Conclusion: Since the density of the substance ($2.5\,\text{g cm}^{-3}$) is greater than the density of water ($1\,\text{g cm}^{-3}$), the substance will sink in water.

Given:
- Mass of packet, $m = 500\,\text{g}$
- Volume of packet, $V = 350\,\text{cm}^3$
- Density of water, $\rho_{\text{water}} = 1\,\text{g cm}^{-3}$

Step 1: Find density of packet:
$$\rho_{\text{packet}} = \frac{m}{V} = \frac{500}{350} \approx 1.43\,\text{g cm}^{-3}$$

Step 2: Compare with density of water:
\rho_{\text{packet}} = 1.43\,\text{g cm}^{-3} > \rho_{\text{water}} = 1\,\text{g cm}^{-3}

Conclusion: Since the density of the packet is greater than the density of water, the packet will sink.

Step 3: Mass of water displaced:

When the packet sinks, it displaces water equal to its own volume ($350\,\text{cm}^3$).
$$\text{Mass of water displaced} = \rho_{\text{water}} \times V = 1 \times 350 = \mathbf{350\,g}$$

Answers:
- The packet will sink in water.
- Mass of water displaced $= 350\,\text{g}$.