Suppose I have a large spinning spherical body with a vertically-oriented rocket motor at one point on the equator. If I fire the rocket continuously, what is the trajectory of the spinning body over the course of multiple rotations? I seem to remember that it would move off on sort of a crabbing, irregular trajectory not quite in line with the orientation of the motor at the time of ignition, but I may be confusing this scenario with the effect of an off-axis rocket motor. In my scenario the motor does not change the spin rate of the larger body like an off-axis motor would.
As I see it, at 0 degrees the rocket accelerates the spinning body in a particular direction, then at the 180 degree mark it cancels out that initial acceleration; rotate back to 0 degrees and and accelerate, back to 180 degrees and decelerate, etc. This is true for every pair of points 180 degrees apart. The final effect is that of a net push over the first 180 degrees of rotation that is cancelled over the next 180 degrees, then the cycle repeats. As long as the rocket fires, the larger body sort of bumps along following a weird curve. Every full 180 degrees of rotation cancels the acceleration of the previous full 180 degrees.
I suspect that the velocity of the spinning body goes to zero every time the rocket gets back to the 0 degree mark, but with the spinning body displaced from its original position. But this is all based on intuition, not calculation. What really happens, mathematically?
For some reason Google is not helping me in finding treatments of off-axis or equatorial rocket motors.
After some mind maths, which I’m only semi-proficient at, I think spherical body will move in perfect circles. You’re correct that the velocity along any diagonal cancels out after 180 degrees, but by the same symmetry the displacement also cancels out as you go through the next 180 degrees.
The acceleration in the x and y directions will be sin(ot) and cos(ot). I assumed, like you that we’d get some kind of overall displacement, but manipulating those two convinced me otherwise.
Are we assuming that the mass of the body is much greater than the mass of the propellant, such that mass change can be neglected? In that case, then in a reference frame where the body was initially at rest, the path will be a cycloid.
To note, this is called a non-ballistic or powered orbit. In this case, the centroid is on a focus laying of the line perpendicular to the thrust axes, the distance of which is determined by the rotation rate of the body. In a sense, this is a described as a cylindrical cam in reverse; that is, the force defines the motion rather than the controlled motion defining the application of force.
Of course, if we are assuming a very large (e.g. planetoid large enough to have signiicant surface gravity), if the effective exhaust velocity (v*[SUB]e[/SUB]) of the propellant is less than the escape velocity of the body the net motion of the system (body and propellant) will be zero, as the propellant will fall back onto the body, cancelling out the momentum transfer from the rocket. (It cannot go into orbit because it will not have a significant tangential velocity.) Even if v[SUB]e[/SUB]* is greater than escape velocity if it isn’t much greater you’ll still lose a significant amount of effective thrust due to the gravitational coupling, so you really have to calculate the effective exhaust velocity at the point at which it exceeds the characteristic energy (C[SUB]3[/SUB]). The impact this has on motion will depend on the mass and velocity of the propellant, but it will be an impulse distributed over time (whereas ejection from the exhaust plane of the rocket is assumed to be instantaneous impulse per unit mass flux) and so you’ll end up with an odd, drunken sort of orbit.
Practically speaking, if the mass change of lost propellant is so small that it is irrelevent to the dynamics of the body, the effective momentum change, and thus resulting motion of the body, will be very small. For example, if you had an effective exhaust velocity (from the whole system) of 100 km/s (the high end for an electric ion or plasma engine) and at mass change of 0.01% (m[SUB]0[/SUB]/m[SUB]1[/SUB] = 1/0.9999) the Tsiolkovsky equation gives a maximum delta velocity of 0.01 km/s. If we use a more extreme case, if you had an effective exhaust velocity of 3,500 km/s (the estimated exhaust velocity of a D-D fusion plasma) and the same mass change the Tsiolkovsky equation gives a delta velocity of 0.35 km/s. I suppose a delta-V of 350 m/s isn’t nothing–it will certainly produce a measureable change in the orbital parameters–but it won’t take you to the orbit of Mars, or is it going to cause the Earth to gyrate like a belly dancer.
It’s not that bad as long as you remain in plane (and don’t have to worry about any external influences). If you do, I believe you’ll find that the trajectory describes a section of a logarithmic spiral with the pitch angle controlled by the mass rate of change.
If the spin just keeps going faster and faster, the movement of the body during each cycle will reduce and basically the sphere will just end up spinning really fast, vibrating a little.
I love learning stuff.