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Velocity $V_s$ of the satellite is given by

$ \large\frac{GMm_s}{r_s^2}=\Large\frac{n_s v_s^2}{r_s}$

$v_s =\sqrt {\bigg(\large\frac{GM}{r_{\Large s}}\bigg)}$

$KE= \large\frac{1}{2} m_s v_s=\large\frac{1}{2}$$ m_s\bigg(\frac{GM}{n_{\Large s}}\bigg)$

$PE= \large\frac{-GMm_{\Large}s}{r_{\Large s}}$

$E$ Total energy $=KE+PE$-----(1)

$\qquad= \large\frac{-GMm_s}{2r_{\Large s}}$

Angular momentum

$L= m_sv_sr_s$

$\quad= m_s \bigg( \large\frac{GM}{r_{\Large s}}\bigg)^{1/2} $$r_{\large s}$

$\quad= (GMm_s^2 r_s)^{1/2}$

Substituting from (1)

$L=(2Em_sr_s^2)^{1/2} $

Hence a is the correct answer.

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