If an object orbiting the Sun sweeps out an area of A in a time t1 at the closest 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?
a
2A
b
A
c
1/2A
d
A^2
The correct answer is b) A.
The area swept out by an object in a given time is proportional to the time spent and the orbital speed of the object. Since the distance between the object and the Sun is the same at both the closest and farthest points of the orbit, the time spent at each point is the same.
Therefore, the object will sweep out the same amount of area at both the closest and farthest points of its orbit.
The area swept out by an object orbiting the Sun is the same at any point along its orbit. Therefore, the area swept out at the farthest point of the object's orbit, t2, would also be A.
To find the area swept out by an object orbiting the Sun at its farthest point, t2, we need to understand the relationship between the areas swept out at the closest and farthest points.
The area swept out by an object in a given time is governed by Kepler's Second Law, also known as the law of equal areas. It states that a line segment drawn from the Sun to the object in a given time will always sweep out the same area, regardless of the object's position in its orbit.
So, regardless of whether the object is at its closest point or farthest point, the area it sweeps out in the same amount of time will be the same.
Therefore, the area swept out at the farthest point, t2, is equal to the area swept out at the closest point, t1.
Based on the options given:
a) This is twice the area swept out at the closest point. (2A)
b) This is the same as the area swept out at the closest point. (A)
c) This is half the area swept out at the closest point. (1/2A)
d) This is the square of the area swept out at the closest point. (A^2)
The correct answer is b) A, as explained earlier that the area swept out is the same at both the closest and farthest points in the orbit.