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?

Question

-2A
-A
-1/2 A
-A^2

Answer 1

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.