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 4 Click here to jump to the problem! fredluis2019-09-17 02:05:02 The solution given above is wrong. Purely qualitatively, by Ampere\'s Law, B field will be in the +/- Z direction at point P. The energy flux vector S points outward (obviously). Thus E is in the XY plane (Orthogonal Triad). tree removalClick here to jump to the problem!
 5 Click here to jump to the problem! fredluis2019-09-17 02:01:31 Opposite sides of the cube have the same potential, so as we go across the cube we start with the potential at V and end with the potential at V. Since there are no charges inside, there is nothing to change the potential so it must be constant all the way across. tree removalClick here to jump to the problem!
 10 Click here to jump to the problem! fireballs2019-09-13 14:18:36 Here\'s an even simpler solution. You know that for an elliptical orbit, the speed of the body is: $\\sqrt{\\frac{GM}{R}}$<$v$< $\\sqrt{\\frac{2GM}{R}}$ Assume for the sake of argument that Jupiter\'s velocity saturates the lower bound here, i.e. that it is a circular orbit and $v_j$ = $\\sqrt{\\frac{GM}{R}}$. Then the speed of the spacecraft is $\\frac{3}{2} v_j$ = $\\frac{3}{2}\\sqrt{\\frac{GM}{R}}$ = $\\sqrt{\\frac{9GM}{4R}}$ > $\\sqrt{\\frac{2GM}{R}}$\r\n\r\nSo the orbit must be hyperbolic.Click here to jump to the problem!