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?
-2A
-A
-1/2 A
-A^2
The object sweeps out equal areas in equal time intervals, according to Kepler's Second Law of Planetary Motion. This means that the area swept out at the farthest point of the object's orbit, t2, will also be A.
So, the answer is -A.
The area swept out by an object orbiting the Sun in a given time is equal to the area of a sector of the orbit. The formula to calculate the area of a sector is A = (1/2) * r^2 * θ, where r is the radius of the orbit and θ is the angle swept by the object.
At the closest point of the object's orbit, the radius is the shortest, resulting in a smaller angle swept. Therefore, the area swept out at the closest point (t1) would be smaller than the area swept out at the farthest point (t2).
Since the area swept out at t1 is A, the area swept out at t2 would be larger. Therefore, the correct answer is -2A.
To determine the area swept out by an object in its orbit at the farthest point, we need to understand Kepler's Second Law of Planetary Motion.
Kepler's Second Law states that an imaginary line joining a planet to the Sun sweeps out equal areas in equal time intervals. This means that the rate at which the object sweeps out area is constant throughout its orbit.
Given that at the closest point of the object's orbital rotation, it sweeps out an area A in a time t1, we can assume that the rate at which the area is swept out is constant.
Therefore, at the farthest point of the object's orbit, it will also sweep out an area A in the same time interval t1.
Hence, the object will sweep out an area of A at both the closest and the farthest points of its orbit. So, the answer is A.