Question

The satellite of mass m revolving in a circular orbit of radius r around the Earth has kinetic energy E. Then its angular momentum will be.
(A) $\sqrt {\dfrac{{\text{E}}}{{{\text{m}}{{\text{r}}^{\text{2}}}}}}$
(B) $\dfrac{{\text{E}}}{{{\text{2m}}{{\text{r}}^{\text{2}}}}}$
(C) $\sqrt {{\text{2Em}}{{\text{r}}^{\text{2}}}}$
(D) $\sqrt {{\text{2Emr}}}$

Answer 1

Try to find a relation between linear momentum and kinetic energy E and then proceed to find angular momentum

Complete step-by-step solution:
As we all know that kinetic energy of a body of mass m having velocity v is:
${\text{K = }}\dfrac{{{\text{m}}{{\text{v}}^{\text{2}}}}}{{\text{2}}}$
Also, linear momentum of a body of mass m moving with velocity v is:
${\text{p = mv}}$
Manipulating the equation as,
${\text{K = }}\dfrac{{{{\text{p}}^{\text{2}}}}}{{{\text{2m}}}}$
${\text{p = }}\sqrt {{\text{2mK}}}$
Angular momentum of a body moving in a circular orbit is defined as the cross product of the radius of path with the linear momentum of the body. Therefore,
$L = r \times p$

$${\text{L = r}}\sqrt {{\text{2mK}}} \\\ {\text{L = }}\sqrt {{\text{2mK}}{{\text{r}}^{\text{2}}}} \\\$$

So the correct answer is option D.

Note: One can also just observe the options. Eliminating the obvious terms can help. Eg. Here all the options other than D have unnecessary dimensions. Dimension of angular momentum = ${\text{[}}{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^{ - 1}}{\text{]}}$. Only option D has the same dimensions.