# How many times do you have to circle the Earth to break orbit?

My science teacher asked us this, and nobody in my class knew. We guessed it would be three, but with no evidence.

(Edit: The asker has not returned to clarify if they meant leaving the Earth-Moon system, or re-entry from orbit. As it has been a few days, and this question has received lots of attention, I have edited the title to reflect the common interpretation of 'leave orbit', and the content of most answers. Note that answers below also address the possibility it refers to re-entry. It would be a shame to close this, and so I think this minor modicum of interpretation is justified.)

• The question your teacher asked doesn't really make sense... – PearsonArtPhoto Jan 17 '18 at 20:20
• It could be a trick question. 'To leave orbit' implies that you are IN orbit (i.e. falling towards the Earth but travelling laterally quickly enough that you keep missing). To be definitively in orbit, presumably you have to circle the Earth at least once. So that is your answer.... 1. Of course, to leave the Earth behind you just need to reach escape velocity, which doesn't require any orbiting at all. – DrMcCleod Jan 18 '18 at 11:34
• This is equivalent to asking "How many times do you have to circle the block before leaving the neighborhood?" – Nuclear Hoagie Jan 18 '18 at 13:43
• It's three for the same reason it takes 3 licks to reach the center of a Tootsie pop. – chepner Jan 18 '18 at 18:34
• "We guessed it would be three, but with no evidence." That makes me sad. – RonJohn Jan 18 '18 at 20:08

This depends on how much thrust you have available. With enough thrust, you don't need to be in Earth orbit at all: you can launch straight into an escape trajectory.

New Horizons did this, more or less: after launch it did about 1/4 orbit before the second stage was ignited again and insertion into its trajectory towards Jupiter began.

With very little thrust, it can take months: SMART-1, one of the first missions to use an ion engine took 14 months to get into Moon orbit. The same principle can be used to leave Earth orbit.

• Comments are not for extended discussion; this conversation has been moved to chat. – called2voyage Jan 18 '18 at 20:59
• This is the answer to most problems in space flight and aviation- "given sufficient thrust.." – tedder42 Jan 24 '18 at 19:10

## Zero

See at: https://en.wikipedia.org/wiki/Escape_velocity for theory.

Once you build enough velocity to surpass gravitational attraction, you will leave planetary orbit. A spacecraft simply circling the earth in orbit is not inherently doing anything to contribute to escaping that orbit.

Think of this as a slightly different question, and the answer becomes more clear.

How many times do you have to circle the Sun to leave orbit?

The Earth has been orbiting our Sun for about 4.5 Billion years with each year being one orbit. The Earth is expected to stay in orbit around the Sun for the next few billion years.

Also consider the moon has been orbiting the Earth for about the same 4.5 billion years. ~12 orbits a year for 4.5 billion is ~54 billion orbits and it is still there.

The number of orbits is not relevant to leaving orbit.

In addition to @Hobbes answer, even if you have high thrust you can use the fuel more efficiently by accelerating when the rocket is moving the fastest due to the Oberth effect.

Explanation in terms of work

Rocket engines produce the same force regardless of their velocity. A rocket acting on a fixed object, as in a static firing, does no useful work at all; the rocket's stored energy is entirely expended on accelerating its propellant in the form of exhaust. But when the rocket moves, its thrust acts through the distance it moves. Force multiplied by distance is the definition of mechanical energy or work. So the farther the rocket and payload move during the burn (i.e. the faster they move), the greater the kinetic energy imparted to the rocket and its payload and the less to its exhaust.

The closer you are to orbital periapsis (the closest point to Earth), the faster your rocket moves, and the more efficient the burns. However the burn isn't instant so your rocket will quickly move out of the most efficient zone. To maximize efficiency you'd then circle the Earth many times, each time firing the rocket near the periapsis.

• This would be the best answer to the teacher but I doubt he/she meant this. Nice to have explained this to the Op because other answers did not mention it. Plus 1 – Alchimista Jan 19 '18 at 12:20
• Of course this has to be balanced against the fact that shutting down and restarting rockets isn't free. – Peter Green Jan 19 '18 at 14:16
• You could burn straight up and still use the Oberth effect, However, you want to head as much horizontally as you can to minimize gravity loss. – Loren Pechtel Jan 21 '18 at 0:56

Here is an example of going to deep space without orbiting the Earth once.

As can be heard in this video SpaceX's launch of the DSCOVR spacecraft to Sun-Earth L1 went there almost directly. The first secondary engine cut-off (SECO-1) was at about T+ 00:08:40, and after what the announcer says (at about T+ 00:09:50) would be "in about 21 minutes", the secondary stage's second burn would put it in a Heliocentric orbit on its way to the SE L1. Considering that a LEO orbit is about 90 minutes, DSCOVR was in Earth orbit for roughly half of one Earth orbit.

There is more to read in the answers to the question Why would a mission to Sun-Earth L1 have an instantaneous launch window? The fractional Earth orbit was used only to minimize the required Δv based on launching from Florida and getting to L1.

below: From Spaceflight 101's Deep Space Climate Observatory  • – uhoh Jan 18 '18 at 15:46

Note that "leave orbit" can mean achieving escape velocity (to go to the Moon, Mars, or beyond), or it can mean slowing down and re-entering the atmosphere and eventually landing (or splashing down, or in extreme cases, lithobraking). Or, to be ridiculously pedantic, you can leave one orbit to enter another (i.e., GTO to GEO).

What matters is what your teacher meant by "circle the Earth". As currently phrased, it's not clear. Orbital maneuvers are expressed in terms of change in velocity (ΔV), not the number of times it takes to go around the planet. To go up (to higher orbit or escape velocity), most chemical rockets do it in less than a single orbit. Rockets or spacecraft using electrical propulsion (ion drives) may take several dozen (or several dozens of) orbits to do it.

To hit a specific target, you may have to go into a parking orbit briefly and wait for things to line up (such as on the Apollo missions). But if all you care about is "leaving orbit", regardless of direction, then it only depends on how much oomph you have in your rocket.

To come back down, most rockets and spacecraft thrust just enough to bring their perigee into the uppermost atmosphere, then let drag and friction take over. That usually takes less than a single orbit as well for chemical rockets.

or

• "In orbit" around what? The Earth? The Sun? The center of the Milky Way?
• Is he/she counting on you "orbiting" the center of the Earth by virtue of standing on its surface, and the Earth spinning on its axis (which isn't really an orbit for a number of reasons, but maybe he/she is using a "horseshoes and hand grenades" definition of orbit)?
• You're right about rockets deorbitting fast. "Number of times..." makes more sense with orbital decay though. – fectin Jan 20 '18 at 18:28

Orbiting in circles cannot get you away from earth. To leave earth's orbit requires accelerating away from the orbit. The acceleration is usually by firing rocket motors that burn fuel expanding hot gases at high speed from a nozzle.

• This answer is exactly right! – uhoh Jan 19 '18 at 0:01

# It doesn't matter how many times you circle. If you go faster than the velocity calculated, you will escape the object's gravitational field.

## Part 1: formula derivation

We can calculate escape velocity using the formula:

G = $$6.67*10^-^1^1$$ $$N m^2/kg^2$$

M = mass of planet

m = mass of sattelite

r = the distance between both masses

K = Kinetic energy, U = Potential Energy

$$K + U = 0$$

$$-GMm/r + 1/2 mv^2 = 0$$

$$2GM/r = v^2$$

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

## Part 2: solving for escape velocity

So let's plug in our numbers:

M = mass of earth, $$5.98*10^2^4$$ $$kg$$

r= radius of earth, $$6.38 * 10^6$$

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

$$v = \sqrt {2* 6.67*10^-11 N m^2/kg^2 * (5.98*10^24 kg)/6.38 * 10^6}$$

$$v = 1.12 * 10^4 m/s$$

Therefore, it doesnt matter how many times you circle, you need to be going faster than $$1.12*10 ^4$$ m/s or about 11.2 km/s to escape the earth's gravitational field.

Doesn't have anything to do with it. The critial parameter is your (parabolic) escape velocity... for earth it's appr. 11,2 km/s. As soon as you exceed this velocity you will leave earth's gravitational field...

I'm no professional but I think I have a clear answer. I would say the simple answer is a fraction of one orbit. Exiting orbit would require enough thrust to escape earth's gravitational pull. You could try going out without some form of orbit but I think it would be a waste of fuel.

The question is at what time do you engage the thrust to send the object on the right path to its destination. If you missed your opportunity to exit orbit, you may let it go around for a second orbit and try again once you're absolutely sure it's headed in the right direction.

It's like football. If you're trying to send something from Earth to Mars, Earth is like the quarterback and Mars is like the wide receiver. You need to know the route the receiver is running so you can place the ball where he's going to be, not where he's at upon release. Once You figure that out, you'll know the trajectory in which the object should leave orbit. You use the momentum from its short orbit around the earth with the extra thrust to help escape orbit at just the right time. Without the thrust, the object will just continue orbiting or fall back to the earth if it loses speed.