In the discussion of Kepler's Second Law of Planetary Motion, the expression $\frac { d A } { d t }$ can be expressed as

A) $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$
B) $\| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$
C) $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$
D) $\| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$
E) $\mathbf { r } ( t ) \bullet \mathbf { v } ( t )$

Question

In the discussion of Kepler's Second Law of Planetary Motion, the expression $\frac { d A } { d t }$ can be expressed as

A) $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$
B) $\| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$
C) $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$
D) $\| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$
E) $\mathbf { r } ( t ) \bullet \mathbf { v } ( t )$

Quizplus · Accepted Answer

Kepler's Second Law, also known as the Law of Areas, states that the line joining a planet and the Sun sweeps out equal areas during equal intervals of time. The rate of change of the area $A$ swept by the radius vector $\mathbf{r}(t)$ with respect to time $t$, denoted as $\frac{dA}{dt}$, is given by $\frac{1}{2} \| \mathbf{r}(t) \times \mathbf{v}(t) \|$, where $\mathbf{r}(t)$ is the position vector of the planet relative to the Sun, and $\mathbf{v}(t)$ is the velocity vector of the planet. This expression comes from the area of the sector of an ellipse and the cross product represents the area of the parallelogram formed by the position and velocity vectors, with the factor of $\frac{1}{2}$ accounting for the fact that we are interested in the area of the triangle (or sector) swept out.

In the Discussion of Kepler's Second Law of Planetary Motion dAdt\frac { d A } { d t }dtdA​

In the Discussion of Kepler's Second Law of Planetary Motion $\frac { d A } { d t }$