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Gravitational Physics

Here we lay out Newton's law of gravity and crack open the universe of consequences that spring from it.

Escape velocity

                           

Due to their strong gravitational fields, large celestial bodies are able to trap objects, such as hobby rockets on Earth, or the planets of the Solar System in orbits around the Sun. As we showed in Atmospheric Thickness, a planet's gravity can keep and organize a layer of gas around a planet.

However, our analysis there made the implicit assumption that there is a stable atmosphere in the first place. As we know, there are heavenly bodies like Mercury or our Moon which have no significant atmosphere to speak of.

In this quiz, you'll derive strict limits on the kinds of particles that can escape a planet or star. Then, you'll turn the calculation on its head, and find objects from which no particles can escape. Toward the end, you'll put it all together and make some informed predictions about the fate of our Sun.

But first, let's meet our friend Walter.

Our friend Walter is all alone, with nobody to play catch with. Tired of waiting around, he decides to play catch with himself by throwing his baseball high into the air.

Suppose the ball leaves his hand, going straight up, with speed \(v_0 = \SI[per-mode=symbol]{30}{\meter\per\second}.\)

How high (in \(\si{\meter}\)) will it go before falling back to Walter, the loneliest boy in the world?

Assumptions: the gravitational field is given by \(\mathbf{g}(r) = -g\hat{\mathbf{z}},\) all the way up, with \(g = \SI[per-mode=symbol]{10}{\meter\per\second\squared}.\)

With so much time to himself, lonely Walter gets quite good at throwing the ball, hurling it with incredible speed. Walter sees that as he throws the ball faster and faster, it goes higher and higher before returning to him at the surface. This gets him thinking:

"Can I throw the ball so fast that it never comes back?"

Without specifying a form for the gravitational field \(\mathbf{g}(r),\) what must be true about it if the ball is to escape the Earth?

In the first problem, you found the relationship between the ball's velocity and its trajectory through the atmosphere \[m\dot{v} = -mg(r).\] So, for Walter's ball to escape it needs enough initial momentum to outlast the pull of Earth's gravitational field: \[v_f = v_i - \int g(r) dt.\] As we know, Earth's gravity is not uniform all the way up (as in the first problem), but actually the inverse square form \(g \sim 1/r^2.\) Which of the following general forms for \(v(r)\) satisfy the equation of motion (either of the differential or integral forms above)?


Hint: you can either check which form satisfies the equation of motion, or carry out the integral with the help of the chain rule, \(\frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt}.\)

You now have the functional form for the speed of lonely Walter's ball \(v\) as a function of its height above the Earth \(r,\) but we still have some undetermined constants (\(A, B,\) etc.).

It may seem like we're at an impasse, but you have some more information about the behavior of the system, namely its initial conditions (\(r_0, v_0\)).

Exploit the initial conditions to find the value of the constant \(B\).

With the constant field approximation, we found that lonely Walter's throws are forever trapped by Earth's gravity. No matter the initial speed \(v_i,\) it eventually turns negative and the ball returns to the surface. But in the real, inverse square field, we showed that it is possible to escape the pull of Earth.

Mathematically, this corresponds to taking the ball's position \(r\) out to \(\infty.\)

Use this insight to solve for \(v_0\) in terms of \(r.\) What do you find for \(v_0\) (in \(\si[per-mode=symbol]{\meter\per\second}\))?

The escape velocity you just found can be rewritten in the transparent form \[\begin{align} v_\textrm{esc} &= \sqrt{2gR_\textrm{planet}} \\ &\propto \sqrt{g}\cdot\sqrt{R_\textrm{planet}} \end{align}\] Quite simply, escape velocity increases with the square root of fold changes in both the strength of surface gravity \(g\) and the size of the planet \(R_\textrm{planet}.\)

For residents of a given planet, this is useful insofar as they may need to calculate the minimum requirements for interplanetary travel. But more interestingly, escape velocity helps to determine the stability of stellar bodies, as their radii and surface gravities are in continual flux due to various internal processes.

Snapshots of the Sun's life cycle as it progressed from a cloud of dust, to a nebulae, to its current state, and its future existence as a Red Giant, before collapsing to a White Dwarf.

Snapshots of the Sun's life cycle as it progressed from a cloud of dust, to a nebulae, to its current state, and its future existence as a Red Giant, before collapsing to a White Dwarf.

In the second half of this quiz, we're going to focus on stellar life cycles and use the form of the escape velocity to reason about the coming fate of the Sun.

At the start of its formation, the Sun was a ball of dust that coalesced over time into a hot, dense nebulae under the attraction of its own gravity. This is perhaps the earliest phase in which a star begins to exhibit some kind of structure.


Suppose we have a spherically symmetric cloud of gas of mass \(M_\textrm{cloud}\) that undergoes a collapse process such that its radius shrinks while its total mass stays constant.

How will its escape velocity change over time?

Eventually, the Sun reached its current state as a so-called "main sequence" star with a hot, dense core capable of nuclear fusion.

As you read this question, the Sun's core is burning down its supply of fuse-able Hydrogen, becoming hotter in the process. Eventually, the core is expected to become so hot that it ignites fusion in the outer, heavy-element layers of the Sun, which in turn will cause the Sun to heat and expand significantly, as it enters its phase as a Red Giant.

What do you expect to happen to the mass of the Sun during the expansive phase?

Hint: how is its escape velocity changing?

Ignoring potential complications from general relativity, calculate the so-called event horizon—the distance below which no object can escape a star.

Hint: as the escape velocity rises, it can't exceed the speed of light.

Details: the star's temperature is \(T,\) its mass is \(M,\) and \(k_B\) is Boltzmann's constant.

Ignoring any change in mass during these transitions, how small would our Sun have to shrink in order to convert to a black hole?

Details: the solar mass is \(\SI{2e30}{\kilo\gram}.\)

In stars like the Sun, the compressive force of self-gravitation is balanced by a quantum effect, called the degeneracy pressure. For this reason, current estimates expect the Sun will never compress below the Schwarschild radius.

Instead, after its expansive phase peters out, the Sun is expected to collapse once again as it radiates much of its energy away, sheds its outer layers, and eventually become a White Dwarf.

In heavier stars, self-gravitation can swamp out the degenerative pressure which leads to core collapse, and formation of a neutron star, or if massive enough, a black hole!

Note: Over an extremely long period of time, the Sun is expected to radiate away all of its non-gravitational energy, until it eases into retirement as a virtually non-radiating black dwarf. The timescale over which a white dwarf is expected to progress toward the black dwarf state is \(T_\textrm{w.d.}\sim\SI{e15}{yr},\) roughly \(\num{e5}\) times the current age of the universe.

In this quiz, you integrated the equation of motion to calculate a so-called escape velocity—the minimum launch speed required to liberate an object from the gravitational clutches of a celestial body.

You then turned this quantity on its head to reason about the accretion and ejection of mass from the a star during its life cycle. Finally, you applied limiting cases, to make a naive approximation for the event horizon of a black hole.

In fact, in making this calculation, we stumbled upon what is known as the Schwarschild radius, and its value is one of the rare cases where the naively application of classical reasoning obtains the same answer as a more involved, logically sound relativistic calculation. Throughout this course, we'll have more occasions to compare and contrast the results of Newtonian gravity with those of general relativity.

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