The orbit of comet Halley around the Sun is a long thin ellipse. At its aphelion (point farthest from the Sun), the comet is 5.30 × 1012 m from the Sun and moves with a speed of 10.8 km/s. What is the comet’s speed at its perihelion (closest approach to the Sun) where its distance from the Sun is 8.90 × 1010 m? Gravitational constant is 6.674 × 10−11 N·m2/kg2 and the mass of the Sun is 1.987 × 1030 kg. (km/s)

To solve for the comet Halley's speed at perihelion, we can use the principle of conservation of mechanical energy in the context of orbital mechanics. The total mechanical energy (kinetic + potential) of an object in an elliptical orbit around a much larger mass, such as the Sun, remains constant.

The total mechanical energy per unit mass \(E\) for a comet orbiting the Sun is given by:

\[
E = \frac{1}{2} v^2 - \frac{GM}{r}
\]

Where:
- \(v\) is the speed of the comet.
- \(G\) is the gravitational constant.
- \(M\) is the mass of the Sun.
- \(r\) is the distance of the comet from the Sun at a given point.

At both the aphelion and the perihelion, the total mechanical energy is conserved. Therefore, the mechanical energy at the aphelion must equal the mechanical energy at the perihelion:

\[
\frac{1}{2} v_a^2 - \frac{GM}{r_a} = \frac{1}{2} v_p^2 - \frac{GM}{r_p}
\]

Where:
- \(v_a = 10.8 \, \text{km/s}\) is the speed of the comet at aphelion.
- \(r_a = 5.30 \times 10^{12} \, \text{m}\) is the distance of the comet from the Sun at aphelion.
- \(v_p\) is the unknown speed of the comet at perihelion.
- \(r_p = 8.90 \times 10^{10} \, \text{m}\) is the distance of the comet from the Sun at perihelion.
- \(G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2\) is the gravitational constant.
- \(M = 1.987 \times 10^{30} \, \text{kg}\) is the mass of the Sun.

We'll first rearrange the equation to solve for \(v_p\), the speed at perihelion:

\[
v_p^2 = v_a^2 + 2GM \left( \frac{1}{r_p} - \frac{1}{r_a} \right)
\]

Finally, we can calculate \(v_p\). Let's compute this step by step.The speed of comet Halley at its perihelion (closest approach to the Sun) is approximately \(55.2 \, \text{km/s}\).

Would you like more details on this calculation or have any additional questions?

Here are 5 related questions you might find interesting:
1. How does the shape of Halley's comet's orbit affect its speed at different points?
2. Why is the speed at perihelion significantly higher than at aphelion?
3. How does the conservation of angular momentum play a role in cometary motion?
4. What would happen to the comet's orbit if the mass of the Sun were different?
5. How do other forces, such as radiation pressure, affect a comet’s orbit over time?

**Tip:** For elliptical orbits, the object moves faster when it's closer to the central mass (like the Sun) due to the conservation of angular momentum.

The orbit of comet Halley around the Sun is a long thin ellipse. At its aphelion (point farthest from the Sun), the comet is 5.30 × 10^12 m from the Sun and moves with a speed of 10.8 km/s. What is the comet’s speed at its perihelion (closest approach to the Sun) where its distance from the Sun is 8.90 × 10^10 m? Gravitational constant is 6.674 × 10−11 N·m²/kg² and the mass of the Sun is 1.987 × 10³⁰ kg. (km/s)

Calculate comet Halley's speed at its perihelion, given its aphelion speed and distance from the Sun. This problem utilizes the principle of energy conservation in orbital mechanics, involving key formulas and gravitational constants. Suitable for advanced high school and college-level physics students.

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