Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is \frac{1}{4}   the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis.  If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then :

 

  • Option 1)

    t1 = t2

  • Option 2)

     t1 = 2t2

  • Option 3)

     t1 = 3t2

  • Option 4)

     t1 = 4t2

     

 

Answers (1)

As we learnt in

Kepler's 3rd law -

T^{2}\: \alpha\: a^{3}

From fig.

AB=AF+FB

2a=r_{1}+r_{2}

\therefore\; a=\frac{r_{1}+r_{1}}{2}

a= semi major Axis

r_{1}= Perigee

- wherein

Known as law of periods

r_{2}= apogee

T^{2}\: \alpha \: \left ( \frac{r_{1}+r_{2}}{2} \right )^{3}

{r_{1}+r_{2}= 2a

 

 \frac{t_1}{t_2}=\frac{A/2+A/4}{A/2-A/4} =\: \frac{3A/4}{A/4}\: =\:3

\therefore t_1=3t_2

 


Option 1)

t1 = t2

Incorrect

Option 2)

 t1 = 2t2

Incorrect

Option 3)

 t1 = 3t2

Correct

Option 4)

 t1 = 4t2

 

Incorrect

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