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Solve it, The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

  • Option 1)

    K_{B}<K_{A}<K_{C}

  • Option 2)

    K_{A}>K_{B}>K_{C}

  • Option 3)

    K_{A}<K_{B}<K_{C}

  • Option 4)

    K_{B}>K_{A}>K_{C}

 
Answers (2)
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Option 2

As we have learned

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

 

V_{A}> V_{B}> V_{C}(vr = constant)

K_{A}> K_{B}> K_{C} 

 

 

 


Option 1)

K_{B}<K_{A}<K_{C}

This is incorrect

Option 2)

K_{A}>K_{B}>K_{C}

This is correct

Option 3)

K_{A}<K_{B}<K_{C}

This is incorrect

Option 4)

K_{B}>K_{A}>K_{C}

This is incorrect

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