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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A satellite of mass $m_s$ revolving in a circular orbit of radius $r_s$ round the earth of mass M has a total energy E. Then its angular momentum will be

\[(a)\;(2Em_sr_s^2)^{1/2} \quad (b)\;(2Em_sr_s^2) \quad (c)\;(2Em_sr_s)^{1/2} \quad (d)\;2Em_sr_s \]
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1 Answer

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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.


answered Aug 24, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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