You don't have to orbit, you can just use a rocket to stay put. All observers that can communicate with infinity for all time agree about the infalling object. It gets frozen and redshifted at the horizon.

### EDIT: in response to question

The issue of two objects falling in one after the other is adressed in this question: How does this thought experiment not rule out black holes? . The answers there are all wrong, except mine (this is not an arrogant statement, but a statement of an unfortunate fact).

When you are near a black hole, in order to stay in place, you need to accelerate away from the black hole. If you don't, you fall in. Whenever you accelerate, even in empty Minkowksi space, you see an acceleration event horizon behind you in the direction opposite your acceleration vector. This horizon is a big black wall that follows you around, and you can attribute the various effects you see in the accelerating frame, like the uniform gravitational field and the Unruh radiation, to this black-wall horizon that follows you around.

When you are very near a black hole, staying put, your acceleration horizon coincides with the event horizon, and there is no way to tell them apart locally. This is the equivalence principle, in the form that it takes in the region by the horizon where there is no significant curvature.

The near-horizon Rindler form of the metric allows you to translate any experiment you can do in the frame near a black hole to a flat space with an accelerating observer. So if you measure the local Hawking temperature, it coincides with the Unruh temperature. If you see an object fall and get redshifted, you would see the same thing in empty space, when accelerating.

The point is that the acceleration you need to avoid falling in is only determined globally, from the condition that you stay in communication with infinity. If you stop accelerating so that you see the particle cross the horizon, the moment you see the particle past the horizon, you've crossed yourself.

Consider the non-relativistic problem of a particle falling into a potential well and releasing all its energy in there. The quantity which is conserved during the infall is the total energy $E = T + V + E_{internal}$, where $T$ is kinetic energy, $V$ is the potential energy, and $E_{internal}$ is some internal "chemical" energy of the particle.

Now we assume that the particle starts at rest at infinity where the potential is zero so that $E = 0 + 0 + E_{internal} = E_{internal}$. As the particle starts falling to the potential well, nothing happens to internal energy, $V$ becomes negative and since energy is conserved, that must be countered with a positive $T$. When we arrive all the way in the potential well, no matter what happens with the kinetic and potential energy, we always have $E= E_{internal}$ and when the energy release in the well comes about, it is exactly $E= E_{internal}$ which is released.

You can think similarly about the black hole along with the realization that $E_{internal} = m_0 c^2$, where $m_0$ is the object's rests mass. In other words, if a particle of (rest) mass $m_0$ falls into a stationary black hole starting at rest at infinity, the black hole will receive exactly $m_0 c^2$ in terms of energy, no matter what happens to the kinematic or potential parts of the energy.

**EDIT**

Of course, the full relativistic problem has to be considered more carefully. First of all, what is exactly the mass of a black hole? One of the postulates of relativity is that a freely falling observer never feels gravity - so a freely falling observer will judge the black hole to be probably weightless and thus there cannot exist any *local*, *frame-independent* notion of the black hole mass. The black hole mass must be, in fact, defined by some coordinated measurements of privileged observers.

Furthermore, notice that no frame is privileged and thus *there is no notion of big or small velocity*! To define a notion of velocity, you also need a privileged frame with respect to which you are measuring it!

Sometimes it so happens that the black hole is in such a state that there exists a family of observers at infinity who collect measurements in which the black hole field appears as stationary. In such a case, we call the black hole stationary and the time in which these observers measure any physical process will be our privileged notion of time throughout the space-time.

It is also these observers through which we define the notion of mass. Since they define a notion of rest and the space-time around them is almost flat, they feel the gravity of the black hole in the weak-field, Newtonian limit. The mass of the black hole is defined exactly and only as the apparent Newtonian mass $M$ in the $\approx -M/r^2$ gravitational force these observers at infinity feel. In other words, you could understand the mass more as *total gravitating energy* as felt by observers at infinity.

Since the background is stationary with respect to this time, there will be a respective integral of motion for the evolution of test bodies moving in this space-time due to Noether's theorem. This integral is the temporal component of four-momentum $p_t$. Now, at infinity the space-time is asymptotically Minkowski space-time and if the test particle starts there at rest, we will just have $p_t = -m$ and other component of four-momentum equal to zero.

As the particle then falls into the black hole, $p_t = -m$ will never change and will play exactly the same role as $-E$ in the argument given above. Similarly, the growth of coordinate velocity (as measured by the observers at infinity) as a compensation of some kind of potential energy can also be traced to the conservation of $p_t$, at least in the Newtonian limit.

Of course, one could argue that this does not automatically make $-p_t$ the contribution to the black hole mass once the particle is absorbed by the central, mathematically ill-behaved singularity. It is an elegant and favorable candidate, because the sum of $-p_t$s from many particles should not be possible to annihilate by Noether's theorem and by some kind of assumption of adiabaticity of the black hole growth, so the $-p_t$ must go "somewhere". But this just tells us that the contribution to the black hole should be a linear function of $-p_t$.

The argument why it is exactly $-p_t$ to increase the black hole mass can come only from the consideration of the infalling body as a *non-test* object, that is an object which perturbs the black hole background with its own gravitational field. From such arguments we know that a part of the infalling energy almost always gets radiated away by gravitational radiation, but if the body gets sufficiently light, the contribution of mass to the resulting black hole indeed converges to $-p_t$.

## Best Answer

The Schwarzschild coordinates (which seem to suggest that no object ever crosses the event horizon when viewed from far outside) were derived for stationary case: no matter flows onto the black hole, the black hole has constant mass. In fact, Schwarzschild was assuming zero stress-energy tensor (vacuum solution).

However if you start adding a lot of mass to the black hole, the situation changes. Imagine you throw a little object towards the event horizon. It "seems" to freeze on the surface of the horizon (it actually visually disappears due to the red shift). Later on, there is a huge amount of material streaming to the black hole. It is thousands of times more mass than the original mass of the black hole. At this point the conditions under which Schwarzschild found his solution no longer stand, because the stress-energy tensor is far from being zero. The event horizon will grow, since it forms wherever the gravitational potential reaches certain value. By adding more mass you unavoidably enlarge the volume where the potential has the required value to form the event horizon.

The case of non-constant mass is described by the Vaidya metric. Mathematically this is described on pages 133-134 of this book.