Question

Exercise 5, parabolic and hyperbolic orbits (10pt) The solution to the radial function for an inverse-square-law force, see for example Taylor equation (8.59) or the equation above, is (6) TH For c = 1 (or the energy E = 0) the orbit reduces to a parabola as we saw in the previous exercise, while for e> (or energy positive) the orbit becomes a hyperbola. The equation for a hyperbola in For a hyperbola, identify the constants a B and 8 in terms of the constants and a forro

Answer 1

LPAPERWTTI - Date : -PAPERWEIZ Lou E = - K law here is no Itecoste where bz e2 un e-kle ank Ht Q El + QEb I The motion time is t z pe de TE + ķ put & weg amea y (9 mk z J L te cos &] 2 mi mk2 4 anolo [cita² da ezl for putabolid ad = e3 orbit és (tandt I tun? 0/) сам к2. amx 2 for general case hins geometry of problem there is a useful angle which allero this hot general case which is eccast í anomaly o is truly anomaly. The used angle maker all chean from figure we have tx=akes Ee) y=bsine afersine est +t-afrat -र) J: Smhfe - eso typerbolą s2 = IIP-E2 y = do E - Ellipse ez paraboly

Date : Page ne. itx betap Q--ķ Semineyaraxis for hyperbola xt-alloshe-e) we write these cas y=ate2 es/ Sinhe - 872 2 0 x2 pr weknow that latae - aloshe tojen sind L y - asinhe tez sub bucting sequening and cleling we get Tatae)² = a costhe T y .- a2sinhE les ) atae)?-_y? = atsink het cost?E) ye = h (atae) az ū - atle? - ke Compau with te² 732lattezt Is ke - 1 1 , B 2 det & -ke QE २

-PAPERWELL. Date : Page no. we know that er e² Eb a k = QEb '+ q Epl k sk put this at Ballet Test e 2, E moeten e there pa e in question I g z ec e²- t = = 1 e2UBE