How does this diagram illustrate Kepler's second law?
A.
The areas of triangle CSD and ASB are equal, but the time to "sweep" those areas is greater for t1 than t2.
B.
The area of triangle CSD is greater than the area of triangle ASB, which means the time it takes to "sweep" those areas is greater for t2 than t1.
C.
The areas of triangle CSD and ASB are equal, and the time to "sweep" those areas is also equal (t1=t2).
D.
The area of triangle CSD is less than the area of triangle ASB, but the time it takes to "sweep" those areas is greater for t2 than t1.

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