A planet moves in an elliptical orbit about the Sun, with theSun at one focus of

Question

A planet moves in an elliptical orbit about the Sun, with theSun at one focus of the ellipse, as in Figure 10-48. A.)Waht is the torque about the center of the Sun due to thegravitational force of attraction of the Sun on the planet? B.) At position A, the planet has an orbital radius rsub1 andis moving with a speed vsub1 perpendicular to the line from the Sunto the planet. At position B, the planet has an orbital radiusrsub2 and is moving with speed vsub2, again perpendicular to theline from the Sun to the planet. What is the ratio of vsub1 tovsub2 in terms of rsub1 and rsub2? A planet moves in an elliptical orbit about the Sun, with theSun at one focus of the ellipse, as in Figure 10-48. A.)Waht is the torque about the center of the Sun due to thegravitational force of attraction of the Sun on the planet? B.) At position A, the planet has an orbital radius rsub1 andis moving with a speed vsub1 perpendicular to the line from the Sunto the planet. At position B, the planet has an orbital radiusrsub2 and is moving with speed vsub2, again perpendicular to theline from the Sun to the planet. What is the ratio of vsub1 tovsub2 in terms of rsub1 and rsub2?

Explanation / Answer

(A) Gravitational force acts along the line jooining theobjects; Sun and the satellite. So radius vector r is parallel to the forcevector, F. So,the torque exerted, (Vector product)                  = r x F =0. (B). Since = 0,                  = dL/dt = 0 Angular moentum So L = r x p =constant. But the linear moentum, p =m.v So, r x mv = constant . Since, v is at 90 degrees with r.          rx mv = r.mv =constant So, r.v is constant.           r1.v1 = r2.v2 The ratio,               v1/v2 = r2/r1.       

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