Question

at an apse the particle moves at right angle to the radius vector I. e at an apse the radius vector is perpendicular to the tangent​

Answer 1

Answer:

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Answer 2

Answer:

Proved.

Step-by-step explanation:

To Prove:- At an apse the particle moves at right angle to the radius

                 vector i.e. at an apse the radius vector is perpendicular to the

                  tangent​.

Proof:-

From the definition of an apse, the apsidal distance r is maximum or minimum.

∴ $u = \frac{1}{r}$ is also minimum or maximum at an apse.

⇒ $\frac{du}{dv} = 0$

For any curve, we have   $\frac{1}{p^{2} } = u^{2} + (\frac{du}{dv} )^{2}$

where p is the length of the perpendicular from the origin to the tangent at any point to the curve.

∴    $\frac{1}{p^{2} } = u^{2}$                 [∵ $\frac{du}{dv} = 0$]

⇒ $\frac{1}{p^{2} } = \frac{1}{r^{2} }$                   [∵ $u = \frac{1}{r}$]

⇒ $p^{2} = r^{2}$

⇒ $p = r$

⇒ r sin Ф = r

⇒ sin Ф = 1

⇒ Ф = 90°

Thus at an apse the particle moves at right angles to the radius vector i.e. at an apse the radius vector is perpendicular to the tangent​.

Hence, proved.

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