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For a satellite given an initial speed and position above a planet, I'd like to know whether it will experience an elliptical, circular, parabolic or hyperbolic trajectory, or crash into the planet. The intitial position $(x,y)$ relative to the centre of Mars, and the velocity $(v_x,v_y)$ are known.

I feel as though the orbit will be circular if effective potential $U^*(r) = L^2/{2mR^2} - GMm/R $ is at a minimum (with respect to r), where L=angular momentum. Differentiating, $\frac{dU^*}{dr} = \frac{GMm} {r^2} - \frac{L^2}{mr^3} = 0$. And hence if it undergoes a circular orbit, radius is $\frac{L^2}{GMm}%$, is this correct?

If air resistance is ignored, how much information is necessary to determine whether a satellite will crash into a planet or not, is there a minimum energy below which it will always spiral in and collide?

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    $\begingroup$ It will never "spiral in and collide" while "air resistance is ignored" unless you invoke general relativity or multi-body effects. In Newtonian mechanics all two-body "orbits" in a $r^{-1}$ potential---open or closed---are conic sections. $\endgroup$ – dmckee Jan 4 '16 at 21:49
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    $\begingroup$ @dmckee: Well...it could still crash into the planet if the small axis of the elliptic orbit is smaller than the radius of the body it's orbiting around. $\endgroup$ – ACuriousMind Jan 5 '16 at 14:46
  • $\begingroup$ @ACuriousMind - yes, but that's hardly "spiraling in". $\endgroup$ – Floris Jan 6 '16 at 20:01
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The vis viva equation is

$$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$

This tells us that if we know the velocity and radial distance of a satellite at any one point in the orbit, we can know it for any other point. We can use this to answer your question.

First - the satellite will escape if the parameter $a$ (the semimajor axis of the orbit) diverges (so $\frac{1}{a}\rightarrow 0$), which happens when:

$$v > \sqrt{\frac{2GM}{r}}$$

The trajectory will be parabolic when the velocity is exactly equal to the escape velocity (eccentricity = 1):

$$v = \sqrt{\frac{2GM}{r}}$$

And for any velocity less than this, the trajectory will be elliptical. Whether the orbit will dip below the surface of the planet (i.e., whether it will crash) depends on the value of $a$ and the eccentricity $e$. We know that $a$ is related to the specific energy $\epsilon$ of the orbit by

$$\epsilon = -\frac{2GM}{a} = \frac12 v^2 - \frac{GM}{r}$$

Thus, it is enough to know the velocity and instantaneous radius to know the parameter $a$:

$$ a = \frac{2GM}{\frac{GM}{r}-\frac12 v^2}\\ =\frac{2r}{1-\frac{v^2 r}{2GM}}$$

This will only be valid for $v<v_{\rm{escape}}$.

Now we need to use this to find the distance of closest approach. Here we need to use angular momentum. Borrowing inspiration from this answer, we write

$$|M|=|\vec{r}\times\vec{v}|=\sqrt{GMa(1-e^2)}$$

We can solve this for $e$. Then we use

$$\frac{r_p}{r_a} = \frac{1-e}{1+e}$$

which you combine with $r_p + r_a = a$ to solve for $r_p$. And if that is less than the radius of the planet, it will crash.

I will leave the details of that calculation as an exercise. There's probably a shortcut I forgot.

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  • $\begingroup$ You using $e$ and $\epsilon$ above. To avoid confusing you should use one or the other but not both. $\endgroup$ – ja72 Jan 6 '16 at 18:57
  • $\begingroup$ @ja72 - they are not the same thing. $e$ is the eccentricity, $\epsilon$ is the characteristic energy. These are the symbols commonly used. Do you have a suggestion for alternative symbols? $\endgroup$ – Floris Jan 6 '16 at 19:33
  • $\begingroup$ @ja72 I do - in the paragraph right after the equation for escape velocity (starting with "And for any velocity less than this..."). Unless "eccentricity $e$" isn't enough definition for you (I also give the relationship between eccentricity and perigee/apogee). $\endgroup$ – Floris Jan 6 '16 at 20:01
  • $\begingroup$ Another way to find the eccentricity is to take the magnitude of the eccentricity vector. $\endgroup$ – David Hammen Jan 6 '16 at 21:24
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For a more conceptual approach: If the total mechanical energy ($K+U$) is positive, the orbit will be hyperbolic. (Remember that $U$ will be negative.)

If $K+U = 0$, the orbit will be parabolic.

If $K+U < 0$, the orbit will be some type of ellipse.

If $K+U=\frac{U}{2}$, the elliptical orbit will be circular (eccentricity, $e$ = 0).

If the orbit is elliptical and non-circular, you need to find the periapsis: $$r_p=a(1-e).$$ Use the approach that Floris gives for that calculation.

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