# Orbital mechanics: hyperbolic trajectory: periapsis

The question is formulated as follows:

A satellite is some distance, r, from the centre of the earth, traveling at a speed $2v_e$, where $v_e$ is the escape velocity. The angle between the velocity and the satellite-earth line is 45 degrees. Express the distance of closest approach between the satellite and the earth centre.

My thoughts.

First I assume only Earth's gravitational acceleration is acting on the satellite. Then, I conclude the trajectory is a hyperbolic trajectory because the speed is larger than the escape velocity.

I can calculate the semi-major axis $a$ of the Kepler orbit using the vis-viva equation (as we know the velocity at a distance $r$ for body earth with gravitational parameter $GM$)

The closest distance to Earth occurs at periapsis $r_p = -a(e-1)$ where $e$ denotes the eccentricity.

Now my question is how to determine the eccentricity?

Is the following perhaps a good method? As the angle $\theta = 45$ degrees between the velocity vector and the radius vector is known, we assume that the angle between the hyperbola asymptotes is equal to $2\theta = 90 ^{\circ}$. In that case we can use $e = 1/cos(\theta)$ to calculate the eccentricity.

Combining all this together with the fact that $V_{esc} = \sqrt{2GM/r}$ we can find that $r_p = r/6$.

I am unsure about the eccentricity calculations. Could anyone advise me on this one?

An alternative to calculate the eccentricity would be using the eccentricity vector, but then I am do not obtain the elegant solution of $r_p = r/6$.

While calculating the orbital parameters works in this case, a simpler method is just to use energy and angular momentum conservation (as mentioned by fibonatic.) First, from energy conservation we have $$-\frac{GMm}{r} + \frac{1}{2} m v^2 = -\frac{GMm}{r_p} + \frac{1}{2} m v_p^2$$ Second, we have angular momentum conservation. At perigee, the satellite will have $dr/dt = 0$; this means that its velocity and radial vector will necessarily be at right angles, and therefore $L_p = m v_p r_p$. Thus, $$m v r \sin (45^\circ) = m v_p r_p$$ This is a system of two equations in two unknowns ($v_p$ and $r_p$), and therefore can be easily solved. That said, I don't get $r_p = r/6$ when I solve these equations, so there's an error somewhere (either in my derivation or in yours.)
It is possible to calculate the eccentricity from the angle of the velocity and other information. So your assumption will not be needed and can only approximate the eccentricity. The orbital eccentricity can be derived from this equation, which uses specific orbital energy, $\epsilon$, and angular momentum, $h$, $$e=\sqrt{1+\frac{2\epsilon h^2}{\mu^2}},$$ where $\mu$ is the gravitational parameter. You could also use other relations, such as $$h=\sqrt{\mu a\left(1-e^2\right)}.$$ I will leave the rest up to you. If you have any further questions or problems, feel free to ask.
• How do you express the specific angular momentum $h$? You think for this problem we should use $\boldsymbol{h} = \boldsymbol{r} \times \boldsymbol{V}$ and then saying $h = |\boldsymbol{h}|$? Dec 5, 2014 at 19:40
• @JohnArbor yes, but remember that $\times$ means the crossproduct, such that $h$ would be the radius times the angular component of the velocity. Dec 5, 2014 at 20:58