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 of a planet during a given time period, we can make use of Kepler's Second Law of Planetary Motion. According to this law, the rate at which the radius vector sweeps out equal areas in equal time periods is constant for a given planet.
Given that the radius vector sweeps an area A in 45 days when the planet is at its slowest orbital speed, we can say that the rate of area swept is A/45.
Now, we need to determine the area swept during a 40-day time period when the planet is at its fastest orbital speed. Since the rate of area swept is constant, we can use the following proportion:
(Area swept in 45 days) / (Time taken in 45 days) = (Area swept in 40 days) / (Time taken in 40 days)
Plugging in the values, we have:
A/45 = x/40
To find the value of x (the area swept in 40 days), we can cross-multiply and solve for x:
40 * A = 45 * x
x = (40 * A) / 45
x = (8/9) A
Therefore, the area swept by the radius vector during a 40-day time period at the planet's fastest orbital speed is (8/9) A. Comparing this result to the original area A, we can see that (8/9) A is less than A.
Hence, the answer is D) Less than A.