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  • The Figure shows elliptical orbit of a planet m about the sum S. The shaded area SCD is twice the shaded area SAB. If t<s

The Figure shows elliptical orbit of a planet m about the sum S. The shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet of move from C to D and t2 is the time to move from A to B then:

Option: 1

t1 = 4t2


Option: 2

t1 = 2t2


Option: 3

t1 = t2


Option: 4

t1 > t2


Answers (1)

best_answer

As we learned in 

Kepler's 2nd law -

Area of velocity = \frac{dA}{dt}=\frac{1}{2}\frac{\left ( r \right )\left ( Vdt \right )}{dt}=\frac{1}{2}rV= \frac{L}{2m}

here L\rightarrow Angular momentum

 

So we have

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

I.e if  Angular momentum is conserved then \frac{dA}{dt}= \frac{L}{2m}=constant

or we can say that equal areas are swept in equal time

and this is also known as The Law of Equal Areas in Equal Time.

As

If t_{1} is the time for the planet to move from C to D and t_{2} is the time to move from A to B then:

And it is given that (area of SCD) = 2*(Area of SAB)

SO

\frac{dA}{dt}= \frac{L}{2m}=constant\\ \frac{\Delta SCD}{t_1}=\frac{\Delta SAB}{t_2}\\ \Rightarrow t_1=2t_2

 

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