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Consider a planet orbiting the Sun, and let { P }_{ 1 }, { P }_{ 2 }, { P }_{ 3 }, and { P }_{ 4 } be the planet’s position at four corresponding time instants t_{ 1 }, t_{ 2 }, t_{ 3 }, and t_{ 4 } such that t_{ 2 }-{ t }_{ 1 }={ t }_{ 4 }-{ t }_{ 3 }. Letting O denote the position of the Sun, determine the ratio between the areas of the orbital sectors P1OP2 and P3OP4. Hint: (1) The area of triangle OAB defined by the two planar vectors \overrightarrow { c } and \overrightarrow { d } as shown is given by Area(ABC)=\left| \overrightarrow { c } \times \overrightarrow { d } \right| (2) the solution of this problem is a demonstration of Kepler’s second law

Step-by-step

Gravity is a central force so { \overrightarrow { M } }_{ O } = 0 and there is conservation of angular momentum.
{ r }_{ 1 }{ \widehat { u } }_{ r }\times m{ \overrightarrow { \upsilon } }_{ 1 }={ r }_{ 2 }{ \widehat { u } }_{ r }\times m{ \overrightarrow { \upsilon } }_{ 2 }
The term \overrightarrow { \upsilon } dt represents the differential displacement of the planet so we can multiply both sides by dt to get
\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ { r }_{ 1 }{ \widehat { u } }_{ r }\times m{ \overrightarrow { \upsilon } }_{ 1 }d } =\int _{ { t }_{ 3 } }^{ { t4 }_{ } }{ { r }_{ 2 }{ \widehat { u } }_{ r }\times m{ \overrightarrow { \upsilon } }_{ 2 }dt }
Using the hint we see that these integrals represent the differential area of an orbital sector. Thus we see that for t_{ 2 }-{ t }_{ 1 }={ t }_{ 4 }-{ t }_{ 3 } the ratio of the orbital sectors is equal to 1.
The ratio of the orbital sectors is equal to 1.

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