Question

A planet rotates in an elliptical orbit with a star situated at one of the foci. The distance from the center of the ellipse to any foci is half of the semi-major axis. The ratio of the speed of the planet when it is nearest (perihelion) to the star to that at the farthest (aphelion) is \rule{1cm{0.15mm}. (in integer)}

Hint:

For any orbital motion under a central force, angular momentum is conserved. This means the product of speed and distance is constant at the points of closest and farthest approach (perihelion and aphelion). The planet moves fastest when it is closest to the star and slowest when it is farthest away.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
For a planet in an elliptical orbit around a star, its angular momentum is conserved. This principle, a consequence of Kepler's second law, relates the planet's speed to its distance from the star. The points of nearest and farthest approach are the perihelion and aphelion, respectively.
Step 2: Key Formula or Approach:
1. Let \(a\) be the semi-major axis and \(c\) be the distance from the center of the ellipse to a focus. 2. The perihelion distance (nearest) is \(r_p = a - c\). 3. The aphelion distance (farthest) is \(r_a = a + c\). 4. Conservation of angular momentum between perihelion and aphelion implies \(m v_p r_p = m v_a r_a\), which simplifies to \(v_p r_p = v_a r_a\). 5. The ratio of speeds is therefore \(\frac{v_p}{v_a} = \frac{r_a}{r_p}\).
Step 3: Detailed Explanation:
We are given that the distance from the center to the focus is half the semi-major axis: \[ c = \frac{a}{2} \] Now, we calculate the perihelion and aphelion distances: \[ r_p = a - c = a - \frac{a}{2} = \frac{a}{2} \] \[ r_a = a + c = a + \frac{a}{2} = \frac{3a}{2} \] Using the conservation of angular momentum, we find the ratio of the speeds: \[ \frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{3a/2}{a/2} = 3 \] Step 4: Final Answer:
The ratio of the speed at perihelion to the speed at aphelion is 3.