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A cosmic by A moves to the Sun with velocity v_0 (when far from the Sun) and aiming parameter l the arm of the vector v_0 relative to the centre of the Sun (figure). Find the minimum distance by which this body will get to the Sun. |
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Answer» Solution :At the minimum separation with the Sun, the cosmic body's velocity is perpendicular to its POSITION vector relative to the Sun. If `r_(min)` be the SOUGHT minimum distance, from conservation of angular momentum about the Sun (C). `mv_0l=mvr_(min)` or, `v=(v_0l)/(r_(min))` (1) From conservation of mechanical energy of the system (sun + cosmic body), `1/2mv_0^2=-(gammam_sm)/(r_(min))+1/2mv^2` So `v_0^2/2=-(gammam_s)/(r_(min))+(v_0^2)/(2r_(min)^2)` (using 1) or, `v_0^2r_(min)^2+2gammam_sr_(min)-v_0^2l^2=0` So, `r_(min)=(-2gammam_s+-SQRT(4gamma^2m^2+4v_0^2v_0^2l^2))/(2v_0^2)=(-gammam_s+-sqrt(gamma^2m_s^2+v_0^4l^2))/(v_0^2)` Hence, taking positive ROOT `r_(min)=(gammam_s//v_0^2)[sqrt(1+(lv_0^2//gammam_s)^2)-1]` |
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