A satellite is orbiting the earth in a circular orbit of radius r. A small impulse is given to the satellite such that its velocity increases by a factor of alpha, where alpha is slightly greater than 1. What is the eccentricity of the new elliptical orbit?

A. ^2 - 1 — Using the conservation of angular momentum and energy at the perigee, the eccentricity is derived from the orbital parameters. For a circular orbit, the perigee distance is r, leading to e = alpha^2 - 1.

JEE Main · PhysicsHard

$A satellite is orbiting the earth in a circular orbit of radius $r$ with velocity $v_0$. A small impulse is given to the satellite such that its velocity increases to $v = \alpha v_0$, where $\alpha > 1$. What is the eccentricity $e$ of the new elliptical orbit?$

A. $\alpha^2 - 1$
B. $2(\alpha - 1)$
C. $\alpha - 1$
D. $\sqrt{\alpha^2 - 1}$

Show correct answer & step-by-step solution

Correct answer: A — $\alpha^2 - 1$

Solution

At perigee,

L = m r v = m r (α v_{0})

and

E = \frac{1}{2} m (α v_{0})^{2} - \frac{GM m}{r}

. Using

v_{0}^{2} = \frac{GM}{r}

, we get

E = \frac{GM m}{r} (\frac{α ^{2}}{2} - 1)

. The eccentricity formula

e = 1 + \frac{2 E L ^{2}}{m ( GM m ) ^{2}}

yields

e = α^{2} - 1

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