# Escape velocity? Orbiting Speeds? Bacon Ice cream?

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I don't seem to understand this no matter how many explanations I look at online. I've heard different numbers over the years as to how fast you need to get into space. So I think I have it right now. It's 18,000mph to get into low earth orbit, and 25,000mph to break free of earth's pull entirely. I don't understand how these numbers are useful however. So i have many questions are:

1. If I could design a rocket with adequate fuel that maintained an upward speed of 1mph, it would just keep going would it not? And if so, how far away from earth do I need to maintain a 1mph thrust until I can shut off the engine and not get pulled back to earth?

2. As I understand it, while we are moving away from the earth, we are still going around the sun. So once we escape earth's orbit, we are still stuck going around the sun yes? Or does the sun start pulling us in when we are far enough away from earth?

3. It took 8+ minutes for the shuttle to accelerate to 18,000mph. Is the time in which we reach 18,000mph of any particular importance if we remove fuel limitations from the equation?

4. If I fired a shell from a cannon straight up, and it accelerated to 18,000mph almost immediately, like a few seconds after it fired, does it automatically get into space regardless if it slowed down below 18,000mph while still in the atmosphere? In other words, does an object only need to reach that speed once at any given time along it's path?

5. I assume space rockets tilt into an angle during launch so they can start working their way into the angle they need to get into orbit. Is it one gradual process or do they do more angling and thrusting once they are in space to settle into an orbit.

6. Why does the ISS have to keep adjusting it's course but the moon doesn't? And does adding weight modules or cargo to the station slow it down?

7. What is the margin of adjustment you can make in your orbit and not fall out of orbit? My assumption is if I was in a steady orbit, then left my ship, I would stay in orbit, at least for a long while. So If I chucked a screwdriver (made of tungsten) at the earth from the space station, would it keep going until it hit?

8. Can the earth fall out of orbit if we remove too much material from it?

Thanks guys!

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Escape speed is the value you need at the surface, with no other forces acting, to get arbitrarily far away. From that and the mass you can calculate the energy. If you have a force (thrust) then you can escape at a lower speed. Escaping the earth is not the same as escaping the sun

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1. If I could design a rocket with adequate fuel that maintained an upward speed of 1mph, it would just keep going would it not? And if so, how far away from earth do I need to maintain a 1mph thrust until I can shut off the engine and not get pulled back to earth?

It would take implausible amounts of energy to do it that way.

You might want to take a look at this: https://what-if.xkcd.com/58/

But getting to space is easy. The problem is staying there.

To avoid falling back into the atmosphere, you have to go sideways really, really fast.

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I don't seem to understand this no matter how many explanations I look at online. I've heard different numbers over the years as to how fast you need to get into space. So I think I have it right now. It's 18,000mph to get into low earth orbit, and 25,000mph to break free of earth's pull entirely. I don't understand how these numbers are useful however. So i have many questions are:

1. If I could design a rocket with adequate fuel that maintained an upward speed of 1mph, it would just keep going would it not? And if so, how far away from earth do I need to maintain a 1mph thrust until I can shut off the engine and not get pulled back to earth?

But, that is of course not the way they launch rockets because it is inefficient. What they do is get it up to the speed they want as quickly as possible. When someone speaks of escape velocity, they are saying that if you reach that velocity, you will never have to fire your rockets again and you will never fall back to the body you have left.

I don't know the exact distance you would have to be from earth to shut off your engines while traveling 1 mph to not get pulled back to earth, but it is the place where the escape velocity is 1 mph. In other words, since the earth's gravity is strongest at the surface of the earth and weaker as you move away from the surface of the earth, the escape velocity at the surface is 25,000 mph, and the further from the surface that you start your trip from, the lower the escape velocity. There is some location far out in space where your escape velocity is only 1 mph.

2. As I understand it, while we are moving away from the earth, we are still going around the sun. So once we escape earth's orbit, we are still stuck going around the sun yes? Or does the sun start pulling us in when we are far enough away from earth?

Yes, you are still stuck around the sun. There is an escape velocity to escape the sun too (or any body for that matter).

3. It took 8+ minutes for the shuttle to accelerate to 18,000mph. Is the time in which we reach 18,000mph of any particular importance if we remove fuel limitations from the equation?

No, not for these purposes.

4. If I fired a shell from a cannon straight up, and it accelerated to 18,000mph almost immediately, like a few seconds after it fired, does it automatically get into space regardless if it slowed down below 18,000mph while still in the atmosphere? In other words, does an object only need to reach that speed once at any given time along it's path?

The shell will begin to slow down as soon as it leaves the barrel of the cannon, but if it leaves at 18,000 mph, it will reach space.

6. Why does the ISS have to keep adjusting it's course but the moon doesn't? And does adding weight modules or cargo to the station slow it down?

Friction from the atmosphere slows it down, which causes its orbit to degrade.

7. What is the margin of adjustment you can make in your orbit and not fall out of orbit? My assumption is if I was in a steady orbit, then left my ship, I would stay in orbit, at least for a long while. So If I chucked a screwdriver (made of tungsten) at the earth from the space station, would it keep going until it hit?

The screwdriver would eventually hit the earth just as you would, only a bit sooner. And remember that most of the motion of the screwdriver is around the earth, not toward it from you throwing it. So it is still going to orbit the earth for quite a while.

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I don't seem to understand this no matter how many explanations I look at online. I've heard different numbers over the years as to how fast you need to get into space. So I think I have it right now. It's 18,000mph to get into low earth orbit, and 25,000mph to break free of earth's pull entirely. I don't understand how these numbers are useful however. So i have many questions are:

1. If I could design a rocket with adequate fuel that maintained an upward speed of 1mph, it would just keep going would it not? And if so, how far away from earth do I need to maintain a 1mph thrust until I can shut off the engine and not get pulled back to earth?

If you are only considering Earth's gravity, a little more than 2.5 trillion miles, or better than 4/10 of a light year. (better pack a good lunch, because at 1 mph, that will take 285 million years.)

However, there is the Sun's gravity to consider, and if you traveled for 580,000 miles, the Sun's gravity would take over and prevent you from falling back. This should only take you 66 yrs at 1 mph

Of course as pointed out, you would need lots and lots of fuel to pull this off, a whole lot more than that needed just to get you up to escape velocity.

2. As I understand it, while we are moving away from the earth, we are still going around the sun. So once we escape earth's orbit, we are still stuck going around the sun yes? Or does the sun start pulling us in when we are far enough away from earth?

Once you escape the Earth's gravity, you will still be in orbit around the Sun, the exact nature of that orbit will depend on how much velocity you have left after escaping the Earth. But don't worry about falling into the Sun. To do that, you would have to have nearly 17 miles per sec (60,750 mph) left over, and be heading in the opposite direction than the Earth is orbiting

3. It took 8+ minutes for the shuttle to accelerate to 18,000mph. Is the time in which we reach 18,000mph of any particular importance if we remove fuel limitations from the equation?

If you remove fuel considerations, then no, it doesn't matter. With fuel considerations, then yes. The faster you get up to speed, the less fuel you use. Look at it this way, if you could burn all your fuel at the instant of take off, you would only need to use enough fuel to get the ship up to 18,000 mph. If you accelerate the ship slowly, then the fuel you are burning at any given moment also has to accelerate the mass of the fuel you will be burning later. Thus the total fuel needed increases.

4. If I fired a shell from a cannon straight up, and it accelerated to 18,000mph almost immediately, like a few seconds after it fired, does it automatically get into space regardless if it slowed down below 18,000mph while still in the atmosphere? In other words, does an object only need to reach that speed once at any given time along it's path?

A shell fired straight up at that speed will reach an altitude of 4145 miles, and then will fall back to the Earth. Since that is far enough to be considered "space" then yes. However it won't stay there. While technically it is in an orbit, it is an orbit that intersects the body of the Earth. speed alone does not make an Earth circling orbit.

I've run out of time for now, if no one else addresses them before I can get to them, I'll try and deal with the others later.

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If I throw a rock up in the air, it falls down.

If I throw it faster, it takes longer to fall back down.

I can plot out how long it takes to fall vs how fast I threw it.

The velocity which I have to throw it in order to get the time to be "forever" is the escape velocity.

I could, in principle, put a really long ladder into space and climb it as slowly as I liked- I'd get there in the end.

There would also be a point where, because the earth is spinning, I would be flung off the ladder up, rather than fall down if I let go of it.

but that's a different question.

If the earth were not spinning (or if I put the ladder at one of the poles) then, if I let go of the ladder, I'd always fall back to earth.

The further I fell from, the faster I would hit the ground (ignoring air resistance).

But as I got further up, each additional step would make less difference to how fast I hit the ground. The speed I landed at would tend towards a limit.

That limit is also the escape velocity.

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If you are only considering Earth's gravity, a little more than 2.5 trillion miles, or better than 4/10 of a light year. (better pack a good lunch, because at 1 mph, that will take 285 million years.)

However, there is the Sun's gravity to consider, and if you traveled for 580,000 miles, the Sun's gravity would take over and prevent you from falling back. This should only take you 66 yrs at 1 mph

Of course as pointed out, you would need lots and lots of fuel to pull this off, a whole lot more than that needed just to get you up to escape velocity.

Janus, how did you come up with 2.5 trillion miles? I'm getting a very different number around 3.8 million kilometers.

3. It took 8+ minutes for the shuttle to accelerate to 18,000mph. Is the time in which we reach 18,000mph of any particular importance if we remove fuel limitations from the equation?

Potentially, although maybe not the push to LEO. There are a few things to consider for the flight profile of Shuttle. One is the crew comfort - it's designed so that at no point during the ascent the g-load goes above 3g so that astronauts don't pass out and can react if something goes wrong. The Thrust-to-Weight ratio is another limiting factor during the take-off. Also the Shuttle+EFT+boosters is very awkward aerodynamically so while atmosphere is still present it shouldn't go above a certain acceleration.

4. If I fired a shell from a cannon straight up, and it accelerated to 18,000mph almost immediately, like a few seconds after it fired, does it automatically get into space regardless if it slowed down below 18,000mph while still in the atmosphere? In other words, does an object only need to reach that speed once at any given time along it's path?

It might get into space, but it won't go into orbit.

6. Why does the ISS have to keep adjusting it's course but the moon doesn't? And does adding weight modules or cargo to the station slow it down?

There's a lot of space junk that ISS has to evade from time to time. The Earth's atmosphere extends to around 10000 kilometers, so as long as you're below this altitude you'll be getting atmospheric drag.

7. What is the margin of adjustment you can make in your orbit and not fall out of orbit? My assumption is if I was in a steady orbit, then left my ship, I would stay in orbit, at least for a long while. So If I chucked a screwdriver (made of tungsten) at the earth from the space station, would it keep going until it hit?

No, the screwdriver would just move to a slightly lower orbit. You would only be able to throw it possibly at ~20 m/s which is a very small adjustment compared to ~7200 m/s LEO velocity. Eventually, after a long-long time, the orbit of the screwdriver will deteriorate and it will fall back to the Earth, but that will take a long time.

The screwdriver would eventually hit the earth just as you would, only a bit sooner. And remember that most of the motion of the screwdriver is around the earth, not toward it from you throwing it. So it is still going to orbit the earth for quite a while.

I rather doubt it. The difference in orbital velocities will be tiny, but your spacecraft will experience way more drag than screwdriver will.

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I rather doubt it. The difference in orbital velocities will be tiny, but your spacecraft will experience way more drag than screwdriver will.

In GrandMasterK's scenario, he was no longer in the spacecraft when he threw the screwdriver. Do you suppose the difference in size between him and the screwdriver will still result in enough additional drag to cause him to hit the earth prior to his screwdriver? I would have thought that the drag on the astronaut (or the spacecraft for that matter) would not have been enough to cause him to catch up to the screwdriver that is traveling at 20m/s toward the earth faster than he is.

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In GrandMasterK's scenario, he was no longer in the spacecraft when he threw the screwdriver. Do you suppose the difference in size between him and the screwdriver will still result in enough additional drag to cause him to hit the earth prior to his screwdriver? I would have thought that the drag on the astronaut (or the spacecraft for that matter) would not have been enough to cause him to catch up to the screwdriver that is traveling at 20m/s toward the earth faster than he is.

There's some confusion here, I guess. So, he's out of the craft on LEO, travelling anywhere around 7800 m/s and throws the screwdriver at 20 m/s. From my layman understanding of how orbits work and depending on the direction it's thrown the screwdriver will have increased it's orbital velocity and orbital eccentricity (thrown prograde), decreased orbital velocity and increased eccentricity (retrograde), increased orbital velocity and slightly shifted apogee/perigee along the orbit (thrown to/from Earth) or changed inclination slightly (thrown north/south direction). Obviously there's a bajillion of intermediate options.

Whatever he does he's unlikely to be able to supply the screwdriver with enough delta-v to hit the planet like very soon. It will just very-very slightly change it's orbital parameters. Anyway, both the astronaut and screwdriver will stay in orbit for a long-long time until their orbits deteriorate enough for re-entry.

It was just a guess that astronaut's orbit will deteriorate faster due to drag, but let's do some proper calculations. Let's assume the orbit is 150 km and mostly circular. The orbital velocity is around 7800 m/s. Atmospheric density at this height is around 2.5*10-9 kg/m3 . Body area of an astronaut facing the direction of travel is about 1m2 (with the spacesuit I'll assume something like 1.7m2). Reference area of the screwdriver is somewhere around 0.07 m2. CD for a person is around 1.1 and I'll approximate the shape of the screwdriver with a rectangle with CD of ~0.9.

Then by the drag equation: $F_D=1/2\rho*v^2C_DA$ for a person in a spacesuit we get 0.142 N and for the screwdriver we get about 0.00476 N. I used the worst-case scenario, i.e. screwdriver thrown retrograde and it's velocity is now 7780 m/s.

Then the deceleration due to drag will be (assuming a person+spacesuit weighs 100 kg and screwdriver 0.3 kg) 0.00142 m/s2 for the person and 0.01586 m/s2 for the screwdriver... which explicitly shows how useless common sense and hunches are when talking physics.

I was wrong. The screwdriver will indeed fall first.

Edited by pavelcherepan

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Janus, how did you come up with 2.5 trillion miles? I'm getting a very different number around 3.8 million kilometers.

I double checked to make sure I didn't make a conversion error along the way. But I couldn't find one.

1 mph = 1.6kph = 1600 meters/hr = 0.444 meters/sec

Escape velocity is found by

$V= \sqrt{\frac {2 \mu}{d}}$

solved for d:

$d = \frac{2 \mu}{V^2}$

mu = 3.987e14 m^3/s^2 for the Earth,

so this gives a distance of ~4e15 m for the distance where V = 0.444m/s (1 mph)

That's 4e12 km or ~2.5e12 miles

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I double checked to make sure I didn't make a conversion error along the way. But I couldn't find one.

1 mph = 1.6kph = 1600 meters/hr = 0.444 meters/sec

Escape velocity is found by

$V= \sqrt{\frac {2 \mu}{d}}$

solved for d:

$d = \frac{2 \mu}{V^2}$

mu = 3.987e14 m^3/s^2 for the Earth,

so this gives a distance of ~4e15 m for the distance where V = 0.444m/s (1 mph)

That's 4e12 km or ~2.5e12 miles

Yeah, but if you launch straight up and keep going straight up won't you get lateral velocity once you're well and truly out of the atmosphere? Assuming you launch from equator you should get ~460 m/s of lateral velocity, so I solved the same equation but for v=460 m/s (I disregarded the extra 1 mph because it's too small to matter much).

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7. What is the margin of adjustment you can make in your orbit and not fall out of orbit? My assumption is if I was in a steady orbit, then left my ship, I would stay in orbit, at least for a long while. So If I chucked a screwdriver (made of tungsten) at the earth from the space station, would it keep going until it hit?

You never really fall out of orbit, instead if you make enough of an adjustment, your new orbit will intersect with body you are orbiting for the ISS for example, you would need to decrease its orbital speed by some 143 m/s to get it to graze the Earth.(ignoring atmospheric drag)

8. Can the earth fall out of orbit if we remove too much material from it?

When the difference between the orbiting object and the body it orbits is very large (like it is for the Sun and the Earth)the mass of the object has little effect on its orbit. (a tennis ball placed in the Earth orbit in its place would orbit in almost the same way as the Earth.)

Thanks guys!

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You had me at bacon flavoured ice cream .

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3. It took 8+ minutes for the shuttle to accelerate to 18,000mph. Is the time in which we reach 18,000mph of any particular importance if we remove fuel limitations from the equation?

Yes. Too rapid (de)acceleration can kill crew.

After exceeding 5-6 g typical human is losing consciousness.

Astronauts and air jet pilots wear special suits to increase their limits (and have regular training).

Edited by Sensei

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Yes. Too rapid (de)acceleration can kill crew.

After exceeding 5-6 Mach typical human is losing consciousness.

Astronauts and air jet pilots wear special suits to increase their limits (and have regular training).

https://en.wikipedia.org/wiki/High-G_training

With respect to the sun, every human on Earth is exceeding Mach 5 because the orbital velocity of the Earth is about 30 km/sec (near Mach 100)

Velocity doesn't trouble us at all. it's acceleration that matters.

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I meant 5-6 g of course.

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A brief enough exposure to 5g wouldn't be nice, but it probably wouldn't give rise to serious damage.

https://en.wikipedia.org/wiki/John_Stapp

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Yeah, but if you launch straight up and keep going straight up won't you get lateral velocity once you're well and truly out of the atmosphere? Assuming you launch from equator you should get ~460 m/s of lateral velocity, so I solved the same equation but for v=460 m/s (I disregarded the extra 1 mph because it's too small to matter much).

The fact that there would be such a small difference between including the 1 mph and not doing so ( ~3 km)is pretty much the point. The Earth's rotation speed overwhelms the speed given in the question. (Besides, that adjustment changes with latitude and drops to 0 at the poles)

In addition, I tried to answer the question in the spirit in which it was asked, with the ship maintaining a constant "upward" speed. "Upward" meaning away from the center of the Earth. (If you try and include the rotation of the Earth's added effect, you will not be maintaining a constant upward speed.)

This is a case where trying to be "too correct" in accounting for all the factors can just end up confusing the questioner. For example, the 4145 mile altitude I gave in one of my answers was not exactly correct either if you take Earth rotation into account, But working out the exact answer involves a number of orbital mechanics equations, and wouldn't really benefit the questioner at this point of the learning curve..

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Thanks for all the answers guys! I am wondering why you can't orbit below a certain point? Also, what determines how far something can go before it starts coming down? Baseball pitches stay level until they get to the batter. Bullets stay at a steady height for a good distance. So if I shot a bullet at 1,000,000MPH would it make it around the earth without dipping into the ground? (assuming nothing is in the way of course)

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Newton used exactly that example (well, not baseball) to explain orbits: http://www.daviddarling.info/encyclopedia/N/Newton_cannon.html

Here is an interactive version: https://scratch.mit.edu/projects/11882821/

Basically, your baseball or bullet needs to reach orbital speed: https://en.wikipedia.org/wiki/Orbital_speed

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Thanks for all the answers guys! I am wondering why you can't orbit below a certain point?

Atmospheric drag is probably the main reason.

Also, what determines how far something can go before it starts coming down? Baseball pitches stay level until they get to the batter. Bullets stay at a steady height for a good distance. So if I shot a bullet at 1,000,000MPH would it make it around the earth without dipping into the ground? (assuming nothing is in the way of course)

No. Bullets and baseballs fall, but the distance is a quadratic effect in time so the motion is small at first. But it's there. Absent any aerodynamics, a bullet drops just as fast fired from a gun as from your hand.

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Thanks for all the answers guys! I am wondering why you can't orbit below a certain point? Also, what determines how far something can go before it starts coming down? Baseball pitches stay level until they get to the batter. Bullets stay at a steady height for a good distance. So if I shot a bullet at 1,000,000MPH would it make it around the earth without dipping into the ground? (assuming nothing is in the way of course)

Once a bullet leaves the end of the barrel, it falls to the ground at the exact same rate as a bullet you dropped from your hand. The fact that the bullet travels so quickly simply gives you the impression that it is at a steady height.

EDIT: Sorry, cross posted with swansont.

Edited by zapatos

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Thanks for all the answers guys! I am wondering why you can't orbit below a certain point? Also, what determines how far something can go before it starts coming down? Baseball pitches stay level until they get to the batter. Bullets stay at a steady height for a good distance. So if I shot a bullet at 1,000,000MPH would it make it around the earth without dipping into the ground? (assuming nothing is in the way of course)

To be quite correct, every object you throw or shoot out of a gun etc, does orbit the Earth ( or rather it's center of mass), it is just that most of that orbit would have to pass through the body of the Earth. The path the object follows will be an ellipse,with the most of it under the surface. The faster the object is traveling when it leaves you, the more circular this ellipse becomes, and a larger portion of it remains above the surface. At the surface of the Earth, it takes a velocity of ~17,840 mph to make an orbit that will stay above the surface for the whole time in a circular path (assuming, as already noted, that we can ignore atmospheric drag which would slow the object down).

If it goes any faster, the orbit gets more elliptical again and will actually climb away from the surface before returning to you at the height at which you threw it You will be the low point of now of the ellipse(when before, you were the high point). Once again, I am ignoring any effects caused by the rotation of the Earth and assuming that the objects initial path was parallel to the ground.

The faster it leaves you, the higher it will climb, right up to when it is moving at ~25226 MPH as it leaves you, At which point, the path is no longer an ellipse, but becomes a parabola with you at the vertex. The object will no longer return to you and will just continue to fly into space Any faster and the trajectory becomes a hyperbola.(in reality, you never see a parabolic trajectory since that requires an exact speed; the slightest bit slower give you an ellipse and the slightest bit faster gives you a hyperbola. Also, at a certain point, the gravity of other bodies has to be taken into account..

Up to now, I've ignored the effect of the Earth rotating. All the velocities mentioned above are measured with respect to the center of the Earth.

Since the Earth rotates, the surface is moving with respect to the center, from 0 mph at the poles to ~1044 mph at the equator. So this velocity, whatever it is where you throw the ball or shoot the gun must be added to the projectiles velocity to work out its exact orbit.

This can be a very complicated process especially if the projectiles motion with respect to the surface is not directly in the same direction as the motion of rotation. Luckily, when it comes to throwing baseballs or the shooting range, this does not present any problem, because the corrections needed to compensate for this is so small we can safely ignore them.

This is not the case with long range artillery or ballistic missiles. With them, you do have to take into account how the Earth's rotation effects the orbital trajectory of the projectile ( for instance firing with the rotation of the Earth will cause it to spend a longer time before it hits the surface again than when firing against the rotation.) and, since once the projectile is launched it moves independently of the Earth's rotation, you have to work out how much the Earth turns while it is in flight to work out exactly what point of the Earth's surface it will strike. So hitting a target with long range artillery or a ballistic missile takes quite a bit of calculation.( for example, when firing a long range artillery shell to the North or South, its path, as measured with respect to the ground, will curve in the East-West direction.)

For a simple example of this, imagine yourself on the equator, and you fire a gun with a muzzle velocity of 17,840 mph. First to the East and then to the West.

The Eastward bullet will be traveling at the muzzle velocity + the surface speed at the equator. This adds up to 18,884 mph or greater than the speed needed for a circular orbit. This means that disregarding obstacles and air resistance. it will return to where you fired it(with respect to the center of the Earth) in ~1.7 hrs. It will also climb to several hundred miles above the Earth's surface before returning to the height you fired it. But is 1.7 hrs, the Earth will have rotated and carried you some 1851 miles Eastward. So when the Bullet returns to the point of its original height, you won't be there. By the time the bullet does catch up with you, it will have climbed again, and will pass over your head.

The Westward bullet will be traveling at the muzzle velocity - the surface speed at the equator. This comes out to being less than that needed fro a circular orbit, and the bullet's path will intersect and hit the surface, burying itself in the ground after traveling for some distance. Two very different results from the same gun fired in different directions.

Edited by Janus

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Great break down Janus! I understand it now. This makes Kerbal Space Program a lot easier to understand.

So if I escape Earth's influence, I am still orbiting the sun. So I'm floating through space like a jackass being carried around the sun at 18.6m/s or something like that form 93 million miles away. From around there I see on wikipedia that my escape velocity for the solar system is 26m/s. So i'm going around at 18.6m/s but I'm going away from the sun at 0m/s. So lets say I steal Sandra Bullock's trusty fire extinguisher and point it at the sun and fire it. I get up to a speed of say 50ft/s going away from the sun (whats the term for moving away from the sun/an object?).

1.When does the sun stop me?

2. If I had a big enough extinguisher tank could I use it to leave the solar system at a crawling pace since there is no drag in space?

3. If I hold on to a rocket in space, I never stop accelerating until I run out of fuel right? So, Is getting up to speeds of a million miles per hour doable in theory if we just piggy back rockets one at a time that were already placed in space?

Thanks guys. You rock!!!!

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Great break down Janus! I understand it now. This makes Kerbal Space Program a lot easier to understand.

So if I escape Earth's influence, I am still orbiting the sun. So I'm floating through space like a jackass being carried around the sun at 18.6m/s or something like that form 93 million miles away. From around there I see on wikipedia that my escape velocity for the solar system is 26m/s. So i'm going around at 18.6m/s but I'm going away from the sun at 0m/s. So lets say I steal Sandra Bullock's trusty fire extinguisher and point it at the sun and fire it. I get up to a speed of say 50ft/s going away from the sun (whats the term for moving away from the sun/an object?).

1.When does the sun stop me?

After you get ~39 km further away from the Sun, but then you will begin to fall back in until you get ~39 km closer to the Sun then you are now. (what you have done is put yourself in a more elliptical orbit around the Sun*) If you want to make the most of that 50 ft/sec, you need to apply it in the direction that you are already orbiting in, that will get you to a maximum of ~153,000 km further from the Sun.

2. If I had a big enough extinguisher tank could I use it to leave the solar system at a crawling pace since there is no drag in space?

Drag is not really the issue. (while you are leaving Earth, it is an issue but not the overriding one). The issue is climbing against gravity. If you try and use the slow crawl starting at the surface of the Earth, you are burning the vast majority of your fuel just to keep the rocket from falling back to Earth. (it you cut your engines, you fall back to the ground). But while in orbit, your orbital speed is already countering the Sun's gravity, so anything you add just lifts you into a higher orbit. So even a small continuous thrust will move you away from the Sun. The only real advantage to this is that it allows you to use high efficiency/low thrust engines like Ion thrusters, rather than high thrust/low efficiency one's like chemical rockets leading to lower fuel to payload ratios.

3. If I hold on to a rocket in space, I never stop accelerating until I run out of fuel right? So, Is getting up to speeds of a million miles per hour doable in theory if we just piggy back rockets one at a time that were already placed in space?

Not really. Let's assume that you have a chemical rocket, Its empty mass (no fuel) is 1 ton. This mass includes the payload, engines, fuel tanks, superstructure to hold it all together, etc. We also assume that it can carry 2 tons of fuel. Thus the mass of the fully fueled ship is 3 times the mass of the unfueled ship. This is called the mass ratio. There is a formula, which I won't go into here, that tells you how much of a change in velocity you can get with a given mass ratio and the exhaust speed of your rocket engine. ( chemical rockets can achieve exhaust speeds of around 2.8 miles per second.)using this you can work out that this rocket can reach a speed of ~3 miles per sec (11,123 mph).

So using your suggestion, let's piggy back this rocket on to another rocket. If we use the same type of rocket with the same amount of fuel, it has to push not only its own mass but the mass of the piggy backed ship. So the total mass of the two ships together is the mass of the first ship (3 tons), plus the mass of the second rocket (3 tons) for at total of 6 ton. After our booster rocket use up its fuel, the total mass is 4 tons. 6 ton/4 tons = 1.5, and that is the mass ratio we use to figure out how fast our tandem of ships is moving after the booster rocket burns out. this works out to being ~1.13 miles per sec. We detach the booster and fire the piggy-backed ship which adds another 3 mile/sec for a total of 4.13 miles/sec. we used twice as much fuel and gained only ~38% more speed.

If we piggy back this pair onto a third rocket, then for the first stage we get a mass ratio of 11/9, and after the this stage has burned out the remaining two ships are moving at 0.56 m/sec. The second stage adds another 1.13 mps and the last stage 3 mps for a total of 4.69 mps.

As you can see, each added layer of piggybacking adds less and less to the final speed. While you could theoretically keep adding stages increasing the final speed bit by bit until it equaled 1,000,000 mph, it would take an implausibly large number of stages to do so.

A better approach is to finding ways of propelling our ship by more efficient engines. Ion thrusters, as mentioned earlier are a possible solution. With them you can possibly get exhaust velocities in the range of 75 mps.

Thus with the same 3 to 1 mass ratio with had with our last stage above, we can get a velocity change of better than 82 mps (29,6625 mph), 30 mps(108964 mph) with the second stage and 15 mps(54,000 mph) with the first stage for a total of 459,589 mph, or nearly 46% of that 1,000,000 mph in three stages. Granted, each additional layer of piggy-backing adds less and less to the final velocity and you are going to end up with an impractical number of stages before you reach your target of 1,000,000 mph, but at least we are getting closer.

To give you an idea of what we are up against, let's assume that we do away with all the individual stages with all there extra mass in terms of engines and fuel tanks etc, and just imagine that somehow we can get our original 1 ton spaceship to carry all the fuel it needs to get up to 1,000,000 mph without having to increase its 1 ton empty weight. How much fuel would it have to carry using the Ion engine?

It turns out to be ~39.5 tons of fuel. ( on the other hand, the same calculation assuming the chemical rocket gives and answer of 1.2e43 tons. That's some 3,683 times the upper mass estimate for the entire Milky way galaxy.)

This is pretty sobering considering that 1,000,000 mph is still only 0.15% of the speed of light. Even with a Ion engine, You would need nearly 60,000,000,000 tons of fuel to get our 1 ton up to even 1% of the speed of light.

Thanks guys. You rock!!!!

*For this problem I am assuming that you start in a perfectly circular orbit around the Sun at average Earth orbital distance. I also had to work out the orbital speed to a higher accuracy first, as your figure's inaccuracy was greater than 50 ft/sec to start with.)

P.S. as a side note, you can use the above rocket equation to work out just how big a fire extinguisher you would need to give you that 50 ft/sec velocity in the first question. Fire extinguishers can exhaust as high as ~328 ft/sec. If you, the empty extinguisher, your spacesuit etc, weighed 180 lbs, this means the extinguisher would have to have contain ~30 lbs of propellant. (that's a fairly hefty extinguisher), and to get you to escape the sun completely, it would need to be many, many, many times the mass of the entire galaxy. 328 ft/sec is just way too small an exhaust velocity to be able to generate much in the way of a total velocity change.

Edited by Janus

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