Gravitation

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Answer:

Here are the key differences between mass and weight:

On Mars vs Earth:

• Mass: Since the amount of matter does not change with location, the mass of any object remains the same on both Earth and Mars.
• Weight: However, weight depends on the gravitational pull acting on the body. Because the gravitational acceleration (g) is not the same on Earth and Mars, the weight of the object differs on the two planets.
• Formula: Weight = Mass × g
• On Mars, g ≈ 3.7 m/s² (about 38% of Earth's g = 9.8 m/s²), so weight on Mars would be approximately 38% of weight on Earth.

Answer:

(i) Free Fall:

A body is said to be in free fall when it moves only under gravity's influence, with no other external force acting on it (except gravity). In such motion:

• The object starts with zero initial velocity (if simply dropped)
• Air resistance affects it in reality, so perfect free fall is possible only in a vacuum
• All objects in free fall accelerate at the same rate regardless of their mass (Galileo's principle)

Examples:

• A stone dropped from a table falls due to gravity alone
• A fruit dropping from a tree
• Astronauts in orbit experience "weightlessness" which is continuous free fall around Earth

Not free fall: Standing on the ground or sitting in an aircraft is not free fall because several forces act besides gravity.

(ii) Acceleration due to gravity (g):

This is the acceleration produced in an object due to the Earth's gravitational pull.

• SI unit: m/s²
• Represented by symbol: g
• Value on Earth's surface: approximately 9.8 m/s² (varies slightly with altitude and latitude)
• It is a vector quantity directed toward the Earth's centre
• On the Moon: 1.63 m/s² (about one-sixth of Earth's value)
• On Mars: 3.7 m/s²
• Formula: g = GM/R² where G is gravitational constant, M is Earth's mass, R is Earth's radius

(iii) Escape Velocity:

Escape velocity is the minimum initial speed required by an object to break free from a planet's gravitational force and never return (reach infinite distance with zero final velocity).

• On Earth: 11.186 km/s (approximately 11.2 km/s)
• Formula: vₑ = √(2GM/R)
• On Moon: 2.38 km/s (much lower due to smaller mass)
• Satellites and rockets are launched with speeds equal to or greater than escape velocity so they can leave Earth's gravitational influence
• Escape velocity depends on the mass and radius of the celestial body but is independent of the mass of the escaping object

(iv) Centripetal Force:

This is the force that pulls an object toward the centre of the circular path in which it moves, keeping it in circular motion.

• It is directed inward toward the rotation centre
• Formula: F = mv²/r or F = mω²r
• In planetary motion, gravitational force provides the centripetal force that keeps planets in orbit

Example: A stone tied to a string moves in a circle because the tension in the string provides the centripetal force. If the string breaks, the stone flies off tangentially (Newton's First Law).

Answer:

Johannes Kepler (1571-1630) analyzed the precise astronomical observations of Tycho Brahe and found that planetary movements follow certain mathematical rules. Based on these observations, he proposed three fundamental laws known as Kepler's Laws of Planetary Motion:

How these helped Newton:

Sir Isaac Newton (1643-1727) used Kepler's laws as crucial evidence for his Universal Law of Gravitation:

1. From Kepler's third law (T² ∝ r³) and assuming circular orbits for simplicity
2. Using centripetal force formula: F = mv²/r
3. Orbital velocity: v = 2πr/T
4. Combining these: F ∝ m/r² (for same central body)
5. By symmetry and Newton's Third Law: F ∝ M/r² (for same orbiting body)
6. Thus: F ∝ Mm/r² or F = G Mm/r²

Newton thus mathematically showed that the gravitational force between two bodies decreases with the square of the distance between them (inverse-square law). He also proposed that this is a universal law applying to all masses in the universe.

Proof:

Let the stone move upward with initial velocity u and rise to maximum height h.

Time taken to rise (t₁):

Using the first equation of motion:

At the topmost point (maximum height):

• Final velocity v = 0
• Acceleration a = -g (gravity acts downward, opposite to upward motion)

Time taken to fall (t₂):

For downward motion from height h:

• Initial velocity = 0 (starts from rest at top)
• Acceleration = +g (downward direction, same as motion)
• Distance = h

Using second equation of motion:

Relation between u and h:

Using third equation of motion for upward path:

Substitute u into t₁:

Compare t₁ and t₂:

Thus, the ascent time equals the descent time: t₁ = t₂

This result assumes no air resistance and constant g. In reality, air resistance makes descent slightly longer due to reduced net acceleration.

Answer:

The force needed to lift or pull an object horizontally against friction depends on its weight:

Original situation:

Weight of object: W₁ = mg

Force of friction (which opposes motion): F_friction = μN = μmg

Where:

• μ = coefficient of friction
• N = mg = normal force (equal to weight for horizontal surface)

Force needed to pull: F₁ = μmg (to overcome friction)

When g doubles:

New weight: W₂ = m(2g) = 2mg

New frictional force: F_friction_new = μ(2mg) = 2μmg

Force needed to pull: F₂ = 2μmg = 2F₁

Thus, when g becomes two times greater, the required pulling force also becomes double, making it twice as difficult to pull a heavy body across the floor.

Additional insight: This applies to:

• Lifting objects vertically: Force needed = weight = mg, so doubles when g doubles
• Pulling horizontally: Force needed to overcome friction ∝ normal force ∝ weight ∝ g
• Even the effort to hold an object stationary increases proportionally with g

Explanation:

The value of gravitational acceleration g depends on:

1. The Earth's mass (M) that effectively attracts the object
2. The distance (R) from the Earth's centre to the object

Formula: g = GM/R² (outside or on surface)

As we move toward the Earth's centre:

1. Effective Mass Decreases:

Only the mass inside the spherical shell at your current radius contributes to gravity. The mass in outer shells exerts no net gravitational force on an object inside them (shell theorem proved by Newton).

2. Distance Decreases:

As R decreases, the 1/R² factor would normally increase g, but the decreasing effective mass counteracts this.

Mathematical Derivation:

For uniform density Earth (approximation):

Thus inside Earth: g ∝ R

At different positions:

• At surface (R = R_earth): g = 9.8 m/s²
• At half radius (R = R_earth/2): g ≈ 4.9 m/s²
• At quarter radius (R = R_earth/4): g ≈ 2.45 m/s²
• As R → 0: g → 0

At the exact centre (R = 0):

• All Earth's mass is symmetrically distributed around you
• Gravitational pulls from all directions cancel each other completely
• Net gravitational force = 0
• Thus g = 0 at Earth's centre

Real Earth consideration: Earth's density increases toward the core, so g initially increases slightly before decreasing to zero, but still reaches zero at centre.

Proof:

According to Kepler's third law of planetary motion:

or

where k is a constant for planets orbiting the same star.

Given:

• Initial distance: r₁ = R
• Initial period: T₁ = T
• New distance: r₂ = 2R
• New period: T₂ = ?

Applying Kepler's third law:

Cross-multiplying:

Taking square root:

Thus, at twice the distance, the orbital period becomes √8 times (approximately 2.83 times) the original period.

Verification: This makes sense intuitively - farther planets move slower and take much longer to complete orbits. For example, Mars (1.5 AU from Sun) takes 1.88 Earth years, while Earth (1 AU) takes 1 year.

Solution:

Step 1: Write Newton's Law of Universal Gravitation

Step 2: Rearrange for Sun's mass (M_sun)

Step 3: Substitute given values

Step 4: Calculate numerator

Step 5: Calculate denominator

Step 6: Final calculation

Answer: The mass of the Sun is approximately 1.96 × 10³⁰ kg.

This is about 330,000 times the mass of Earth, which matches known astronomical values.