4
$\begingroup$

The consensus on the internet seems to be that radial and normal burns don't change the total energy of the orbit, since you're thrusting perpendicular to your motion. I'm having trouble squaring that with the following scenario:

Imagine a satellite orbiting a body at 4m/s. It then performs a radial impulse burn of 3m/s. It's final speed is the 4m/s prograde, plus the 3m/s radial = 5m/s. It's speed has increased, and since the burn was instant, it hasn't changed its position. Thus, its gravitational energy (a function of position) is the same, and its kinetic energy (a function of speed) has increased. Therefore, different orbital energy.

Where am I going wrong?

$\endgroup$
2
$\begingroup$

I think you are taking some general ideas that hold reasonably well in most situations and finding they don't apply beyond those areas.

The consensus on the internet seems to be that radial and normal burns don't change the total energy of the orbit, since you're thrusting perpendicular to your motion.

If your burn is both "radial" and "perpendicular to your motion", then your motion must be circular. So a radial burn in a circular orbit does no work. As long as the burn is small compared to the existing velocity of the satellite, the orbit doesn't change much and we can consider it to still be circular. You can burn this way constantly and it won't change the KE.

However if you burn a lot at once, your orbit will change and it will no longer be circular. Radial burns are no longer perpendicular to motion. In such a situation, radial burns will now do work. You can't wave this away just by declaring the burn to be impulsive.

$\endgroup$
  • $\begingroup$ "If you burn a lot at once". Do you mean because the jerk is nonzero, or perhaps because the particle mass is decreasing? Otherwise it's hard for me to see why the orbit would no longer be circular. $\endgroup$ – AGML Sep 30 '16 at 16:50
  • 1
    $\begingroup$ Or do you just mean that the initially radial direction is no longer radial, if you allow the burn to proceed over finite time? $\endgroup$ – AGML Sep 30 '16 at 16:55
  • 2
    $\begingroup$ Yes, its because the first part of the burn causes an acceleration that changes the orbit. The burn after that point has to either turn to remain perpendicular, or it will have a component along the velocity vector so that it does work and changes the KE. $\endgroup$ – BowlOfRed Sep 30 '16 at 17:47
  • $\begingroup$ The distinction between brief and extended burns is (a) almost always left out of the discussion and (b) one that space-flight assumed to be resolved to 'as brief as possible' for a long time because the rockets involved were low-ISP chemical things and you had to do everything as close to minimum energy as possible. However the use of practical ion drives has changed that. $\endgroup$ – dmckee Aug 2 '17 at 22:15
0
$\begingroup$

I think for that consensus you found you should always consider a burn with finite thrust. So each applied change in velocity will take some positive amount of time to achieve, during which the attitude of the craft will continuously be adjusted, such that the thrust always points perpendicular to the current velocity. This basically means that you are only changing the direction the velocity is pointing and not its magnitude (this could still happen during a burn, but happens due to gravity).

$\endgroup$
-1
$\begingroup$

What a radial thrust will not increase is the kinetic energy of the satellite. The work-energy theorem $\delta K~=~\delta W$ $=~\vec F\cdot d\vec r$ does indicate that if the displacement and force are perpendicular there is no work. Clearly since $d\vec r~=~\vec vdt$ a radial thrust will not increase the kinetic energy.

A radial thrust will increase the potential energy. This will mean the total energy will increase, even though the velocity of the satellite $v~=~\sqrt{GM/r}$ actually decreases with increased radius.

$\endgroup$
  • $\begingroup$ If there is a force in a particular direction, the movement of the body will be in that direction. That's just Newton's law of motion. Your equation about F.dr doesn't make any sense. There is no possible way that any thrust or other force can increase the gravitational potential energy directly. That also makes absolutely no sense. The potential energy can only increase if the satellite moves higher, which means that some kinetic energy is converted to potential energy. $\endgroup$ – Suzu Hirose Sep 30 '16 at 2:15
  • 2
    $\begingroup$ There is no possible way a thrust or force can change gravitational potential? So an elevator lifting a mass does not change its gravitational potential --- really? $\endgroup$ – Lawrence B. Crowell Sep 30 '16 at 9:57
  • $\begingroup$ An elevator lifting a mass first has to move it from a lower to a higher position. If it's moving then it has gained kinetic energy, which then turns to potential energy. This is very basic physics indeed. $\endgroup$ – Suzu Hirose Sep 30 '16 at 10:38
  • 1
    $\begingroup$ An object being lifted by an elevator moving at a constant velocity has constant kinetic energy. $\endgroup$ – Lawrence B. Crowell Sep 30 '16 at 15:35
  • $\begingroup$ An object being lifted by an elevator has zero kinetic energy before it started moving. This site is rather entertaining for the lack of knowledge of very basic things and attempt to blindside people with specious claims. $\endgroup$ – Suzu Hirose Oct 1 '16 at 0:14
-5
$\begingroup$

If you increase the speed of the object, or otherwise change its velocity, its trajectory will change, so its orbit will change, and its kinetic energy will increase.

The consensus on the internet seems to be that radial and normal burns don't change the total energy of the orbit, since you're thrusting perpendicular to your motion.

That's hocus-pocus.

$\endgroup$
  • 3
    $\begingroup$ "Increase the speed" is not the same as "change its velocity". The velocity can change without the speed increasing. KE does not increase when the acceleration does not increase the speed. $\endgroup$ – BowlOfRed Sep 30 '16 at 5:12
  • $\begingroup$ A force perpendicular to the direction of motion of course does no work because it does not change the speed. KE depends only on the speed, not the velocity as a vector. $\endgroup$ – AGML Sep 30 '16 at 16:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.