When a planet is at its slowest orbital speed its radius vector sweeps an area, A, in 45 days. What area will the radius vector for this planet sweep during a 40-day time period while at its fastest orbital speed?
A) A
B) More than A
C) 0.5 A
D) Less than A
To determine the area swept by the radius vector during a specific time period, we can use the concept of "conservation of angular momentum" in orbital motion. According to this concept, the product of the radius vector and the angular momentum remains constant throughout the planet's orbit.
In this scenario, we are given that the initial radius vector sweeps an area, A, in 45 days. Let's assume the initial radius vector is r1, and the initial orbital speed is v1.
Using the conservation of angular momentum, we have:
r1 * v1 = r2 * v2
where r2 is the radius vector during the 40-day time period, and v2 is the fastest orbital speed. We need to determine the area, A2, swept by the radius vector during this time period.
Since the angular momentum remains constant, we can rewrite the equation as:
r1 * v1 = r2 * (2v1) [Using the relation between fastest and slowest orbital speeds]
Simplifying the equation, we find:
r2 = (r1 * v1) / (2v1) = r1/2
This means the radius vector during the 40-day time period is half the length of the initial radius vector.
Now, the area swept by the radius vector is directly proportional to the square of its length. Therefore:
A2 = (r2)^2 = (r1/2)^2 = (1/4) * r1^2 = A/4
Therefore, the area swept by the radius vector during the 40-day time period at the fastest orbital speed is 1/4th of the initial area (A).
So, the correct answer is:
D) Less than A