JEE Main 2016 · 9 Apr · Shift 1 · Physics Q6

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PhysicsSingle correct+4 / -1Question 6 of 90

Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is $\dfrac14$ the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If $t_1$ is the time taken for planet to go over path abc and $t_2$ for path taken over cda then :

An ellipse with major axis db and minor axis ca; the sun S is on the major axis left of the centre, and triangle csa is marked. The planet moves a → b → c → d.

t_1=t_2

t_1=2t_2

t_1=3t_2

t_1=4t_2

Answer and worked solution

Official answer: C

By Kepler's second law time is proportional to area swept. Over abc the radius from S sweeps half the ellipse plus triangle csa: $\frac12+\frac14=\frac34$ of the area. Over cda it sweeps $\frac14$ . So $t_1=3t_2$ .

Topics:Kepler's laws

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JEE Main 2016 · 9 Apr · Shift 1, Physics question 6