# A proposal to reduce the delta-V to orbit.

1. Jun 18, 2010

### RobertGC

To reach orbit you have to have your vehicle have a horizontal, i.e., tangential, orbital velocity, of about 7,800 m/s and have sufficient altitude, say at least 100 km, the altitude considered to be "space". To get to this altitude you have to have a separate velocity in the vertical direction. The usual way to estimate this vertical velocity is by using the relation between kinetic energy and potential energy. It gives the speed of v = sqrt(2gh) to reach an altitude of h meters. At 100,000 m, v is 1,400 m/s. So then it is common to estimate the required delta-V to orbit to be 1400 + 7800 = 9,200 m/s.
However, it seems to me you can reduce this by traveling in a straight-line path. If you travel at an angle to the horizontal so that your vertical velocity component is 1,400 m/s and your horizontal component is 7,800 m/s then you only actually need sqrt(7800^2 + 1400^2) = 7,925 m/s delta-V. Actually if you add on the ca. 460 m/s velocity you get for free from the Earth's rotation you might be able to reduce this to sqrt(7400^2 + 1400^2)
= 7,531 m/s. So what I'm trying to investigate is if it is possible for a rocket, without using wings or lifting surfaces, to travel at such a straight-line trajectory at an angle from lift-off so that the achieved velocity will be in this range.
The question is: if you angle the rocket from the start with the thrust vector along the center line with the trajectory angle such that the vertical component of the thrust equals the rocket weight could you have the rocket travel at a straight-line all the way to orbit? I'm inclined to say no because the gravity is operating at the center of gravity of the rocket not at the tail where the thrust is operating. This would certainly work if you had a
point particle, but I'm not sure if it would work when your body has some linear extent.
This method for traveling at a straight-line at an angle for some or all of the trip would make my calculations easier. However, I'll show in a following post there is another way to do it even if this first method doesn't work. The second method though would require some modification to the usual design of rockets and is more computationally complicated.

The question I'm asking can be boiled down to this: imagine you have a long cylindrical object, could be a pencil, could be broom stick. You can give it an initial thrust at the bottom and push it away at an angle. It will then follow a curved trajectory with its center of mass following a parabolic arc, disregarding air drag.
Then what I'm asking is will it work to supply a continual push at the bottom with the force maintained at the bottom at a fixed angle to the horizontal so that the vertical component of this force is the cylindrical body's weight?
Will the body maintain a continual straight-line trajectory at this set fixed angle?

This is really actually a question in continuum mechanics, sometimes called solid mechanics. In physics we often idealize a body subject to forces as a point particle. Idealized this way, the thrust force applied to the rocket would add as a vector to the force of gravity so it would cancel gravity no matter what the angle of the trajectory.
But in continuum mechanics you have to consider the physical extent of the body and where on the body the forces are applied. I imagine this is a common type of problem addressed in mechanical engineering and civil engineering.
A force applied at one position on the body won't have the same effect as when it is applied to another point for instance in regards to the torque produced. Torque is measuring the turning "force" on the body. But it's defined as the cross product of the force vector times the radial vector to the center of rotation.
Intuitively, what we have to worry about is rotation of the rocket with the thrust applied only at the tail. But if the rocket were to rotate it would be about the center of gravity. However, if we make it so the thrust is always along the center line, the radial vector to
the cg and the force vector are parallel, resulting in a 0 torque vector. That would mean there would be no rotation around the center of gravity so we should be able to maintain our straight-line trajectory.
This would be valid if the center of gravity were fixed. But the cg is accelerating as the thrust is applied. So I'm not sure if this argument applies in that case.

Bob Clark

2. Jun 18, 2010

### Janus

Staff Emeritus
Space craft launched from Earth use a "gravity turn" trajectory. In essence, as soon as possible after launch, they pitch over slightly . Gravity then "steers" the ship. As the ship continues to slowly pitch over due to gravity, the ships trajectory levels out more and more. By the time it reaches orbit it is traveling entirely in the horizontal. This method prevents the need for stealing thrust to steer the ship (something you would have to do to maintain your "straight line" trajectory.)

Rest assured that the professionals in charge of these launches have thought of and done everything possible to get the most out of their rockets as they can.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook