The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then new angular momentum will be

The correct option is 4L.

To find the new angular momentum of the satellite, we can establish the relationship between the orbital angular momentum of a satellite and its orbital radius.

The orbital velocity $v$ of a satellite of mass $m$ revolving around a planet of mass $M$ at a distance $r$ from the center is given by:
$v=\sqrt{\frac{GM}{r}}$
where $G$ is the universal gravitational constant.

The orbital angular momentum $L$ of the satellite is defined as:
$L=mvr$

Substituting the expression for orbital velocity $v$ into the equation for angular momentum, we get:
$L=m\sqrt{\frac{GM}{r}}r=m\sqrt{GMr}$

Since the mass of the satellite $m$, the mass of the planet $M$, and the gravitational constant $G$ remain constant, the angular momentum is directly proportional to the square root of the orbital radius:
$L\propto \sqrt{r}$

We can write the ratio of the new angular momentum $L_2$ to the initial angular momentum $L_1$ as:
$\frac{L_2}{L_1}=\sqrt{\frac{r_2}{r_1}}$

Given that the initial distance is $r_1=r$, the initial angular momentum is $L_1=L$, and the new distance is increased to $r_2=16r$:
$\frac{L_2}{L}=\sqrt{\frac{16r}{r}}=\sqrt{16}=4$

Therefore, the new orbital angular momentum is:
$L_2=4L$