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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}

• Option 2)

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

• Option 3)

$K_{A}

• Option 4)

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

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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}

This is incorrect

Option 2)

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

This is correct

Option 3)

$K_{A}

This is incorrect

Option 4)

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

This is incorrect

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