If an object orbiting the Sun sweeps out of an area of A in a time t1 at the closet point of the object's orbital rotation. How much area does the object sweep out at the farthest point of the object's orbit t2?
According to Kepler's second law of planetary motion, the object sweeps out equal areas in equal time intervals. Therefore, the object will sweep out the same area A at both the closest and farthest points of its orbit.
To determine how much area the object sweeps out at the farthest point of its orbit (t2), we can use Kepler's Second Law, also known as the Law of Equal Areas.
According to Kepler's Second Law, an imaginary line connecting the Sun and the object (known as the radius vector) sweeps out equal areas in equal periods of time. This means that the rate at which an object sweeps out area is constant throughout its orbit.
Given that the object sweeps out area A in a time t1 at the closest point of its orbit, we can assume that it will sweep out the same area A in the same time t1 at the farthest point of its orbit.
Therefore, the object will sweep out an area of A at the farthest point of its orbit (t2), just as it did at the closest point of its orbit (t1).
To determine the area an object sweeps out at different points of its orbit around the Sun, we can make use of Kepler's Second Law, also known as the Law of Equal Areas. According to this law, a line that connects a planet or object with the Sun sweeps out equal areas in equal times. This means that the area swept by the object over a given time period remains constant, regardless of its position in its orbit.
Now let's consider the scenario given in the question. At the closest point of the object's orbit, the object sweeps out an area labeled as A in a time t1. We can say that this area is constant because of Kepler's Second Law.
At the farthest point of the object's orbit, we need to calculate the area swept out, labeled as A'. To find A', we need to determine the time it takes for the object to sweep out this area.
To do this, we can use the concept of orbital periods. The time it takes for an object to complete one full orbit around the Sun is called its orbital period. Let's denote the orbital period as T.
Since the area swept out is constant, we can write the following equation:
A = A'
Now, considering that the object orbits the Sun, we can determine its orbital period using Kepler's Third Law. Kepler's Third Law states that the square of the orbital period is proportional to the cube of the average distance between the object and the Sun.
Therefore, we can write:
T1^2 / T2^2 = r1^3 / r2^3
In this equation, T1 is the time period at the closest point of the orbit (t1), T2 is the unknown time period at the farthest point (t2), r1 is the distance between the object and the Sun at the closest point, and r2 is the distance between the object and the Sun at the farthest point.
To calculate the area A', we need to find t2. Rearranging the equation, we can solve for t2:
T1^2 / T2^2 = r1^3 / r2^3
T2^2 = (r2 / r1)^3 * T1^2
T2 = sqrt((r2 / r1)^3 * T1^2)
Now, knowing T2, we can calculate the area A' swept out at the farthest point:
A' = A = (1/2) * base * height
The base can be taken as the average distance of the object from the Sun, which is (r1 + r2) / 2. The height can be taken as the distance traveled by the object in time t2, which is the circumference of its orbit at that distance.
A' = (1/2) * (r1 + r2) / 2 * (2 * π * (r1 + r2) / 2 * t2)
Simplifying the equation:
A' = (1/2) * π * (r1 + r2) * t2
Now, by substituting the calculated value of t2, we can find the area A'.