# What are speed limits for free fall, orbital trajectory, and unbound trajectory?

This might be a beginner's question, but I cannot find a full answer.

I'd like to clarify the theoretical speeds that change the nature of the trajectory of an object launched at some altitude on Earth, horizontally and in the equator plane, assuming no atmosphere, without any propulsion mean after the launch. My incomplete reasoning:

1. If the object is launched with an initial horizontal move with a low speed it will fall free (parabolic trajectory)

2. If the speed is increased, the object may start to move in elliptical orbit, such that it will just reach the sea level at the opposite point of the launch (perigee) and return to apogee at the launch point.

3. Increasing further the speed, the orbit should become circular.

4. Increasing again, the orbit will be elliptical again, with the perigee being at the launch point, and the apogee at the opposite point.

5. Increasing again, up to what I believe is the "critical speed", the ellipse change back to a parabolic curve, the trajectory is open, and the object will not return (the null speed being reached only at an infinite distance).

6. Increasing past this speed, the trajectory continue to be open, but the curve is now hyperbolic.

A common illustration of the experience:

To sum up, what I think true, and what I don't know about launch speeds:

• $0 \leq s < s_1$, free fall / parabolic
• $s_1 \leq s < s_2$, elliptical, apogee at launch point
• $s = s_2$, circular
• $s_2 \leq s < s_3$, elliptical, perigee at launch point
• $s = s_3$, parabolic, open
• $s_3 < s$, hyperbolic, open

Question: What are the names and values (for Earth) of $s_1$, $s_2$ and $s_3$?

The kind of information I've got, which have been useful: At Purdue, at University of Virginia, at ScienceBlogs.

• Orbital speed $s_2$, the speed for injection into a circular orbit, depends on the altitude above the sea level: 7.91 km/s at sea level, 7.73 at 300 km.
• Escape velocity ($s_3$) is the speed at which an object will not orbit the Earth, but escape it along a parabolic trajectory. 11.18 km/s at sea level, 10.93 at 300 km/s.

Below escape velocity, the object will move along an ellipse, if ellipse doesn't intersect the ground surface (free fall in this case).

The circular orbit is a particular case of elliptical orbit. We learned at school that the free fall trajectory is a parabola, this is not valid in real life where the Earth is not flat, and the gravitational force varies with the distance to the Earth center.

For orbits, below $s_2$, the launch point becomes the apogee and above $s_2$, it becomes the perigee.

Above escape velocity the curve is hyperbolic, with no return.

• $s_3$ in your question is called the escape velocity. – kasperd Dec 21 '14 at 22:51

What they teach in high school physics is wrong. When you toss a penny, it does not follow a parabolic path. Here is a picture of the penny's actual path (neglecting friction from air and rock and neglecting the fact that the Earth is not a point mass at the center):

If you tossed at a 45 degree angle, one focus of the ellipse will be in front of you. The other focus of the ellipse is at earth's center.

The eccentricity of this ellipse is around .9999. The eccentricity of a parabola is 1. So the penny toss is very to close to parabola shaped, especially from the scale of the penny tosser.

I wish they wouldn't keep teaching this. It confused me when I was first studying orbital mechanics. I thought "How can a penny toss be parabolic? it's nowhere near 11.2 km/s".

So from the earth's surface (neglecting friction from air and rocks):

0 to less than 7.9 km/s is elliptical orbit (but with perigee below earth's surface).

7.9 is circular.

More than 7.9 to less than 11.2 is elliptical.

11.2 is parabolic.

More than 11.2 km/s is hyperbolic

• Yes, what they taught in high school physics is technically wrong, but still very useful and correct enough to, for example, aim cannons and guns and hit the targets. Similarly the equations I provided in my answer are also technically wrong once General Relativity is taken into account. However they are still very useful and correct enough to send spacecraft around the solar system and hit their targets. Someday General Relativity, which is useful and correct enough to compute Mercury's orbit and the fate of co-orbiting neutron stars, will also be shown to be technically wrong ... – Mark Adler Dec 21 '14 at 20:46
• It never struck me before today that the ballistic curve was an ellipse indeed. Your demonstration is clear and simple. Thanks for the speed values too. Any idea about how 7.9 km/s is named? – mins Dec 21 '14 at 23:01
• "What they taught you in high school was wrong" - pretty much sums up every course I took in high school, except for the pure mathematics courses. – corsiKa Dec 22 '14 at 0:12
• @mins 7.9 km/s is in this case just orbital speed at a given altitude around a specific mass body. 11.2 is escape velocity (again depends on distance to focus and mass of the body being escaped, in this case from the mean sea-level Earth). The latter shouldn't be confused with delta-v needed to launch into LEO, which, given specific launch parameters, can be fairly close to escape velocity when air resistance and Earth's rotation are accounted for. – TildalWave Dec 22 '14 at 11:03
• @Noordung: Thanks. Also adding the link, very good explanation too. – mins Dec 22 '14 at 11:17

1. This is only approximately a parabola for short falls. It is really an elliptical orbit whose periapsis is below the surface. Therefore the "orbit" comes to an abrupt stop when it intersects the surface. The parabola approximation is for a constant gravitational acceleration on a flat Earth. However the gravitational acceleration is not constant, and diminishes as $1/r^2$, which gives an elliptical path. Also Earth is not flat.

2. Also an elliptical orbit that just barely misses the surface of the hypothesized airless body.

3. Also an elliptical orbit that is circular.

4. Also an elliptical orbit. All of 1 through 4 have a negative total energy relative to the body.

5. This one is in fact parabolic, but not at all like the parabola that is the approximation for short falls. This intermediate state between an ellipse and a hyperbola is not useful, since it has zero measure. It only exists when the total energy is exactly zero. Once it has been derived as one of the solutions, it can be safely ignored from then on.

6. Now the total energy is positive, so the trajectory is a hyperbola.

All of the above can be represented in a single equation, where $\mathcal{E}$ is the total specific energy of the object in flight, i.e. the energy per unit mass in units of distance squared per time squared:

$$\mathcal{E}={v^2\over 2}-{\mu\over r}$$

$\mathcal{E}$ is a conserved quantity that is the same at all points in the ellipse or hyperbola, so long as there are no impacts with atmospheres or surfaces, the object isn't firing any rockets, and there are no other forces on the object except for the gravity of the body. At any given point in the trajectory, $v$ is the magnitude of the velocity at that point relative to the body, and $r$ is the radius, or distance from the center of the body at that point. $\mu$ is the gravitational parameter for the body, $\mu=GM$, where $M$ is the mass of the body and $G$ is Newton's gravitational constant. Aside from the notational convenience, $\mu$ is what is actually measured, and can almost always be determined much more accurately than $M$, which is limited by our knowledge of $G$. (The exception is when we can see nothing else affected by the gravity of the body, in which case crude estimates of $M$ are made using the size of the object and assumptions about its density.)

When $\mathcal{E}$ is negative, you have your cases 1..4. When $\mathcal{E}$ is zero, you have your case 5. when $\mathcal{E}$ is positive, you have your case 6.

For $\mathcal{E}$ negative, you have:

$$\mathcal{E}=-{\mu\over r_p+r_a}$$

where $r_p$ is the periapsis radius and $r_a$ is the apoapsis radius of the elliptical orbit. Radii are measured from the center of the body.

For $\mathcal{E}$ positive, you have:

$$\mathcal{E}={v_\infty^2\over 2}$$

Where $v_\infty$ is the velocity magnitude at infinity, i.e. how fast the departing object will be moving relative to the body once it is far enough away that the gravitational acceleration of the body is negligible.

When $\mathcal{E}=0$, you have just reached escape velocity.

You now have all you need to compute all of your velocities with just a calculator. (Assuming that the calculator has square root.) For Earth, $\mu_e\approx 398600\,\mathrm{km^3/s^2}$ and the radius of the surface at the equator is $r_e\approx 6378\,\mathrm{km}$. Go to town.

• Great answer. Regarding the common parabolic approximation for small-scale ballistic arcs (as in word problems about baseballs), you might clarify that not only the magnitude of the acceleration is usually held constant, but the direction. – Russell Borogove Dec 21 '14 at 18:21
• +1 good answer. I also volunteered an answer because the "parabolic path" of objects thrown from earth's surface is one of my pet peeves. I am hoping my picture of a nearly parabolic ellipse is helfpul. – HopDavid Dec 21 '14 at 20:00
• Yes, that helps. +1. – Mark Adler Dec 21 '14 at 20:35
• Thanks for the full explanation. For an height of 300 km above the sea level, I find an escape velocity of 10.93 km/s. By another source I found that the speed for a circular orbit is v²=µ/r (I don't think it can be deduced from your answer). At 300 km ASL, this is 7.73 km/s. At sea level I do indeed find @HopDavid values. Very good! – mins Dec 22 '14 at 0:37
• Everything and more can be found with the equations I provided. For a circular orbit, $r_p=r_a=r$. Solve for $v$. To get your other speeds for various elliptical orbits, you want $v$ at either $r=r_p$ or $r=r_a$. – Mark Adler Dec 22 '14 at 1:32