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A planet of mass m moves along an ellipse around the Sun so that its maximum and minimum distances from the Sun are equal to `r_1` and `r_2` respectively. Find the angular momentum M of this planet relative to the centre of the Sun. |
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Answer» As the planet is under central force (gravitational interaction), its angular momentum is conserved about the Sun (which is situated at one of the focii of the ellipse) So, `mv_1r_1=mv_2r_2` or, `v_1^2=(v_2^2r_2^2)/(r_1^2)` (1) From the conservation of mechanical energy of the system (Sun+planet), `-(gammam_sm)/(r_1)+1/2mv_1^2=-(gammam_sm)/(r_2)+1/2mv_2^2` or, `-(gammam_s)/(r_1)+1/2v_2^2(r_2^2)/(r_1^2)=-((gammam_s)/(r_2))+1/2v_2^2` [Using (1)] Thus, `v_2=sqrt(2gammam_sr_1//r_2(r_1+r_2))` (2) Hence `M=mv_2r_2=msqrt(2gammam_sr_1r_2//(r_1+r_2))` |
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