## Question:

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.