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? What is the formula answer?

Answer 1

The object orbiting the Sun sweeps out equal areas in equal time intervals, according to Kepler's Second Law of Planetary Motion. This means that the object will sweep out the same area at both the closest and farthest points of its orbit.

Therefore, the area swept out by the object at the farthest point of its orbit (t2) would also be A.

Mathematically, we can represent this as:

Area at t1 = Area at t2 = A