A satellite of mass M / 2 is revolving around earth in a circular orbit at a height of R / 3

Question

A satellite of mass \(\frac{M}{2}\) is revolving around earth in a circular orbit at a height of \(\frac{R}{3}\) from earth surface. The angular momentum of the satellite is \(M \sqrt{\frac{G M R}{x}}\). The value of x is _____, where M and R are the mass and radius of earth, respectively.

(G is the gravitational constant)

Answer 1

Answer is : 3

  

\(r = R+ \frac{R}{3}\) 

\( = \frac{4R}{3}\) 

\(V_0 = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM\times 3}{4R}}\) 

\(L = \frac{M}{2}. \sqrt{\frac{3GM}{4R}}.\frac{4R}{3}\)

\( = M\sqrt{\frac{3GM}{4R \times 4} \times \frac{16R^2}{9}}\)

\( =M\sqrt{\frac{GMR}{3}}\) 

\(x = 3\)