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The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t_{1} is the time for the planet to move from C to D and t_{2} is the times to move from A to B then:

  • Option 1)

    t_{1}=4t_{2}

  • Option 2)

    t_{1}=2t_{2}

  • Option 3)

    t_{1}=t_{2}

  • Option 4)

    t_{1}>t_{2}

 

Answers (1)

best_answer

 

Kepler's 2nd law -

Area of velocity = \frac{dA}{dt}

\frac{dA}{dt}=\frac{1}{2}\frac{\left ( r \right )\left ( Vdt \right )}{dt}=\frac{1}{2}rV

\frac{dA}{dt}\rightarrow Areal velocity

dA\rightarrow  small area traced

- wherein

Simiar to Law of conservation of momentum

\frac{dA}{dt}= \frac{L}{2m}

L\rightarrow Angular momentum

Known  as Law of Area

 

 Equal areas are swept in equal time t_{1}, the time taken to go from C and B = 2t_{2}

where t_{2}  is the time taken to go from A to B 

As it is given that area SCB = 2SAB


Option 1)

t_{1}=4t_{2}

Incorrect option

Option 2)

t_{1}=2t_{2}

Correct option

Option 3)

t_{1}=t_{2}

Incorrect option

Option 4)

t_{1}>t_{2}

Incorrect option

Posted by

prateek

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