Question

4. Show that the separation vector rmn=(XM - Xm) of two point masses m and M moving under their mutual gravitational attraction obeys the same acceleration equation as a star of small mass (i.e., a test particle of negligible mass) attracted by a mass (m+M); i.e., drmn/dt2=-G (m+M) rmn/I'mn. 20 points.

Answer 1

4. Answer Tomin (7m-im) m I'm beim consider o point as origin. The position vectors of me and M with bespect to a point are Rim and XM and the position vector of their centre of mass (G point) wisit o point is Ŕ. Now if we take the centre of mass (6) of the system as origin, and let z'm and The be position vectors of the particles in and M respectively pelative to this origin. Their separation vector is : Tomin = (xu - Am ). The force acted on them is only goavitational atisaction force. ... mxm = Feet = - Geming namin --- the extennal and and mñón = Fest = - Gemme Fomin ----(2) ext = G.MM Now, from figure, and Zn = xm Ř oth = xm - mxm + MXM Again I know R mm m + M M& M then I'm = xm - mxm + MXM I'm = m . ( xm - Pm) > Tóm =- the nomin where M= = m Im = – M Domin Deduced mass of system i mim =- 18 .-- (3) mm 1 mtu [ Hesse minus signus indicates the dissection of Smin is opposite , otherwise in scelar foom can write mig Mini Tาท just

and similarly, emas = Kom - Ř m = xm - on xm + mim M& M s m +M m ton = monto Tin – Km) mxm = suomion + min = ju õimin ---- (4) Foom , (2), 3, 4 I can write , ет m м minn → mm mt Po minha 12- nomin = – Gemm min dt2 domin – Ge (m + M) smin dt2 which is nothing but the acceleration equation as a star of small mass (i.e, a test pastice of negligible massim) atroacted by a mass (m+m). [ mdamin - - GIM+M) m! Jom (Prooved) 072