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Equation of the trajectory of a typical space object around any planet, in polar coordinates $(r, \theta)$ (i.e. a general conic section geometry), is given as follows. ( $h$ is angular momentum, $\mu$ is gravitational parameter, $e$ is eccentricity, $r$ is radial distance from the planet center, $\theta$, is angle between vectors $\vec{e}$ and $\vec{r}$.
(A) $r=\frac{\left(h^{2} / \mu\right)}{1-e \cos \theta}$
(B) $r=\frac{\left(h^{2} / \mu\right)}{e-\cos \theta}$
(C) $r=\frac{\left(h^{2} / \mu\right)}{1+e \cos \theta}$
(D) $r=\frac{\left(h^{2} / \mu\right)}{e+\cos \theta}$



Question ID - 155505 | SaraNextGen Top Answer

Equation of the trajectory of a typical space object around any planet, in polar coordinates $(r, \theta)$ (i.e. a general conic section geometry), is given as follows. ( $h$ is angular momentum, $\mu$ is gravitational parameter, $e$ is eccentricity, $r$ is radial distance from the planet center, $\theta$, is angle between vectors $\vec{e}$ and $\vec{r}$.
(A) $r=\frac{\left(h^{2} / \mu\right)}{1-e \cos \theta}$
(B) $r=\frac{\left(h^{2} / \mu\right)}{e-\cos \theta}$
(C) $r=\frac{\left(h^{2} / \mu\right)}{1+e \cos \theta}$
(D) $r=\frac{\left(h^{2} / \mu\right)}{e+\cos \theta}$

1 Answer
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Answer Key / Explanation : (C) -

(C) $r=\frac{\left(h^{2} / \mu\right)}{1+e \cos \theta}$

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