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Escape Velocity

Figure shows ...

Question

Figure shows the orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse. If t $_{1}$ is the time taken by the planet to travel along ACB and t $_{2}$ the time along BDA, then :

t_{1} = t_{2}

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t_{1} > t_{2}

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t_{1} < t_{2}

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nothing can be concluded

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Solution

The correct option is B $t_{1} > t_{2}$
Here, the angular momentum is conserved i.e. $m v r$ is constant. Where $r$ is the distance from center of the sun to center of the planet.

Now, consider one point each from both the mentioned paths.(Shown in fig.)

Applying conservation of angular momentum for these points we get

$m v_{1} r_{1} = m v_{2} r_{2}$ . Simplifying this, $\frac{v_{1}}{v_{2}} = \frac{r_{2}}{r_{1}} < 1$

Time period are given by

$t_{1} = \frac{L}{v_{1}}, t_{2} = \frac{L}{v_{2}}$

Comparing them by taking the ratio

$\frac{t_{1}}{t_{2}} = \frac{L}{v_{1}} \times \frac{v_{2}}{L}$

So, $\frac{t_{1}}{t_{2}} = \frac{v_{2}}{v_{1}} > 1.$ Thus $t_{1} > t_{2}$

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Figure shows the orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse. If t1 is the time taken by the planet to travel along ACB and t2 the time along BDA, then :

Figure shows the orbit of a planet P round the sun S. AB and CD are the minor and major axes of the ellipse. If t $_{1}$ is the time taken by the planet to travel along ACB and t $_{2}$ the time along BDA, then :