[thedoc (OP)]

Leslie Wolf  asked the Naked Scientists:

My question is about orbiting bodies, and Earth's orbit in particular.

Is this statement correct: All orbiting bodies in the vacuum of space are perpetually free-falling faster and faster while encircling the larger body, but maintaining a constant distance from the larger body, because of their sideways speed, which, together with gravity, creates the circular orbit.

Many thanks for your answer,
Leslie Wolf.

What do you think?

[evan_au]

Is this statement correct: All orbiting bodies in the vacuum of space are perpetually free-falling faster and faster while encircling the larger body, but maintaining a constant distance from the larger body, because of their sideways speed, which, together with gravity, creates the circular orbit.

I can modify it slightly to become a more accurate statement:

orbiting bodies in the vacuum of space are perpetually free-falling while encircling the larger body, but maintaining a constant distance from the larger body, because of their constant sideways speed, which, together with gravity, creates the circular orbit.

Where it went off track:

• all orbiting bodies: Most orbiting bodies are not in circular orbits (with the larger body at the center), but an ellipse (a squashed circle), with the larger body at one of the two focal points of the ellipse. Even if you launched a satellite on a circular orbit around the Sun, it would not stay that way for very long, because the tug of other planets in the solar system would distort it into an ellipse.
  
• free-falling faster and faster: If you dropped an object off a high building (in a vacuum), the longer it fell, the faster it would go. In this case, it is free-falling faster and faster. But this is a poor model for an an object in a circular orbit - for one thing, as it falls, it is getting closer to the Earth.
  
• faster and faster: An object in a circular orbit maintains a constant speed as it progresses around its circular orbit.
  
• faster and faster: Perhaps recently you saw an animation of two black holes colliding? In this case, they did go faster and faster as they spiraled inwards, but this only occurs in extremely strong gravitational fields, and does not apply to any detectable amount in the Solar system (or anywhere else I would like to be). But this is not a circular orbit - it is a spiral.

[rmolnav]

Strange enough, there is another site with this question (even with audio podcast)
https://www.thenakedscientists.com/articles/questions/how-do-objects-orbit-space
I consider it has a basic error. I tried to refute it sending a post, but apparently unsuccessfully.
It was as follows.
They say:
"David - If you fall sunwards, you start traveling faster and therefore, you can find yourself traveling fast enough to stay in orbit
Andrew - So it self-corrects it?
David - I think so, yes.".
Sorry , but that is erroneous. If you just start falling sunwards, you certainly start traveling faster, but what increases is the radial component of velocity vector (previously null), not the tangential component ...
Then gravity increases (proportionally to the dicrease of the square of the distance), getting bigger than the required centripetal force for a rotation at that distance (tangential v to the square/r, multiplied by your mass), which also increases but proportionally only to the decrease of the distance. That "excess" of sunwards force will make radial sunwards velocity component increase more and more ...
"You" would continue falling towards the sun, eventually into the sun... You´d better not try !!"

[Janus]

Sorry , but that is erroneous. If you just start falling sunwards, you certainly start traveling faster, but what increases is the radial component of velocity vector (previously null), not the tangential component ...

After reading the link, it is apparent that they are talking about an object already in orbit with an existing tangential velocity.  In this case, if the body were to begin to fall inward towards the Sun (say to some brief inward acting force), then as its distance to the sun decreased, its tangential speed must increase.   The angular momentum must remain conserved, and thus as the radial vector of the trajectory decreases, the tangential velocity must increase to compensate.

For example, let's say that the object is at Mars' orbital distance of 227.94 million km, and in a circular orbit with a velocity of 24.13 km sec.   It has an 5 km/sec delta V added in the direction towards the Sun.  This gives its a new velocity of 24.64 km at a angle of 11.7 degrees to its original path. This will actually raise the average distance( the semi major axis) of the orbit from the Sun to 238 million km (it may seem odd that an inward push would cause a net increase in orbit, but orbital mechanics can be funny that way. In this case, the added inward delta v increases the total velocity and KE of the body, which in turn increases its  total orbital energy)
 It's Areal velocity( the area its radial line sweeps out per sec) will remain constant at 2.75e 15 m^2/sec. 
Its new orbital period will be 732.85 days (compared to its old period of 686.98 days) from this and the areal velocity, we can work out the eccentricity of the new orbit, because the Areal velocity also is equal to


where a is the semi major axis
and P is the period.

This gives an eccentricity of 0.206 which results in a perihelion of 189 million km.  At which time its orbital velocity, all of which will be tangential, will be 29.1 km/sec.

You can work out the tangential velocity at any distance from the Sun with

Where A is the constant Areal velocity, r is the radial distance and V_t the tangential velocity, Which shows that V_t increases as r decreases for the same free falling object. The only time this would not happen is if the tangential velocity is zero to start with.

The radial velocity can be found by the vector difference between the total velocity at r found by

where k is the Gravitational parameter for the Sun (Gm)
and the tangential velocity.

What you will find is that the radial velocity will be zero at aphelion, will increase to a maximum, then will start to decrease again until it is once more zero at perihelion.  The tangential velocity will vary between a  minimum at aphelion and  a maximum at perihelion.

[PmbPhy]

Leslie Wolf  asked the Naked Scientists:

My question is about orbiting bodies, and Earth's orbit in particular.

Is this statement correct: All orbiting bodies in the vacuum of space are perpetually free-falling faster and faster while encircling the larger body, but maintaining a constant distance from the larger body, because of their sideways speed, which, together with gravity, creates the circular orbit.

Many thanks for your answer,
Leslie Wolf.

What do you think?

I'm not certain if it was Newton who came up with this idea but he's certainly well-known for it. Newton used a thought experiment to explain how a satellite orbits a gravitating body. See:

https://en.wikipedia.org/wiki/Newton%27s_cannonball

[rmolnav]

In this case, if the body were to begin to fall inward towards the Sun (say to some brief inward acting force), then as its distance to the sun decreased, its tangential speed must increase.   The angular momentum must remain conserved, and thus as the radial vector of the trajectory decreases, the tangential velocity must increase to compensate.

Thank you.
I must say I was considering a circular orbit, to simplify. In that case, an inward force toward the sun only could cause an inward acceleration (Newton´s 2nd L.), and tangential velocity could not increase (Newton´s 1st Law).
Keep in mind that Newton´s Laws are in the root of conservation of momentum laws (both linear and angular). And if either of the later seems to contradict the former, we should check our work !!
You speak in terms of angular momentum, which is a concept especially aimed to cases of spinning solids, where particles have a game of different velocities. It is a kind of tool (momentum of inertia concept included) to simplify calculations, that could also be done with linear momentum conservation law in simple cases as ours.
Being the hole mass of rotating object considered to be in a point, its C.G., tangential component of linear momentum has to keep equal to mv unless an impulse (force with not null tangential component, multiplied by time) is given to the object,  … And that is not our case.
I confess not to have carefully gone through your numbers, but there must be an error somewhere.
With elliptical orbit, as far as I can see, it would depend on the orbit point where changes started. Along two fourth of the orbit the tangential component of a force towards the sun would be opposite to initial tangential speed. Then there would be even a tangential deceleration. The opposite along the two other fourths.

[rmolnav]

Is this statement correct: All orbiting bodies in the vacuum of space are perpetually free-falling faster and faster while encircling the larger body, but maintaining a constant distance from the larger body, because of their sideways speed, which, together with gravity, creates the circular orbit.

I find it correct, BUT I wouldn´t say "... are perpetually free-falling FASTER AND FASTER while encircling the larger body" (already said by evan_au)
Being constant centripetal acceleration perpendicular to speed vector, it "changes and changes" speed vector direction, but that doesn´t mean f. & f.
PmbPhy and Leslie Wolf and evan_au: please kindly note that in my first post I didn´t refer to what said in this thread, but what in the linked one,  "twin" headed to this ... (?)

[rmolnav]

In this case, if the body were to begin to fall inward towards the Sun (say to some brief inward acting force), then as its distance to the sun decreased, its tangential speed must increase.   The angular momentum must remain conserved, and thus as the radial vector of the trajectory decreases, the tangential velocity must increase to compensate.

After further ruminating on the issue, I´ve found where the error is.
In terms of angular momentum, if the radial vector of the trajectory
decreases, moment of inertia also decreases but to the square of the former.
But what has to increase to compensate is the ANGULAR speed, NOT the tangential velocity (ang. mom.=mom. of Inertia multiplied by angular speed).
On the one hand, angular speed would have to increase proportionally to the decrease of momentum of inertia, that is to the square of the decrease of the radius …
On the other, a smaller radius would reduce tangential speed in same proportion (for a given angular speed).
Bottom line: tangential speed would not increase due to that given inward impulse.
But that impulse would give some sunward velocity, apart from an increase of sun´s gravity attraction … 

[rmolnav]

On the other, a smaller radius would reduce tangential speed in same proportion (for a given angular speed).

Sorry ... Afterwards I´ve realized that is wrong! For a given angular speed, tangential speed decreases proportionally to the radio decrease, NOT to the square ...
More ruminating necessary !!

[Janus]

In this case, if the body were to begin to fall inward towards the Sun (say to some brief inward acting force), then as its distance to the sun decreased, its tangential speed must increase.   The angular momentum must remain conserved, and thus as the radial vector of the trajectory decreases, the tangential velocity must increase to compensate.

Thank you.
I must say I was considering a circular orbit, to simplify. In that case, an inward force toward the sun only could cause an inward acceleration (Newton´s 2nd L.), and tangential velocity could not increase (Newton´s 1st Law).

And I was considering an body that begins in a circular orbit and is momentarily acted on by an additional inward force.  As long as this momentary force acts more or less instantaneously,  The object will have initially gained an inward component of velocity with out changing its tangential velocity, but from that moment on, and as it continues inward, its tangential velocity will increase.

Keep in mind that Newton´s Laws are in the root of conservation of momentum laws (both linear and angular). And if either of the later seems to contradict the former, we should check our work !!

Nothing I said is in contradiction with Newton's laws, it all is derived from Newton's work.

You speak in terms of angular momentum, which is a concept especially aimed to cases of spinning solids, where particles have a game of different velocities. It is a kind of tool (momentum of inertia concept included) to simplify calculations, that could also be done with linear momentum conservation law in simple cases as ours.
Being the hole mass of rotating object considered to be in a point, its C.G., tangential component of linear momentum has to keep equal to mv unless an impulse (force with not null tangential component, multiplied by time) is given to the object,  … And that is not our case.
Angular momentum can be assigned to any object.  All that is needed is a point that the angular momentum is measured relative to.( even an object moving in a straight line can have an angular momentum relative to any chosen point.)  In the case of an orbiting object this point is the focus of the orbit.  In reality, the conservation of angular momentum of an orbit is just another way of expressing Kepler's 2nd law.  The equation I gave for Areal velocity is the mathematical expression of that law.
I confess not to have carefully gone through your numbers, but there must be an error somewhere.

While it is possible that I could have made an arithmetic error somewhere along the line, the math is good, as is the general conclusions. The point is that once the initial additional influence has acted and an inward velocity component has been added, the object will enter a new elliptical orbit, over which its tangential speed will vary. It will not continue to make increasingly smaller orbits.

With elliptical orbit, as far as I can see, it would depend on the orbit point where changes started. Along two fourth of the orbit the tangential component of a force towards the sun would be opposite to initial tangential speed. Then there would be even a tangential deceleration. The opposite along the two other fourths.

At what point the inward force acts during an elliptical orbit, will determine the exact nature of the new orbit, but unless it is extreme enough to result in a new orbit which has a periapis smaller than the radius of the body it orbits(in which case it will collide with that body some fraction of an orbit later), the result will be a stable orbit.

And while the last analysis dealt with a single brief force, it also can hold for an continued increase in inward working force.
Consider our earlier object stating in a Mar's distance circular orbit.
The orbital speed can be found by the general equation.

k is the gravitational parameter or GM for the body it is orbiting. In this case, the Sun.
r is its present orbital distance
a is it average orbital distance
For a circular orbit a=r, so it simplifies to

Now lets assume that we suddenly double the gravity of the Sun and hold it at this doubled magnitude.  This, in effect make k =2k in the first equation.  r remains the same, as does the present velocity of the object.
By plugging in these into the first equation we can solve for a, the new average distance for the orbit.

r = 227.94 million km = 2.2794e11 m
v=24.13 km/sec = 24130 m/s
k =1.3275e20 m^3/s^2
Which means that the average distance for the new orbit after gravity doubles will be:
151.9 million km.
The present distance of the object is the aphelion of the new orbit.  The perihelion (closest approach to the Sun) will be 75.86 million km (between the orbits of Mercury and Venus). 
Plugging this into the first equation along with the new average distance for the orbit a( while remembering to make k=2k), gives an orbital velocity of 72.47 km/sec  Since the orbital velocity at both perihelion and aphelion is totally tangential, the tangential velocity of the object will vary from 24.13 to 72.47 after we doubled the gravity.

[rmolnav]

As long as this momentary force acts more or less instantaneously,  The object will have initially gained an inward component of velocity with out changing its tangential velocity, but from that moment on, and as it continues inward, its tangential velocity will increase.

Nothing I said is in contradiction with Newton's laws, it all is derived from Newton's work.

Thank you.
I hope now i´ve really found the error … But not in what you say: the error was mine !
Basic laws I was referring to are about vectors (force, acceleration, speed, linear momentum, impulse …). We can applied them separately to their components along perpendicular axis, typical x,y axis of coordinates (if within a plane).
But to apply them to components such as radial (circular movement for simplification) and tangential is erroneous, because their actual directions vary with time …
E.g., in our case the initially given (with the short impulse) inward radial velocity vector, time afterwards will have to be added to previous (when no impulse given) velocity in that moment, which was only tangential. But those vectors won´t be perpendicular to each other any more: given inward velocity will have a tangential component, that added to previous one causes its increase …
And neither in terms of total energy the increase of velocity contradict laws : apart from the energy added with the impulse, potential energy decreases with radius, and that can compensate the kinetic energy increase.
 

[Janus]

As long as this momentary force acts more or less instantaneously,  The object will have initially gained an inward component of velocity with out changing its tangential velocity, but from that moment on, and as it continues inward, its tangential velocity will increase.

Nothing I said is in contradiction with Newton's laws, it all is derived from Newton's work.

Thank you.
I hope now i´ve really found the error … But not in what you say: the error was mine !
Basic laws I was referring to are about vectors (force, acceleration, speed, linear momentum, impulse …). We can applied them separately to their components along perpendicular axis, typical x,y axis of coordinates (if within a plane).
But to apply them to components such as radial (circular movement for simplification) and tangential is erroneous, because their actual directions vary with time …
E.g., in our case the initially given (with the short impulse) inward radial velocity vector, time afterwards will have to be added to previous (when no impulse given) velocity in that moment, which was only tangential. But those vectors won´t be perpendicular to each other any more: given inward velocity will have a tangential component, that added to previous one causes its increase …
And neither in terms of total energy the increase of velocity contradict laws : apart from the energy added with the impulse, potential energy decreases with radius, and that can compensate the kinetic energy increase.
 

If your interested, here's a link to Newton's proof of Kepler's 2nd Law:
https://www.math.ku.edu/~lerner/m500f09/NewtonKepler.pdf

[rmolnav]

Newton used a thought experiment to explain how a satellite orbits a gravitating body. See:
https://en.wikipedia.org/wiki/Newton%27s_cannonball

Thank you. Very interesting. In relation with what said in the wikipedia linked page:
"If the speed is the orbital speed at that altitude (at the top of a very high mountain), it will go on circling around the Earth along a fixed circular orbit, just like the Moon. (C) for example horizontal speed of at approximately 7,300 m/s for Earth” (huge speed!!)
In another site I used that case of some object rotating around Earth near its surface. We were discussing the question “Would we weigh less at the equator?”
Afterwards I said that perhaps the question itself was not sufficiently precise, because one thing is the weight a bath scale shows (commonly called “our weight”), and another the total attraction exerted on our body by Earth …
Perhaps it could interest you. In my example I said:
Let us imagine Earth not rotating, and a kind of 40,000 km highway along the equator, with neither atmospheric air friction nor surface friction (for simplification). And with no solid Earth obstacles.
An object could be rotating (let us choose West to East) as a satellite following the equator, as long as it had sufficient angular speed.
If, starting from that scene, angular rotation decreased more and more, required centripetal acceleration would diminish (to the square of angular speed), with unchanged gravity force. Satellite type motion would end, and the excess of gravity attraction would have to be compensated by ground upward push (Newton´s 2nd Law). Small at the beginning, and increasing as long as angular speed continued decreasing.
When angular speed were 2π radians a day, we would be in the real situation of people standing “still" at the equator. And that ground upward push would be what gauged with a bath scale if situated under our feet, what we commonly call “our weight”. Because the real case with Earth rotating doesn´t change physically anything, just friction becomes actually zero even with atmosphere (no relative movement).
But at poles, even if we were at same distance from Earth´s C.G., no part of gravity attraction would have to produce any centripetal acceleration, because we were at zero distance from axis of rotation …
So, our weight there would be bigger than at the equator. As I say, even if our planet were a perfect sphere …

[PmbPhy]

In another site I used that case of some object rotating around Earth near its surface. We were discussing the question “Would we weigh less at the equator?”

Note: The correct term for that is revolving not rotating. Unfortunately dictionaries confuse these two as meaning the same thing. (sigh)

Afterwards I said that perhaps the question itself was not sufficiently precise, because one thing is the weight a bath scale shows (commonly called “our weight”), and another the total attraction exerted on our body by Earth …

A bath scale will measure the total inertial force acting on the body on the scale. The gravitational force is also an inertial force (do you know how this term is defined?). That fact is how general relativity got started since Einstein argued that the gravitational force is actually an inertial force.
Perhaps it could interest you. In my example I said:

When angular speed were 2π radians a day, we would be in the real situation of people standing “still" at the equator.

We have satellites in orbit around the earth for which that's true. The orbit is referred to as geosynchronous orbit.

But at poles, even if we were at same distance from Earth´s C.G., no part of gravity attraction would have to produce any centripetal acceleration, ..

I'm not sure what you mean. Do you know the difference between centripetal acceleration and centrifugal acceleration?

So, our weight there would be bigger than at the equator.

Yup. In fact there's an article by a well known physicist in the American Journal of Physics who defines weight in a similar fashion, i.e. the force required to hold a body at a specific position, e.g. located on the bathroom floor in that frame of reference.

Its a given that weight varies with latitude. I think its been measured in fact.

[Bill S]

If I remember correctly: you weigh about 0.5% more at the poles than at the equator.  Not an enormous bonus for dieters.

[rmolnav]

If I remember correctly: you weigh about 0.5% more at the poles than at the equator.  Not an enormous bonus for dieters

Its a given that weight varies with latitude. I think its been measured in fact.

For the moment, just a short post about that ... It will take me much more time to deal with other PmbPhy´s questions and comments.
Even in the original article I refuted, that was given as a fact ... But they sad the unique reason of that was that distance to Earth´s C.G. is bigger at the equator ...
And that is only one of the reasons, we could say a static (o geometric) reason ...  The other is dynamic, the fact that it is a quasi-sphere revolving, as I commented in my last post, and parallel diameters decrease from equator to poles.

[rmolnav]

A bath scale will measure the total inertial force acting on the body on the scale. The gravitational force is also an inertial force (do you know how this term is defined?). That fact is how general relativity got started since Einstein argued that the gravitational force is actually an inertial force.

Thank you.
I´m arguing within Newton´s Mechanics, considered sufficient for our cases, whatever the deep explanation of gravity and inertia could be.
So, I consider
"An inertial force is a force that resists a change in velocity of an object. It is equal to—and in the opposite direction of—an applied force, as well as a resistive force. The concept is based on Newton's Laws of Motion, including the Law of Inertia and the Action-Reaction Law”

"But at poles, even if we were at same distance from Earth´s C.G., no part of gravity attraction would have to produce any centripetal acceleration", ..
I'm not sure what you mean. Do you know the difference between centripetal acceleration and centrifugal acceleration?

Don´t you realize we are discussing on centrifugal force in another thread?:
https://www.thenakedscientists.com/forum/index.php?topic=68025.new#new
I won´t enter now that discussion field. I´ll just try to put clearer what said about  "at poles, .. no part of gravity attraction would have to produce any centripetal acceleration”
Not considering gravity pull of moon and sun on us, the unique forces exerted on each of us are Earth´s attraction, and upward push of ground under our feet, which is avoiding our fall with “g” acceleration.
If Earth (and we with it) were not revolving, we would have no acceleration at all. Total forces exerted on each of us would have to be null: upward pull exerted by the ground would have to be equal to Earth´s attraction (Newton´s 1st Law). If a bath scale between ground and our feet, it would show attraction on us exerted by Earth.
But as Earth is revolving, the ground is avoiding our fall with “g” linear acceleration, but we are actually “falling” with the centripetal acceleration there, due to our “rotation”. But that acceleration, added to our actual speed vector, “only” makes us rotate.
According to Newton´s 2nd Law, upward force on us (ground push) has to be smaller than downward one (gravity attraction exerted by Earth), being that difference equal to our mass times our centripetal acceleration.
So, a portion of total gravity attraction is causing our centripetal acceleration, and the rest is what is compressing our feet against the ground, and any possible object in between such us a bath scale.
As required centripetal acceleration diminishes from equator to poles, those compressing forces do the opposite.
I suppose you will bring up the question of inertial reference frames to explain that: it is true that the scale is also rotating …
But as I told you in the linked page, I consider that idea is quite correct when dealing with velocities and accelerations, but not always if dealing with forces … Legs and feet of an astronaut trying to stand in the ISS would experience even no compression at all, due to exactly same reasons given above, in this case being necessary the total gravity attraction to cause the necessary centripetal acceleration.
When back on Earth, if he had not made special exercises, tell him he has actually no bone problems, that it was just something only “apparent”, due to the reference frame … !! 
 

[rmolnav]

According to Newton´s 2nd Law, upward force on us (ground push) has to be smaller than downward one (gravity attraction exerted by Earth), being that difference equal to our mass times our centripetal acceleration.

Just a detail to keep in mind: all those forces are vectors ... At the equator all have the direction from the place we are to Earth´s C.G. (vertical), but at higher latitudes not, because required centripetal acceleration is perpendicular to the axis of rotation ...
That requires the introduction of angular details at each latitude, but it doesn´t change the bottom line: the increase of our "weight" from equator to poles.
Another curious thing: due to that fact, local vertical lines vary (though very, very slightly) with latitude, in the sense that they only go exactly in the direction to Earth´s C.G. at the equator and poles.