Would someone please explain why this maths doesn't work?

[Goody1981]

I haven't checked any of this out in the game because I just thought about it, and not at home. In my head, what I'm about to suggest shouldn't work (and I'm sure it doesn't) but I don't know why!

Kerbin's rotational velocity is 174 m/s, and we all know the reason we launch East is because we already have that horizontal velocity, so why not use it. So even if you launch straight up, with no gravity turn at all, you still have that 174 m/s horizontal velocity.

Now, presumably there is some (very high) altitude where the required orbital velocity is 174 m/s. 

So, if you launched straight up, set apoapsis at that specific altitude, and coasted all the way up to it - when you reached the very peak of your trajectory, your vertical velocity would be zero, and your horizontal velocity would be 174 m/s.

And if you have a horizontal velocity of 174 m/s, and the required orbital velocity at that altitude is 174 m/s... then why wouldn't you be in orbit?!

Edited May 4, 2017 by Goody1981

[Rocketthrust]

I'll have to check that.  Still a good question.

Edited May 4, 2017 by Rocketthrust

[Red Iron Crown]

First, such an altitude would be outside of Kerbin's sphere of influence. By my calculations it would be at an altitude of about 113,000km, Kerbin's SoI is only 84,000km in radius. 

Second, you won't have that 174m/s of velocity when you get there. Your periapsis will still be well below Kerbin's surface, a burn at the target altitude is required to raise it up to the same altitude. The reason for this is that the kinetic energy of the craft is converted into potential energy as it climbs to apoapsis. 

The equation describing this relationship is the vis-viva equation, one of the most important equations in spaceflight. 

[Goody1981]

Thanks @Rocketthrust @Red Iron Crown for replying..

Had a feeling that it would be out of the SOI, so ok, we can't do this in the game.

But say we could get to that altitude.. why wouldn't we still have that horizontal velocity? Isn't one of the most fundamental parts of projectile motion (and as soon as we are coasting, we're a projectile) that horizontal velocity and vertical velocity are separate, and there is no acceleration in the horizontal plane, therefore nothing to change our speed?

I've played thousands of hours in the game, so I know intuitively that the periapsis should be below the surface without a prograde burn at Ap to bring it up, but I still can't see the answer to my question of if orbital velocity is 174m/s horizontally, and we're going 174 m/s horizontally, why aren't we in orbit?

Edited May 4, 2017 by Goody1981

[Red Iron Crown]

2 minutes ago, Goody1981 said:

But say we could get to that altitude.. why wouldn't we still have that horizontal velocity? Isn't one of the most fundamental parts of projectile motion (and as soon as we are coasting, we're a projectile) that horizontal velocity and vertical velocity are separate, and there is no acceleration in the horizontal plane, therefore nothing to change our speed?

Let me stop you right there. You need to cast aside your cartesian plane physics and go full Keplerian. More important than horizontal vs vertical speed is total energy, kinetic plus potential. When in freefall the sum of these is fixed, add to one and you must subtract the same amount from the other.

2 minutes ago, Goody1981 said:

I've played thousands of hours in the game, so I know intuitively that the periapsis should be below the surface without a prograde burn at Ap to bring it up, but I still can't see the answer to my question of if orbital velocity is 174m/s horizontally, and we're going 174 m/s horizontally, we aren't we in orbit?

Run that vis-viva equation on that altitude and speed and you'll find that the SMA would require periapsis to be at nearly the same altitude. You cannot arrive at an apoapsis altitude of an elliptical orbit with circular orbit speed for that altitude, because then you'd already be in a circular orbit!

[1101]

I don't know the maths for this(trying to learn orbital mechanics, not very hard though).  That said, I think if you could (magically) teleport an object from Kerbin surface to 113,000km as per Red Iron Crown, the lateral Kinetic Energy (if kept in the teleport) should keep you in orbit.  I think the difference though is that in the rocket powered ascent you would lose some (all?) of that horizontal K.E. as you travel across (an admittedly small) part of the surface of Kerbin.  I think this would be due to the trajectory moving across the surface at 175m/s, so gradually it would become a more vertical component and be nullified by gravity.  If the game drew the trajectory on the map relative to Kerbin's surface rather than centre, you might get a weird looking trajectory.

Feel free to prove me wrong.

[DrLicor]

@Goody1981, to respond on your last question. If I understand it right. 
The surface velocity isn't your orbital velocity. Those are two completely separate things. 
An (circulair) orbit means that the force that pulls you to the planet equals the you have foward, or the orbital speed. 
Orbit means zero gravity and on earth you don't have a zero gravity.

12 minutes ago, Goody1981 said:

but I still can't see the answer to my question of if orbital velocity is 174m/s horizontally, and we're going 174 m/s horizontally, we aren't we in orbit

On the surface of kerbin, we do indeed go 174m/s, but we don't orbit it. 
And as mentioned earlier, an orbital speed of 174m/s equals the gravity of kerbin at a height of 113.000km.

But let's forget the SOI for a moment, the only way you could achieve that orbit, you have to go prograde on your orbital velocity, not the surface velocity. When going straight up, you'll counter the force going east. 
And you can only maintain that force west, by burning a little bit west and eventually the last second 100% west

Edited May 4, 2017 by DrLicor

[Ten Key]

25 minutes ago, Goody1981 said:

Kerbin's rotational velocity is 174 m/s, and we all know the reason we launch East is because we already have that horizontal velocity, so why not use it. So even if you launch straight up, with no gravity turn at all, you still have that 174 m/s horizontal velocity.

Velocity is not conserved. Angular momentum is conserved. As your rocket increases its distance from the center of rotation, its angular velocity will decrease so that its angular momentum remains constant. 

[MaxL_1023]

The integrated vector sum of Kerbin's gravity will result in you losing most of your horizontal velocity as you coast to Apoapsis. You will be in an elliptical orbit the whole way up, with the perapsis at a value close to what it would be if Kerbin was a point mass and you just started on the pad with 174 m/s of horizontal velocity. 

[Goody1981]

32 minutes ago, Red Iron Crown said:

Let me stop you right there. You need to cast aside your cartesian plane physics and go full Keplerian. More important than horizontal vs vertical speed is total energy, kinetic plus potential. When in freefall the sum of these is fixed, add to one and you must subtract the same amount from the other.

Run that vis-viva equation on that altitude and speed and you'll find that the SMA would require periapsis to be at nearly the same altitude. You cannot arrive at an apoapsis altitude of an elliptical orbit with circular orbit speed for that altitude, because then you'd already be in a circular orbit!

Ahh I think I get what you're saying - being in orbit is about the energy you have. In my example you have 174 m/s purely horizontal only at your Ap. To be in a (circular) orbit you need that horizontal velocity at all parts of your orbit, which takes more energy.. is that getting close? 

Thanks to the other replies too, but I can't see how you can lose that horizontal velocity? Only one force once you are coasting - gravity pulling you straight down. Which only affects vertical velocity. What force changes your horizontal velocity?

Edited May 4, 2017 by Goody1981

[Ten Key]

 

[Reusables]

It's because Kerbin is round. 

First, did you get that the horizontal direction is relative to the ground? It's not a fixed direction on orbital mechanics, which causes some trouble. In detail, it actually exerts a phantom force - called fictitious force or inertial force.

As you move eastward while going up, the ground position directly under you moves eastward as well. This causes horizontal direction to go downside(or fall). Since you are going up, this means you lose horizontal speed - which is horizontal element of velocity. Though this is not actually a loss, as it reappears as a gain on the vertical speed.

This is similar with how an object on circular orbit don't fall: Its horizontal direction falls, matching the moving direction of the object. As a result, it ends up going horizontally all the time.

 

Besides, kerbal aircrafts got automatically pitched up because of this effect. Do you know what happens with IRL planes? 

[JoeNapalm]

 

This is a lot like asking why apples aren't oranges.

Standing on the surface and orbiting are two entirely different physical events.

You're pressed onto the surface by gravity and whisked along with it by inertia, when you're standing on the surface.

When you're orbiting, you're falling at a velocity and angle that allow you to continually miss the ground, while gravity keeps dragging you along after the planet.

If the surface rotation was at orbital velocity, you'd achieve orbit by picking your feet up...you'd also have to be standing on something like a Dyson Sphere that had a radius equal to the radius of a geosynchronous orbit.

 

-Jn-

[Mjarf]

I thought the main reason for launching east is because all the other celestial bodies rotate in that direction? That 174m/s of delta-v isn't very relevant for a trip that requires 5-10km/s or more delta-v.

[Red Iron Crown]

3 hours ago, Mjarf said:

I thought the main reason for launching east is because all the other celestial bodies rotate in that direction? That 174m/s of delta-v isn't very relevant for a trip that requires 5-10km/s or more delta-v.

It's primarily to reduce the cost of achieving orbit around Kerbin. Once in orbit the cost to go to another planet is the same whether you're orbiting eastward or westward. Kerbin's moons are cheaper from an eastward orbit, though. 

[Goody1981]

4 hours ago, JoeNapalm said:

This is a lot like asking why apples aren't oranges.

Standing on the surface and orbiting are two entirely different physical events.

You're pressed onto the surface by gravity and whisked along with it by inertia, when you're standing on the surface.

When you're orbiting, you're falling at a velocity and angle that allow you to continually miss the ground, while gravity keeps dragging you along after the planet.

If the surface rotation was at orbital velocity, you'd achieve orbit by picking your feet up.

Thanks for the reply  but I think you missed my point somewhat. Orbital velocity is dependent on altitude, the lower you are the faster you need to go. I was never saying 174 m/s was orbital velocity at ground level... but it IS an orbital velocity somewhere very high (13,000 km I'm told)

 

5 hours ago, Abastro said:

It's because Kerbin is round. 

-snip-

That makes a lot of sense - thanks! And similar (I now realise) to what @Red Iron Crown was saying about getting rid of my Cartesian plane thinking haha... hard to ditch those high school physics lessons! Thanks all for the replies ...think I get it now  

[XLjedi]

After stumbling upon this fascinating discussion...  reading the OP...  scrolling thru rapidly...  I will now go back to my large rocket, hit it with a hammer, and hope it makes it off the launch pad this time. 

Too many smart people playing this game.

Edited May 5, 2017 by XLjedi

[HebaruSan]

38 minutes ago, Goody1981 said:

think I get it now  

I'm going to beat the dead horse because this question was interesting to try to answer intuitively.

In the square coordinate grid that I think you were imagining, you start off directly above the center of the planet. Gravity points straight "down" according to the grid, you're planning to thrust straight up, and you're already moving eastward at 174 m/s.

Now think where you are 1 minute after lift-off, and assume that you keep your vessel's orientation fixed according to the square coordinate grid, and that your horizontal velocity is constant (it isn't, but we haven't gotten there yet). You move up by some amount depending on your TWR, but your eastward velocity means that you are now 174*60=10440 meters off-center. Gravity no longer pulls straight down according to the grid! It's slightly diagonal now, toward the center of the planet, but your initial velocity is still "horizontal" according to the grid, so gravity is pulling against it, reducing it (and so it wouldn't be quite 10440 meters, since you're not going 174 m/s the whole time).

But even the initial velocity you have left is not fully "horizontal" anymore according to what you need for an orbit! If you orient yourself in the cabin based on where the ground is, part of your initial velocity vector now points "upwards." Combine these two factors together, and you lose some of the 174 m/s to gravity and some of it to vector rotation (which ironically was caused by the velocity itself!).

Edited May 5, 2017 by HebaruSan

[Goody1981]

7 minutes ago, HebaruSan said:

I'm going to beat the dead horse because this question was interesting to try to answer intuitively.

In the square coordinate grid that I think you were imagining, you start off directly above the center of the planet. Gravity points straight "down" according to the grid, you're planning to thrust straight up, and you're already moving eastward at 174 m/s.

Thanks! (And great pic  )

[Archgeek]

Nicely put, @HebaruSan, you beat me to it!

I was going  to say that, as you zip upward and away from Kerbin, drifting eastward with that initial 174m/s component, something strange happens with the concept of "down".  As you drift in the direction formerly known as East, Kerbin starts to become less below you and more behind you, the "surface-east" and "up" vectors intermingling more and more the further you go, with gravity increasingly sapping that inital surface kick as they do so.

[Snark]

On 5/5/2017 at 3:31 PM, HebaruSan said:

But even the initial velocity you have left is not fully "horizontal" anymore according to what you need for an orbit! If you orient yourself in the cabin based on where the ground is, part of your initial velocity vector now points "upwards." Combine these two factors together, and you lose some of the 174 m/s to gravity and some of it to vector rotation (which ironically was caused by the velocity itself!).

Excellent example, HebaruSan!

Just to build on the above quote:  Imagine an even simpler case:  Imagine that you have a spherical, rotating planet... and no gravity at all.  Your ship is on the equator, held in place only by sticky feet.  It lets go and is flung away from the planet (because it keeps moving in a straight line, with no forces on it).

Suppose the planet is rotating such that its equator is moving at 100 m/s.  At the moment the ship lets go, it has a "horizontal velocity" of 100 m/s, and a "vertical velocity" of zero.

Now follow that ship, as it keeps coasting outwards in a perfectly straight line at a uniform 100 m/s.  By the time it's, say a dozen planetary radii out, which direction is it moving as seen from the planet?  Steeply vertically upwards!  And since its total speed is still an unchanged 100 m/s, that means that the "horizontal" component is going to be something very small.

There was no force on the ship.  Nothing accelerated it.  It didn't move faster, or slower, or change direction at all.  But it went from horizontal at 100 m/s to nearly vertical, with very little horizontal component.

What did that?  Geometry. 

To @Goody1981:  as has been observed, there are two conserved quantities in ballistic motion:

• Total energy (kinetic + potential)
• Angular momentum

...and that's it.  You'll note that "horizontal speed" is not in the above list.