# Why should space probes have to orbit the Earth before being launched at other planets?

I'm curious to know why any satellites that need to be sent to other planets should be moving around the Earth rather than directly going towards their destination? Today, India launched one satellite to Mars, and from the news I got to know these lines:

The launch vehicle will stay in the Earth's orbit for nearly a month, building up the necessary velocity to break free from our planet's gravitational pull.

So my question is, why can't we directly send satellites to another planet?

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• The lines you quote explain that it needs to happen in order to escape the Earths gravity. Is there a specific part of this you are struggling with or is your question more about why we need to escape gravity? – RhysW Nov 5 '13 at 11:23
• @RhysW Thanks for replying,My question is why ?We just need 11.2km/s and this is supplied by fuel – javaBeginner Nov 5 '13 at 11:25
• This question is highly related: space.stackexchange.com/questions/637/… The issue is that you get "gravity drag" if you don't immediately get into an orbit, so even if you leave LEO right away, you still attain the LEO orbital parameters at some point. – AlanSE Nov 5 '13 at 14:15
• I asked this recently (5 years after this question was asked): space.stackexchange.com/questions/27904/… and uhoh did a great answer. – Magic Octopus Urn Jun 21 '18 at 21:04

All interplanetary probes that I am aware of were launched into a parking orbit, and then waited some time in that orbit before restarting a stage or igniting another stage to inject on the desired outgoing asymptote. This is done for convenience to allow long launch windows on days in the launch period. It is possible and slightly more efficient to launch directly from the launch pad to the interplanetary trajectory, but you will have an instantaneous launch window and your launch pad needs to cross the plane of the outgoing trajectory to get that efficiency. Instantaneous launch windows are risky, since if there's a problem such as a boat where it shouldn't be, you don't launch that day. With the longer window offered by going first into a parking orbit, often an hour or more, you have time to remedy the problem and still launch.

(As an aside, I worked on a project, which was later cancelled, that was going to use an Ariane 5 to launch to Mars. The problem was that its upper stage did not at the time have a qualified restart capability. As a result, we had to plan to inject directly from the launch pad with no parking orbit. Though the French Guiana launch site was fortuitously close to where we would want it for that opportunity, it wasn't exactly right so we had to incur some inefficiency with a dog-leg maneuver to get going the right way.)

In most cases the wait in the parking orbit is measured in tens of minutes. You are often restarting a second stage with limited battery life, so you don't want to wait for more than an orbit. In order to have a two week or longer launch period, you need to accept some inefficiency in higher injection energies before and after the optimal day.

India is doing something a little different, which has some risk but offers higher efficiency. They have a fixed departure day from Earth orbit with the smallest injection energy. Whenever they launch, they wait in Earth orbit until that blessed day, and then they go. That means they need an upper stage with longer life. On the plus side, breaking up the injection burn into several pieces, executed at each perigee, allows the engine to be much smaller and lighter than a typical upper stage engine that's trying to flow out all the fuel in a single burn near the Earth.

As noted in the other answers, the injection burns whether done 20 minutes later or 20 days later are done at perigee for the maximum energy change per unit of $\Delta V$. I'll throw an equation in here just for fun. In basic physics you learn the equation for kinetic energy: $K={m v^2\over 2}$. If you differentiate that with respect to $v$, you get $dK=m v\,dv$. So your change in energy is proportional to your velocity times the $\Delta V$. The faster you're going, the more change in energy you get for a fixed $\Delta V$. You're going the fastest at perigee. This is known as the Oberth effect.

• Some comments on this answer, which I think is the best so far: 1) The ascent stage is usually tuned for a non-vacuum environment. The injection stage needs to be tuned for a vacuum environment. 2) The paragraph about when to burn (perigee) is describing the Oberth Effect. – Erik Nov 5 '13 at 21:06
• Thanks for your nice explanation +1 for the same.As per your equation dv will be zero because because satelite moves in round orbit(sorry If I am wrong) though speed will not be zero.So my question why satelite is not allowed in the straight path(assuming no obstacles in the path).Longer the displacement higher will be the velocity – SpringLearner Nov 6 '13 at 3:56
• $dv$ is the change in velocity from the engine thrust over a short interval and is not zero. I do not understand your question about "the straight path". – Mark Adler Nov 6 '13 at 17:52
• @MarkAdler sorry for the late response,I did not get any notification for this comment.My question is will the satellite move travel in trajectory? – SpringLearner Nov 13 '13 at 4:14
• @javaBeginner: You need a basic understanding of orbital mechanics, which I cannot convey in comments. Start with the Wikipedia article. Bottom line: to go up relative to a body, you need to thrust 90° from up in the direction of your orbital velocity. So to go to Mars, you need to thrust in the direction of Earth's orbit. Thrusting up won't get you there. I can't answer any more questions here. – Mark Adler Nov 13 '13 at 7:57

As @Ame mentioned, the rocket didn't have enough fuel to put it there in one shot, like most US/Russian rockets do. However, the actual physics behind the orbital maneuver is slightly different than described. Specifically, the physics is called the Oberth effect. The short explanation of this is that a rocket thrust is more effective if done at perigee. Firing the rocket during successive perigee passes will improve the efficiency, ultimately allowing one to have maximum efficiency in one's rocket bursts. A true gravity assist does not use any thrusters, but this effect essentially magnifies the effectiveness of the thrust.

• Thanks for answering +1 for your explanation.I have heard that this mangalyan will orbit round the earth for minimum of 25 days to attain the sufficient speed to reach the destination.So can you please tell me what is the present speed and what will be the final speed. – SpringLearner Nov 6 '13 at 3:45

There are several reasons why Satellites need to orbit Earth before they go interplanetary...

The first reason: The launch site is very rarely in the right position to start an interplanetary flight. Earth rotates on a tilt, so a launch has to be timed when Kennedy Space Center crosses the ecliptic plane (the general plane that most planets orbit on). Also, it has to be in the right season, so that when the probe goes out to orbit it ends up headed the right direction when it leaves Earth's SOI. All of this is possible with a direct ascent profile, it just takes really good timing - but these perfect windows come very rarely.

Any probe that makes ecliptic orbit around Earth, first, has a launch window pretty much every 45 minutes, as apposed to a couple times per year.

The second reason: The amount of Delta-V required to escape Earth's SOI is pretty big. While it is possible to build rockets large enough to do so - the limiting factor is really the efficiency of typical rocket fuel and rocket engines.

To lift a probe out of Earth's orbit with a rocket takes a pretty heavy rocket. That heavy rocket must be lifted into Low Earth orbit, which takes a massive rocket.

One way to improve this fact is to make your rocket much more efficient - but we are already close to the theoretical efficiency limit of a chemical rocket. So NASA started using ION propulsion, which is way more efficient than a chemical rocket - it's also very weak - which is the thirdreason...

The third reason: Now that most probes use Ion propulsion, they don't have the thrust to just eject themselves from earth in a direct ascent - they spend weeks with the Ion thruster thrusting for a little while (a few minutes) at a key point in the orbit. Each time the Ion engine does this, their orbit gets closer and closer to Earth escape velocity.

Once the probe is out of Earth's SOI, it can basically turn the Ion engine On and leave it there for as long as it wants to complete interplanetary maneuvers. Usually, most maneuvers between planets are minor course corrections to take advantage of a planetary fly-by to sling-shot to much higher velocities.

TL;DR? Many reasons: Timing - launch position and orbital orientation mean few good launch windows for direct ascent, reaching orbit first allows for many more options. Too much fuel needed - Getting a probe way from the earth takes a lot of chemical propellant, so we now use Ion engines instead. Ion engines are weak - it takes a long time (weeks!) for these highly efficient engines to do their job.

• Nice explanation,thanks +1 – SpringLearner Nov 6 '13 at 6:10

In the case of India's PSLV, the launch vehicle is not powerful enough to insert the Mars Orbiter Mission probe directly on a route to Mars. In contrast MAVEN is to be launched using the much more powerful Atlas V.

The probe has to use gravity assist to gain enough velocity to travel to Mars. I.e. the probe first goes into a highly elliptical orbit, and uses specially timed short burns to achieve an acceleration due to Earth's gravity and motion around the sun. Especially the motion of the Earth relative to the sun and the probe is here of help, since this impulse is added to the probe's existing impulse. The picture in the Wikipedia article is quite telling:

• Erm, the rocket would already have all the assist of the Earth's rotation around the Sun, since it's launching from the Earth, in fact more since there's some of the angular velocity to be had too due to Earth's rotation on its axis. This is not really the issue, the point is in rocket engine performance and getting in orbit doesn't come even close to achieving velocity required to escape the Earth's gravity well. Case in point, the ISS orbital velocity is at roughly 4.791 miles/s (7.71 km/s). – TildalWave Nov 5 '13 at 13:45
• Hm, yeah, maybe I misunderstood this concept. You argument seems to raise a valid point. When reading the article about the Hohmann orbit (en.wikipedia.org/wiki/Hohmann_transfer_orbit), which the MOM seems to use, it doesn't look like gravitational assist anymore. Just very clever timing of the orbital burns. – Arne Nov 5 '13 at 13:57
• Though this answer gets it wrong, there is a gravity assist benefit to what Mangalyaan is doing. That benefit is not in energy, but in inclination. In the normal approach, a low circular parking orbit needs to be inclined to line up with the outgoing asymptote. Depending on the inclination, that can reduce the launch vehicle mass capability. In this case, they can launch to the most efficient inclination, equal to the launch site latitude, and then use the final Earth perigee approach and injection burn to get the proper outgoing asymptote. – Mark Adler Nov 5 '13 at 17:18
• Interesting. Good clarification! +1 – Arne Nov 5 '13 at 22:07
• good explanation +1.Cant it be possible that instead of moving round the earth to attain the maximum speed,it will go in a linear path – SpringLearner Nov 6 '13 at 3:58

A rocket floating in place (not flying up at all, just hovering over the launchpad) needs to generate enough thrust that would otherwise give it $9.8 m/s^2$ acceleration (1g) in horizontal flight. That's quite a bit, and that's all wasted thrust. If you launch directly up, earth gravitational acceleration is what you have to substract from your acceleration at all times. Every second, no matter what else you spend your fuel on, you need to spend a hefty dose on just not falling down.
• I am not making sense of your comment. You can't avoid getting the Earth's rotation by going straight up. Anyway, my point stands, which comes from the fact that it takes a little $\Delta V$ to circularize your parking orbit, where that $\Delta V$ does not contribute to the departure C3. – Mark Adler Nov 7 '13 at 15:17