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

Hint: 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.