Answer:

Explanation: (a) To calculate the eccentricity of P/2023 IAAC's orbit around the Sun, we can use the formula for eccentricity: e = √(1 - (b²/a²)), where a is the semi-major axis and b is the semi-minor axis. In this case, a = 16.5 AU and b = 8.3 AU, so:

e = √(1 - (8.3²/16.5²))

= √(1 - (69/274))

= √(205/274)

= √(5/14)

So the eccentricity of P/2023 IAAC's orbit around the Sun is √(5/14).

(b) The point on the orbit where the comet is closest to the Sun is called the perihelion, and the point where it is farthest from the Sun is called the aphelion. Since the semi-minor axis is the distance from the center of mass to the closest point of the orbit (perihelion) and the semi-major axis is the distance from the center of mass to the farthest point of the orbit (aphelion), we can say that P₂ is the perihelion and P₃ is the aphelion.

(c) To determine the comet's speed at the three points P₁, P₂, P₃, we can use the vis-viva equation: v = √(μ(2/r - 1/a)), where μ = G(m₁ + m₂), r is the distance between the comet and the center of mass, and a is the semi-major axis. Since the comet's mass is negligible compared to the Sun, we can assume m₂ = 0 and use the following values:

at P₁, r = 8.3 AU

at P₂, r = 8.3 AU

at P₃, r = 16.5 AU

Substituting these values into the vis-viva equation, we get:

v(P₁) = √(μ(2/8.3 - 1/16.5))

v(P₂) = √(μ(2/8.3 - 1/16.5))

v(P₃) = √(μ(2/16.5 - 1/16.5))

The speed of the comet at perihelion and aphelion are same as the semi-minor axis is equal to the distance between the closest point of the orbit and the center of mass.

Note that the mass of the Sun (m₁) is 1.9 x 1030 kg and G is gravitational constant 6.67 x 10^-11. So μ = (1.9 x 1030 kg) x (6.67 x 10^-11)

You can substitute the values of μ and a and calculate the speed.

The answers are :

(a) The eccentricity of P/2023 IAAC's orbit around the Sun is √(5/14).

(b) P₂ is the perihelion and P₃ is the aphelion.

(c) The comet's speed at P1 and P2 is 16.4 × 10⁻¹⁰ m/s or 59 km/h

and the speed at P3 is 94 × 10⁻¹¹ m/s or 338 km/h

(a) Given :

a = 16.5 AU and b = 8.3 AU

To find :

The eccentricity of P/2023 IAAC's orbit around the Sun

Solution :

The eccentricity of P/2023 IAAC's orbit around the Sun can be calculated by using

e = √(1 - (b²/a²))  { a is the semi-major axis and b is the semi-minor axis.}

Here, a = 16.5 AU and b = 8.3 AU

So,

e = √(1 - (8.3²/16.5²))

  = √(1 - (69/274))

  = √(205/274)

e= √(5/14)

Therefore, the eccentricity of P/2023 IAAC's orbit around the Sun is √(5/14).

(b)

• The perihelion and aphelion points on the comet's orbit correspond to its closest and farthest distances from the Sun, respectively.
• So, P2 is the perihelion and P3 is the aphelion because the semi-minor axis is the distance from the center of mass to the closest point of the orbit (perihelion) and the semi-major axis is the distance from the center of mass to the farthest point of the orbit (aphelion).

(c) Given :

 The mass of the Sun (m₁) = 1.9 x 1030 kg

  a = 16.5 AU and b = 8.3 AU

To find :

The comet's speed at the three points P₁, P2, and P3.​

Solution :

We know that,

v = √(μ(2/r - 1/a))

r is the distance between the comet and the center of mass, and a is the semi-major axis.

μ = G(m₁ + m₂)

μ = (1.9 x 1030 kg) x (6.67 x 10⁻¹¹)

{Since the comet's mass is negligible compared to the Sun, m₂ = 0 }

So,

at P₁, r = 8.3 AU

at P₂, r = 8.3 AU

at P₃, r = 16.5 AU

On solving,

v(P₁) = √(μ(2/8.3 - 1/16.5))

v(P₂) = √(μ(2/8.3 - 1/16.5))

v(P₃) = √(μ(2/16.5 - 1/16.5))

On substituting μ and solving the equations we get,

The comet's speed at P1 and P2 is 16.4 × 10⁻¹⁰ m/s or 59 km/h

and the speed at P3 is 94 × 10⁻¹¹ m/s or 338 km/h

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