国际物理林匹克竞赛试题ThSolution.docVIP

Th1 AN ILL FATED SATELLITE SOLUTION 1.1 and 1.2 1.3 2.1 The value of the semi-latus-rectum l is obtained taking into account that the orbital angular momentum is the same in both orbits. That is The eccentricity value is where E is the new satellite mechanical energy that is Combining both, one gets This is an elliptical trajectory because. 2.2 The initial and final orbits cross at P, where the satellite engine fired instantaneously (see Figure 4). At this point 2.3 From the trajectory expression one immediately obtains that the maximum and minimum values of r correspond to and respectively (see Figure 4). Hence, they are given by that is and For, one gets The distances and can also be obtained from mechanical energy and angular momentum conservation, taking into account that and are orthogonal at apogee and at perigee What remains of them, after eliminating v, is a second-degree equation whose solutions are and . 2.4 By the Third Kepler Law, the period T in the new orbit satisfies that where a, the semi-major axis of the ellipse, is given by Therefore For ? = 1/4 3.1 Only if the satellite follows an open trajectory it can escape from the Earth gravity attraction. Then, the orbit eccentricity has to be equal or larger than one. The minimum boost corresponds to a parabolic trajectory, with ? = 1 This can also be obtained by using that the total satellite energy has to be zero to reach infinity (Ep = 0) without residual velocity (Ek = 0) This also arises from or from . 3.2 Due to , the polar parabola equation is where the semi-latus-rectum continues to be . The minimum Earth - satellite distance corresponds to , where This also arises from energy conservation (for E = 0) and from the equality between the angular momenta (L0) at the initial point P and at maximum approximation, where and are orthogonal. 4.1 If the satellite escapes to infinity with residual