If an object falls without friction in a gravitational field, is the average SPEED independent of the path taken?
Consider a marble falling 1 metre in a vacuum under the influence of the Earth's gravity.
Will it have the same average speed (scalar) if a frictionless track causes it to take a less direct path to the same destination?
No. On a frictionless track the average speed definitely depends on the path. The final speed is independent of the path, but not the average speed.
Consider two paths composed of one shallow section and one steep section. In one path it is shallow first and then steep, and the other path is steep first and then shallow. The shallow first path will traverse the shallow section at a low speed, so it will take quite a while, then on the steep section it will briefly get to a high speed. The steep first path will immediately get to a high speed, then it will traverse the shallow section at that high speed. So the average speed will be higher on the steep-first path than on the shallow-first path.
The average speed is not independent of the path taken. In fact, since you have not imposed any restrictions on the track, it can, in theory, be made to take any value between $0$ and $\sqrt{2gh}$ where $h$ is the height of the starting point relative to the ending point.
Consider a (smooth curve) frictionless track that starts with a downhill section of height $h_1$ ($0 \lt h_1 \lt h$), followed by a section of length $L$ at constant height, and finally by another downhill section of height $h-h_1$ to the end point. The time taken for the downhill sections does not depend on $L$, but the time taken on the level section is proportional to $L$ and can therefore be made arbitrarily long by taking arbitrarily large $L$. This means that the average speed approaches $\sqrt{2gh_1}$ as $L\to\infty$.
If the track is allowed to go below the ending point, then $h_1$ can be chosen to be larger than $h$, turning the final downhill into an uphill. In such cases, the average speed can be made even larger than $\sqrt{2gh}$.
Even if the total length of the track is fixed, the average speed can still be changed by fixing $L$ and changing $h_1$.
There is a famous problem called the brachistochrone which asks what is the fastest descent for a similarly frictionless object with a horizontal displacement, and then answers it using graduate level math. Since this problem exists, I should think the answer to your question is obvious.
'Average speed' is a fairly meaningless term. If one doesn't understand calculus, one might naively assume that adding the start and finish speeds and then dividing by 2 would give you an answer, and it would, but it would be a fairly useless answer because the tortoise would have lost to the sleeping hare despite finishing first. The only scenario where this version of average speed has any validity is where the speed changes at a constant rate (remember, $0$ is also a constant). The other, and more correct algebraic approach would be to measure the distance and the time it takes to traverse that distance and then divide. Here, however, we run into a problem - there are no extant frictionless tracks meaning we need to use calculus to find an arc length.
If we allow ourselves to use calculus concepts (without pesky integrals), we can take a time average ($\frac{\vec x_n - \vec x_{n-1}}{\Delta t}$) as many times as it takes to accurately describe the speed along the entire path. We could also define average speed as an average over distance, but I think most high schooler would balk at that.
When people say "average speed", they almost always mean "time-averaged speed". This quantity does depend on the path taken between two fixed points, as the other answers discuss.
But if we instead consider the (much less common!) concept of the height-averaged speed, then this quantity is indeed the same for all (non-strictly dropping) paths that begin and end at the same heights and have the same initial speed $v_0$. This path-independent quantity equals $$\frac{(2g\, \Delta h + v_0^2)^{3/2} - v_0^3}{3 g\, \Delta h},$$ where $\Delta h$ is the difference between the initial and final heights. If the object is released from rest ($v_0 = 0$), then this path-independent average speed simplifies to $\frac{2}{3} \sqrt{2 g\, \Delta h}$. (This formula also holds for paths that rise along certain segments - but only if you negatively weight those rising segments, which I admit is not entirely natural to do.)
If the object orbits in a circle with constant velocity the average speed is the instantaneous speed $\rm v=\sqrt{GM/r}$. If it falls to the ground the average velocity will depend on the initial velocity and the height of the ground and will in most cases not equal the orbital velocity.
If the initial velocity has to be zero there is only one path and therefore no other path to compare that to, so the answer to your question is no.
The increase in kinetic energy $\rm E_k=m v^2/2$ will equal the difference in potential energy $\rm E_p=-GMm/r$ though, so if you were asking about that instead about the average velocity the answer would have been yes.
Yes by the law of energy conservation, the speed at the end is the same on every path if you have no friction, but not the average speed .
I think it's good to see an explicit counterexample here. Suppose we construct a first section of the track $AB$ with base angle $30^{\circ}$ as shown:
Let $AB=1$ and release the mass from rest at $A$. Setting $g=1$ to make computations easier, we know that on this section:
$$a_{AB} = g\sin(30^{\circ}) = \frac{1}{2}, \qquad v_{AB}=\frac{1}{2}t, \qquad x_{AB} = \frac{1}{4}t^2, $$
now since $AB=1$, we get $t_{AB}=2$.
Now, from $B$ onwards the mass will be falling freely, but it has some initial horizontal and vertical velocity components due to the motion on $AB$. They are given by:
$$ v_{_{xB}} = a_{AB}t_{AB}\cos(30^{\circ}) = \frac{1}{2}\times 2\times\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2},$$ $$ v_{_{yB}} = \frac{1}{2}. $$
now, let the mass fall for an additional $1$ units of vertical distance. This uniquely sets the location of point $C$ by demanding $h_2=1$. So we get:
$$ h_2 = v_{_{yB}}t_{BC}+\frac{1}{2}t_{BC}^2$$ $$ \frac{1}{2}t_{BC}^2+\frac{1}{2}t_{BC}-1=0 \Rightarrow t_{BC} = 1.$$
This in turn also gives us $d_2$, the horizontal distance traversed from $B$ to $C$. That is:
$$ d_2 = v_{_{xB}}t_{BC} = \frac{\sqrt{3}}{2} $$
Now finally we must find the arc length on the parabola going from $B$ to $C$. This isn't a problem, we know that:
$$ s_{BC} = \int_{B}^{C} \sqrt{1+\dot{y}^2(t)}\ \mathrm{d}t $$
where on the path from $B$ to $C$ we have $y(t) = \frac{1}{2}t^2 + \frac{1}{2}t$ as we already found, so $\dot{y} = t+\frac{1}{2}$. Hence:
$$ s_{BC} = \int_{0}^{t_{BC}} \sqrt{5/4+t^2+t}\ \mathrm{d}t \approx 1.429 $$
So we have what we need now to find the average velocity on our first path:
$$ \bar{v}_{ABC} = \frac{AB+s_{BC}}{t_{AB}+t_{BC}}$$ $$\boxed{ \bar{v}_{ABC} = \frac{1+1.429}{2+1} \approx 0.81 }$$
now we construct the second path by imagining we get rid of the track segment AB and instead put a direct one from A to C, as indicated by the purple line segment in the diagram. We need to find the base angle $\theta$ this line makes with the horizontal to carry out this computation. This is simply:
$$ \theta = \arctan(AD/CD) = \arctan\left(\frac{h_1+h_2}{d_1+d_2}\right) = \arctan\left(\frac{0.5+1}{\sqrt{3}/2+\sqrt{3}/2}\right) \approx 40.89^{\circ} $$
So now, proceeding as before:
$$ a_{AC} = \sin(40.89^{\circ}) \approx 0.654 $$ $$ v_{AC} \approx 0.654t, \qquad x_{AC}=\frac{1}{2}0.654t^2$$
Now we need to find $AC$ from what we have:
$$ AC = \sqrt{(d_1+d_2)^2+(h_1+h_2)^2} \approx 2.291 $$ $$ \Rightarrow t_{AC} = \sqrt{\frac{2AC}{0.654}}\approx 2.64$$
So finally we compute:
$$ \boxed{\bar{v}_{AC} = \frac{AC}{t_{AC}} \approx 0.865} $$
we see clearly that $\bar{v}_{AC} \neq \bar{v}_{ABC}$. Also note that the free fall section in $BC$ isn't really a problem if we want to think only about actual track segments: we could just as well imagine an actual track lying along the parabola from $B$ to $C$, that just "happens" to coincide with the path taken at free-fall. Since the track is ideally smooth anyway by assumption, this will have no effect on the overall motion in this particular case.
Since there are plenty of places to go wrong in a numeric computation like this, it would be nice to do one "sanity check" which is that the speed (instanteneous now, not average) at $C$ must be the same in both cases. This is due to energy conservation. We expect in fact that:
$$ \frac{1}{2}mv_{C}^2 = mg(h_1+h_2) \Rightarrow v_{C} = \sqrt{2g(h_1+h_2)}$$ and plugging in our numbers (and setting $g=1$ again): $$ v_{C{\text{ (expected)}}} = \sqrt{2(0.5+1)}=\sqrt{3} $$
on the first path, after segment $AB$ we have, as already computed: $$ v_{_{xB}} = \frac{\sqrt{3}}{2}, \quad v_{_{yB}} = \frac{1}{2}, $$ from $B$ to $C$ the horizontal component $v_{_{xB}}$ stays constant, while the vertical one increases by $t_{BC}$ (rather $gt_{BC}$ with $g=1$ again), so that: $$ v_{_{xC}} = \frac{\sqrt{3}}{2}$$ $$v_{_{yC}} = v_{_{yB}} + t_{BC} = \frac{1}{2} + 1 = \frac{3}{2}. $$
so we see that the first path gives us:
$$ v_C = \sqrt{v_{_{xC}}^2+v_{_{yC}}^2} = \sqrt{3/4 + 9/4} = \sqrt{3}.$$
So that one checks out. Now let's check the second, direct path AC, we already found that $v_{AC} \approx 0.654t$, and $t_{AC} \approx 2.64$, so just plug this in:
$$ v_C \approx 0.654 \times 2.64 \approx 1.726 \approx \sqrt{3} $$
one can also easily verify that the error here is entirely due to the rounding, so this does check out. i.e. instead of $a_{AC}=0.654$ and $t_{AC}=2.64$ take the pre-rounded expressions $a_{AC}=\sin(40.89^{\circ})$ and $t_{AC}=\sqrt{\dfrac{2\times 2.291}{\sin(40.89^{\circ})}}$ so:
$$ v_C \approx \sin(40.89^{\circ})\sqrt{\frac{2\times 2.291}{\sin(40.89^{\circ})}} \approx 1.7318 \approx \sqrt{3} $$
which is closer to $\sqrt{3}$, as expected (still not perfect because some rounding errors are present from earlier stages, but you get the idea).
