Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have this dilemma: Suppose you have a space ship somewhere in deep space, where there is no drag force or substantial gravity. If the ship has a single engine situated in such a way that the center of thrust is off-axis with respect to the center of mass, will the ship fly straight when the engine is turned on or will it begin to rotate?

I firstly thought that it would fly straight, since there is no drag or weight to pull the ship on other axis, but I found this question Force applied off center on an object which suggest that I'm wrong, although I don't fully understand the answers there.

Can anyone explain or provide links to material that could shed some light for me in the physics behind this?

share|improve this question
I think you're looking for an explanation of torque. That wikipedia page (especially towards the end) has lots of relevant material. You could also have a look at this page. –  Kitchi Mar 2 '13 at 11:33
Short answer: yes, it would begin to rotate. –  Michael Brown Mar 2 '13 at 12:12
Ok, I found out from wiki that A force F applied to a rigid body at a distance d from the center of mass has the same effect as the same force applied directly to the center of mass and a couple Cℓ = Fd. That's regardless of any weight or friction. Is this because in one semispace around the point at which the force is applied there is more mass, and thus more inertia than in the other? –  user62966 Mar 2 '13 at 12:21
@user62966, Yes those two are equipollent forces –  ja72 Jun 1 '13 at 16:29

2 Answers 2

Here is a brief analysis of the problem which involves:

i) Momentum and the concept of impulse

ii) Analysis of a force into two components

iii) Moment of a force

iv) Moment of inertia and the parallel axis theorem

For the sake of simplicity let us assume we have a spherical spaceship of radius R in the outer space. Let there be a rocket at the surface of the spaceship pointing in a direction that does not pass through the centre of the sphere. Imagine now we are firing the rocket.

The impulse will generate a force, $F$, in the usual way and push the spaceship in the direction of the rocket. The force exerted on the sphere in that direction can be analysed into the tangent and the perpendicular to the surface of the sphere. If $\theta$ is the angle between the rocket direction and the normal to the sphere we have:

Tangent component: $F_T=F\sin(\theta)$

Normal component: $F_N=F\cos(\theta)$.

The normal component is parallel to the radius of the sphere and passes through the centre (CM) and has no moment with respect to the centre. This component will push the sphere in the normal direction.

The tangent component has a moment with respect to the centre


This component would rotate the sphere, should the axis of the sphere be pivoted, but it is not! However, due to the inertia of the mass of the sphere, it would be sufficient to give a pivotal leverage for the tangent force to rotate the sphere. The law of conservation of energy must be written, for a short time interval of application of the force, as it moves the spaceship by a displacement ${\bf x}$, in the form

${\bf F.x}=\frac {1}{2}mv^2+ {\frac {1}{2}I{\omega}^2}$

The first term on the RHS is the kinetic energy due to the linear motion, and the second is the kinetic energy due to the rotational motion.

The spaceship will rotate about the centre of mass of the spaceship, and the axis of rotation will be perpendicular to the plane made between the tangent force and the radius of the sphere. The axis will pass through the centre of the sphere because, according the parallel axis theorem, the moment of inertial is minimum when the axis of rotation passes through the centre of mass. Hence the spaceship will have the minimum energy, and this is the preferred energy state for the spaceship.

share|improve this answer

Here are the applicable laws (Newton-Euler equations)

$$ \sum \vec{F} = m \,\vec{a}_G $$ $$ \sum \vec{M}_G = I \dot{\vec{\omega}} + \vec{\omega}\times I \omega $$

where the total torque applied is $\vec M_G = \vec{r}_G \times \vec{F} $ and $\vec{r}_G$ is the vector of the point of force $\vec{F}$ application relative to the center of mass. $\vec{a}_G$ is the linear acceleration vector at the center of gravity, and $\dot{\vec{\omega}}$ is the angular acceleration.

So if we start with a co-moving inertial frame these come down to

$$ \vec{a}_G = \frac{\vec{F}}{m} $$ $$ \dot{\vec{\omega}} = \frac{\vec{r}_G \times \vec{F}}{I} $$

and when $\vec{r}_G \neq 0$ then there is indeed angular acceleration.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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