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# A Satellite is in an elliptic orbit around the earth with an aphelion of $6 \mathrm{R}$ andperihelion of $2 \mathrm{R}$. Find the eccentricity of the orbit.A. $\dfrac{1}{6}$B. $\dfrac{1}{3}$C. $\dfrac{1}{2}$D. $\dfrac{1}{4}$

Last updated date: 02nd Aug 2024
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Hint: We know that the maximum distance of load from the centre of column, such that if load acts within this distance there is no tension in the column. The maximum distance is called the Limit of eccentricity. Eccentricity measures how much the shape of Earth's orbit departs from a perfect circle. These variations affect the distance between Earth and the Sun. Because variations in Earth's eccentricity are fairly small, they're a relatively minor factor in annual seasonal climate variations. The orbital eccentricity (or eccentricity) is a measure of how much an elliptical orbit is 'squashed'. Elliptical orbits with increasing eccentricity from $\mathrm{e}=0$ (a circle) to $\mathrm{e}=0.95 .$ For a fixed value of the semi-major axis, as the eccentricity increases, both the semi-minor axis and perihelion distance decrease.

Let us consider that $\mathrm{r}_{\mathrm{a}}$ and $\mathrm{r}_{\mathrm{p}}$ denote distance of aphelion and perihelion of the
$\mathrm{r}_{\mathrm{a}}=\mathrm{a}(1+\mathrm{e})$ and $\mathrm{r}_{\mathrm{p}}=(1-\mathrm{e})(\mathrm{a}$ is semi major axis of ellipse)
As $\mathrm{r}_{\mathrm{a}}=\mathrm{bl}$ and $\mathrm{r}_{\mathrm{p}}=2 \mathrm{R}$
$\dfrac{\mathrm{a}(1+\mathrm{e})}{\mathrm{a}(1-\mathrm{e})}=\dfrac{6 \mathrm{R}}{2 \mathrm{R}}=3$ where $\mathrm{e}=\dfrac{1}{2}$
So, eccentricity $=\dfrac{1}{2}$