1] assuming that the moon describes a circular orbit of radius R about the earth in 27 days and that titan describes a circular orbit of radius 3.2R about saturn in 16days compare the masses of the saturn and earth?

2] assuming that the easth is spherical in shape (r=6400km) rotating abt its axis with a period 24hrs ,find hw muh acceleration due to gravity and at poles would differ?

To tackle these questions, we need to apply some principles from physics, particularly those related to orbital mechanics and gravitational acceleration. Let’s break down each question step by step.

Comparing the Masses of Saturn and Earth

First, we need to analyze the orbital characteristics of the Moon and Titan. We can use Kepler's Third Law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (R) of its orbit around a central body. This can be expressed mathematically as:

T² ∝ R³

For the Moon orbiting Earth:

• Orbital period (T₁) = 27 days = 27 × 24 × 60 × 60 seconds = 2,332,800 seconds
• Radius (R₁) = R

For Titan orbiting Saturn:

• Orbital period (T₂) = 16 days = 16 × 24 × 60 × 60 seconds = 1,382,400 seconds
• Radius (R₂) = 3.2R

Applying Kepler's Third Law for both moons, we have:

For the Moon:

T₁² = k * R₁³

For Titan:

T₂² = k * R₂³

Dividing the two equations gives:

(T₁² / T₂²) = (R₁³ / R₂³)

Substituting the values:

(2,332,800² / 1,382,400²) = (R³ / (3.2R)³)

This simplifies to:

(2,332,800² / 1,382,400²) = (1 / 32)

Calculating the left side:

(2,332,800 / 1,382,400)² = (1.687)² ≈ 2.84

Setting the two sides equal gives:

2.84 = 1/32 * (M₁/M₂)

Where M₁ is the mass of Saturn and M₂ is the mass of Earth. Rearranging gives:

M₁/M₂ = 2.84 * 32 ≈ 90.88

This means Saturn is approximately 90.88 times more massive than Earth.

Acceleration Due to Gravity at the Poles

Now, let’s find out how the acceleration due to gravity varies at the poles of Earth. The formula for gravitational acceleration (g) at the surface of a spherical body is given by:

g = G * M / r²

Where:

• G is the gravitational constant (approximately 6.674 × 10⁻¹¹ N(m/kg)²)
• M is the mass of the Earth (approximately 5.972 × 10²⁴ kg)
• r is the radius of the Earth (approximately 6400 km or 6.4 × 10⁶ m)

Substituting the values into the formula gives:

g = (6.674 × 10⁻¹¹) * (5.972 × 10²⁴) / (6.4 × 10⁶)²

Calculating this yields:

g ≈ 9.81 m/s²

However, due to the Earth's rotation, there is a slight reduction in gravitational acceleration at the equator compared to the poles. The centrifugal force caused by rotation is greatest at the equator and zero at the poles. The formula for the effective gravitational acceleration (g') at the equator is:

g' = g - ω² * r

Where:

• ω is the angular velocity of the Earth (2π radians / 24 hours = 7.27 × 10⁻⁵ rad/s)
• r is the radius of the Earth (6.4 × 10⁶ m)

Calculating the centrifugal effect:

ω² * r ≈ (7.27 × 10⁻⁵)² * (6.4 × 10⁶) ≈ 0.034 m/s²

Thus, the effective gravitational acceleration at the equator is:

g' ≈ 9.81 - 0.034 ≈ 9.776 m/s²

At the poles, there is no centrifugal force acting against gravity, so:

g' at poles ≈ 9.81 m/s²

In summary, the acceleration due to gravity is slightly higher at the poles than at the equator due to the Earth's rotation. The difference is about 0.034 m/s², making gravity at the poles approximately 9.81 m/s², while at the equator, it is about 9.776 m/s².