If $\mathrm{A}$ is the areal velocity of a planet of mass $\mathrm{M}$, then its angular momentum is

The areal velocity of a planet is defined as the rate at which the area is swept by the radius vector, given by $\dfrac{dA}{dt} = \dfrac{1}{2} r^2 \dfrac{d\theta}{dt}$.

The angular momentum $L$ of a planet of mass $M$ moving in an orbit is given by $L = mvr = mr^2 \omega$, where $\omega = \dfrac{d\theta}{dt}$.

Substituting $\omega$ into the expression for angular momentum, we get $L = M r^2 \dfrac{d\theta}{dt}$.

Comparing this with the expression for areal velocity $A = \dfrac{1}{2} r^2 \dfrac{d\theta}{dt}$, we see that $r^2 \dfrac{d\theta}{dt} = 2A$.

Therefore, $L = M(2A) = 2MA$.

Answer: $2 \mathrm{MA}$