# Does an orbiting body accelerate?

Wikipedia says that :

"Transverse acceleration (perpendicular to velocity) causes change in direction. If it is constant in magnitude and changing in direction with the velocity, we get a circular motion"

and this is confirmed in some threads, for example here

"...a force acting perpendicular to the velocity will merely change its direction. In an object moving in a low-eccentricity orbit, the gravitational force is always nearly perpendicular to the velocity, so there isn't a large speed change."

But we also know that a perpendicular force always causes an acceleration according to the rule of addition of forces. If the centripetal force is greater :

the resulting vector is near the perpendicular, if the centripetal force is in perfect balance the resulting vector should point exactly at 45°.

Can you please explain why this doesn't apply to an orbiting body? Can you clearly specify what is the real trajectory of the body in a circular orbit, is it a perfect circle or is it more like a sawtooth motion,

and it moves tangentially and then it is brought back on track by the centripetal force?

Edit:

I actually took the image from the question which is considered a duplicate: it shows vector addition between the tangential and centripetal vector, and shows that, when the latter is greater than v^2/r the body accelerates and the resultant direction is at about 70°. I am taking that as the basis of my own question:

If that answer is correct, and we just change the centripetal acceleration to exactly v^2/r, why shouldn't this scheme be appropriate anymore? If we use the same logic and rules, the body will accelerate less (* 1.41) and the resultant will make an angle of 45° with the tangent vector , but it should still accelerate. Since everybody says it doesn't, why so? what is the difference ?

The sawtooth pattern is, of course, infinitesimally small, but yet it should be there, else the vector would not be tangential and we know it is, since if you cut the string of an orbiting body, it flies off at a tangent. Doesn't that imply the the motion of the body is always in a straight line?

• possible duplicate of How can tangential acceleration from a radial force be explained? – Floris Jun 28 '15 at 5:55
• It will be a circle, not a sawtooth. The key is that these vector diagrams actually describe things that happen in an infinitesimally short time - but making the vectors "finite size" makes it easier to grasp. I think that the earlier question I marked may be a sufficient explanation for you (it even contains the same diagram...) – Floris Jun 28 '15 at 5:56
• I think it's "correct" to think of such an orbit as a "sawtooth", i.e. it goes outside of the orbit, then the centripetal force brings it back into the "circle". Now I use quotes because of course it's not correct--to be correct we need calculus, i.e. infinitesimal steps not finite discrete steps (which are what produce your "sawtooth" pattern). – Jared Jun 28 '15 at 6:09
• Is it an acceleration or not depends from the cause of such a movement. A satellite does not accelerate because it is in equilibration and follows a geodesic path, a body on a rope has a acceleration, then faster one rotate the body, then more he will feel this. – HolgerFiedler Jun 28 '15 at 6:20
• @HolgerFiedler A satellite does not accelerate because it is in equilibration and follows a geodesic path, Watch out now, this is not correct and quite confusing in regard to this question. Though the centripetal acceleration might be negligible in a close-up point-of-view, it is there and it is the cause of the circular path/orbit of the satellite. – Steeven Jun 28 '15 at 11:03

But we also know that a perpendicular force always causes an acceleration according to the rule of addition of forces.

All forces cause acceleration. Perhaps you mean specifically tangential acceleration (changes in speed)?

If the centripetal force is greater the resulting vector is near the perpendicular, if the centripetal force is in perfect balance the resulting vector should point exactly at 45°.

This makes no sense to me. If the centripetal force is greater than what? In an orbit, the centripetal (or near centripetal) force is the only one present. There are no other forces for it to be greater than. In a circular orbit, the acceleration vector points at 90°, not 45°.

Because a body in orbit is affected by a net force (gravity) it must accelerate. In the case of a circular orbit, this acceleration is only a change in direction, not a change in speed. The acceleration vector is perpendicular to the velocity.

I actually took the image from the question which is considered a duplicate: it shows vector addition between the tangential and centripetal vector

Actually, it doesn't. There is only one force, $F$. Rather than adding vectors, it is decomposing a vector into orthogonal components, one tangential and one radial.

and shows that, when the latter is greater than v^2/r the body accelerates and the resultant direction is at about 70°. I am taking that as the basis of my own question:

Again, because there is only one force here, it always accelerates the body. If the radial acceleration is greater than $v^2/r$, then the radius will decrease (get closer to the center). If the radial acceleration is equal to $v^2/r$, then the object will remain at the same radius.

If that answer is correct, and we just change the centripetal acceleration to exactly v^2/r,

This part is unclear. What is the centripetal acceleration beforehand? How would you change the acceleration? Are you adding, moving, or changing forces in some way? What two forces do you think are adding together to make an angle of 45°?

• All forces cause acceleration Not quite; the sentence should rather be "The net force causes acceleration". – Steeven Jun 28 '15 at 9:07
• You missed the point of the question, in the quoted image the acceleration F points at 90°, too, but the resultant vector is much greater than the tangential and points at 70°. WHy, if F = v^2/r should there be no (greater) resultant vector pointing at 45°? – user83553 Jun 28 '15 at 9:16
• @Mark, something is unclear in what you ask. There can only be one resultant force vector. And if the acceleration points at 90 degrees then the resultant force will also point at 90 degrees. This is Newton's second law, $F_{res}=ma$ – Steeven Jun 28 '15 at 9:36
• @Steeven, in the quoted image the resulting vector is F and is pointing at 70°. If the tangential and radial vectors balance if should point at 45% and imply that the tangential vector has increased by 1.41. Is it now clear? Isn't F the actual resulting vector in that image? Why shouldn't the joint action of two normal and equal vectors produce an acceleration? – user83553 Jun 28 '15 at 9:44
• @Mark Why shouldn't the joint action of two normal and equal vectors produce an acceleration of course they should, this acceleration is just not at 90 degrees - it is along with $F$ – Steeven Jun 28 '15 at 9:51

If the centripetal force is greater [than the tangential force] the resulting [force] vector is near the perpendicular [or centripetal] [force], if the centripetal force is in perfect balance [with the tangential force] the resulting [force] vector should point exactly at 45°.

Can you please explain why this doesn't apply to an orbiting body?

Who says this doesn't apply to an orbiting body? It sure does.

If you bring a satellite to space and push it so it moves towards the ground, then gravity is not perpendicular. It is parallel, and there will be no orbit. If you instead push it parallel to the ground, gravity is perpendicular to the motion. Then a circular orbit is initiated. In the ideal case it is all about the initial velocity direction.

Can you clearly specify what is the real trajectory of the body in a circular orbit, is it a perfect circle or is it more like a sawtooth motion

In a circular orbit the orbit is a circle. the sawtooth motion you picture is simply a way to understand how the body moves a bit to the side and a bit inwards - if these bits are negligible, this is a circular motion.

I actually took the image from the question which is considered a duplicate: it shows vector addition between the tangential and centripetal vector, and shows that, when the latter is greater than v^2/r the body accelerates and the resultant direction is at about 70°.

when the latter is greater than v^2/r This is not possible. The centripetal acceleration is given by $a_r=v^2/r$.

If you change $a$ (if the pull is harder e.i.) you also change the orbit radius $r$. This expression will always be.

The sawtooth pattern is, of course, infinitesimally small, but yet it should be there, else the vector would not be tangential and we know it is, since if you cut the string of an orbiting body, it flies off at a tangent. Doesn't that imply the the motion of the body is always in a straight line?

else the vector would not be tangential Remember now that the acceleration is actually only in the resulting direction. This resulting acceleration can then be split into its components, which in this situation is a tangential and a radial (centripetal) acceleration. This is simply a mathematical way to view the directions.

Yes, there is a tangential component, but also a radial component, so the acceleration is not tangential, but along the resulting vector.

If you cut the string, then you suddenly remove all acceleration. Nothing is centripetal, so the objects direction doesn't change anymore (the path is straight), and nothing is tangential, so the object doesn't change its speed either. This is the reason for the straight motion of constant speed a body takes if you cut the string (remove the force) that caused the orbit.

If that answer is correct, and we just change the centripetal acceleration to exactly v^2/r, why shouldn't this scheme be appropriate anymore? If we use the same logic and rules, the body will accelerate less (* 1.41) and the resultant will make an angle of 45° with the tangent vector , but it should still accelerate. Since everybody says it doesn't, why so? what is the difference ?

I am sorry, I don't understand this part of your question. and we just change the centripetal acceleration to exactly v^2/r, but the centripetal acceleration is always equal to $v^2/r$. but it should still accelerate, yes of course, if there is acceleration no-matter the direction it will accelerate. Since everybody says it doesn't, why so? Who says it doesn't?