My dad and I thought about a problem we had when discussing Earth's gravitational field. This wasn't the original question but I still would like to know the answer.

So essentially my question is:

Is there a point when you're travelling faster and faster around Earth that you are not going to fall back to Earth, but orbiting away from Earth? And if you keep the speed at that point (maybe not possible), are you going to orbit around Earth in perfect circle-shaped orbits?


1 Answer 1


This is a great question!


From a circular orbit, a little more or less velocity just makes your orbit slightly elliptical. If your orbit happened to be very close to the Earth's surface (LEO or low Earth orbit, like the ISS), then a little less velocity (at a given height) would put you deep into the atmosphere and you'd burn up. But that's not related to orbital mechanics, it's just a little "detail" related to orbiting in LEO. You'll need about 41% more velocity to really escape the Earth completely. Any less than 41% extra, and you'd just be in a high elliptical orbit.

At the height of the International Space Station of 400 km above the Earth's equator of 6378 km, that speed is 7.67 km/s.

"space is hard" + "space is math" → "math is hard"? No way!

Orbital mechanics is not intuitive, so here is a quickie summary of some things.

  1. The Earth's gravity field is pretty close to that of a perfect sphere (within about 0.1%).
  2. The gravity field outside of a perfect sphere is the same as the field of a point at the center of the same mass. (Newton's Shell Theorem)

  3. The vis-viva equation gives you the speed of an object in any elliptical orbit using the semi-major axis $a$ and the current distance from the center $r$:

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

  1. If it's a circular orbit, you can set $r$ = $a$ and get:

$$v^2 = \frac{GM}{a} $$

  1. The Earth's standard gravitational parameter $GM_E$ is about 3.986E+14 m^3/s^2. That's the gravitational constant $G$ times the mass of the Earth $M_E$. Because we monitor Earth satellites very carefully, we know that product much more accurately than we know either $G$ or $M_E$ separately.

  2. The energy per kilogram (specific energy) of an object in orbit is the sum of the kinetic and potential terms:

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

    For a circular or elliptical orbit it will be negative (bound orbit), for a (barely) escaping orbit (parabola) it will be zero. Positive energy also means escape but it also means even at very large distances there will be plenty of "left-over" velocity.

  3. Plugging in the numbers using the equatorial radius of the Earth of 6378 km plus an altitude (say of the ISS) of 400 km giving $a$ of 6,778,000 meters, you get the orbital speed of about 7669 m/s or 7.67 kilometers/second or about 17,100 miles per hour. As "homework" you can calculate the approximate altitude of the ISS when it's speed was 17,500 when this classic sign on the ISS was printed (google "space station speed limit sign"):

    Astronaut Ellen Ochoa going 17,500 mph

    above: "STS110-353-012 (8-19 April 2002) --- Astronaut Ellen Ochoa, STS-110 mission specialist, poses by the speed limit signs in the Unity node on the International Space Station (ISS)." from here. Read more about the sign's history in this "historical" answer.

    By coincidence, NASA just released today a YouTube video using that iconic sign in the title: Space Station Science at 17,500 Miles Per Hour.

  4. If you were in this orbit, but went a little faster or slower, your orbit would just become slightly elliptical. If you were at 6778 km but were going only 7550 m/s instead of 7669 m/s, you can back-calculate the semi-major axis $a$ from the vis-viva equation to be only 6576 km. $a= (r_{peri} + r_{apo})/2$ gives your periapsis or lowest point to be 6375 km or 3 km below Earth's equatorial surface. So your theoretical orbit is still fine but you've re-entered the atmosphere and burned up.

  5. How fast to escape? From any point, the escape velocity is the square root of 2 times the velocity of a circular orbit at that height. You can calculate that by setting the energy (from point #6 above) to be zero.

    $$0= \frac{1}{2}v^2-\frac{GM}{r}$$

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

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

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

    At 6778 kilometers, that's 10,845 m/s, or 1.414 times larger than the circular orbital speed of 7668 m/s which is 41% faster. Anything less, and you'd just go into a high ellipse, but come right back again after each orbital period.


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