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Standard XII

Physics

Kepler's Law

The figure re...

Question

The figure represents an elliptical orbit of a planet around sun. The planet takes time $T_{1}$ to travel from $A$ to $B$ and it takes time $T_{2}$ to travel from $C$ to $D$ . If the area $C S D$ is double that of area $A S B$ , then

T_{1} = T_{2}

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T_{1} = 2 T_{2}

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T_{1} = 0.5 T_{2}

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Solution

The correct option is B $T_{1} = 0.5 T_{2}$
According to Kepler's laws, planet takes equal time to sweep equal areas about the sun. So, the areal velocity of the planet remains constant.
Since, area of CSD is twice the area of ASB, $\Rightarrow T_{2} = 2 T_{1} \Rightarrow T_{1} = 0.5 T_{2}$ .

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The figure represents an elliptical orbit of a planet around sun. The planet takes time T1 to travel from A to B and it takes time T2 to travel from C to D. If the area CSD is double that of area ASB, then

The correct option is B T1=0.5T2According to Kepler's laws, planet takes equal time to sweep equal areas about the sun. So, the areal velocity of the planet remains constant. Since, area of CSD is twice the area of ASB, ⇒T2=2T1⇒T1=0.5T2.

The figure represents an elliptical orbit of a planet around sun. The planet takes time $T_{1}$ to travel from $A$ to $B$ and it takes time $T_{2}$ to travel from $C$ to $D$ . If the area $C S D$ is double that of area $A S B$ , then

The correct option is B $T_{1} = 0.5 T_{2}$
According to Kepler's laws, planet takes equal time to sweep equal areas about the sun. So, the areal velocity of the planet remains constant.
Since, area of CSD is twice the area of ASB, $\Rightarrow T_{2} = 2 T_{1} \Rightarrow T_{1} = 0.5 T_{2}$ .