# Climbing Out Of A Well

Al can’t contain himself. “Wait, it’s gravity!”

Vinnie and I are puzzled. “Come again?”

“Sy, you were going on about how much speed a rocket has to shed on the way to some special orbit around Mars, like that’s a big challenge. But it’s not. The rocket’s fighting the Sun’s gravity all the way. That’s where the speed goes. The Earth’s gravity, too, a little bit early on, but mostly the Sun’s, right?”

“Good point, Al. Sun gravity’s what bends the rocket onto a curve instead of a straight line. Okay, Sy, you got a magic equation that accounts for the shed speed? Something’s gotta, ’cause we got satellites going around Mars.”

“Good point, Vinnie, and you’re right, there is an equation. It’s not magic, you’ve already seen it and it ties kinetic energy to gravitational potential energy.”

“Wait, if I remember right, kinetic energy goes like mass times velocity squared. How can you calculate that without knowing how big the rocket is?”

“Good question. We get around that by thinking things through for a unit mass, one kilogram in SI units. We can multiply by the rocket’s mass when we’re done, if we need to. The kinetic energy per unit mass, we call that specific kinetic energy, is just ½v². Look familiar?”

“That’s one side of your v²=2GM/R equation except you’ve got the 2 on the other side.”

“Good eye, Al. The right-hand side, except for the 2, is specific gravitational potential energy, again for unit mass. But we can’t use the equation unless we know the kinetic energy and gravitational potential are indeed equal. That’s true if you’re in orbit but we’re talking about traveling between orbits where you’re trading kinetic for potential or vice versa. One gains what the other loses so Al’s right on the money. Traveling out of a gravity well is all about losing speed.”

Al’s catching up. “So how fast you’re going determines how high you are, and how high you are says how fast you have to be going.”

Vinnie frowns a little. “I’m thinking back to in‑flight refueling ops where I’m coming up to the tanker from below and behind while the boom operator directs me in. That doesn’t sound like it’d work for joining up to a satellite.”

“Absolutely. If you’re above and behind you could speed up to meet the beast falling, or from below and ahead you could slow down to rise. Away from that diagonal you’d be out of luck. Weird, huh?”

“Yeah. Which reminds me, now we’re talking about this ‘deeper means faster‘ stuff. How does the deep‑dive maneuver work? You know, where they dive a spacecraft close to a planet or something and it shoots off with more speed than it started with. Seems to me whatever speed it gains it oughta give up on the way out of the well.”

“It’s a surprise play, alright, but it’s actually two different tricks. The slingshot trick is to dive close enough to capture a bit of the planet’s orbital momentum before you fly back out of the well. If you’re going in the planet’s direction you come out going faster than you went in.”

“Or you could dive in the other direction to slow yourself down, right?”

“Of course, Al. NASA used both options for the Voyager and Messenger missions. Vinnie, I know what you’re thinking and yes, theoretically stealing a planet’s orbital momentum could affect its motion but really, planets are huge and spacecraft are teeny. DART hit the Dimorphos moonlet head-on and slowed it down by 5%, but you’d need 66 trillion copies of Dimorphos to equal the mass of dinky little Mercury.”

“What’s the other trick?”

“Dive in like with the slingshot, but fire your rocket engine when you’re going fastest, just as the craft approaches its closest point to the planet. Another German rocketeer, Hermann Oberth, was the first to apply serious math to space navigation. This trick’s sometimes called the Oberth effect, though he didn’t call it that. He showed that rocket exhaust gets more effective the faster you’re going. The planet’s gravity helps you along on that, for free.”

“Free help is good.”

~~ Rich Olcott

# Rockfall

<continued>  The coffee shop crowd had gotten rowdy in response to my sloppy physics, but everyone hushed when I reached for my holster and drew out Old Reliable.  All had heard of it, some had seen it in action — a maxed-out tablet with customized math apps on speed-dial.

“Let’s take this nice and slow.  Suppose we’ve got an non-charged, non-spinning solar-mass black hole.  Inside its event horizon the radius gets weird but let’s pretend we can treat the object like a simple sphere.  The horizon’s half-diameter, we’ll call it the radius, is rs=2G·M/c²G is Newton’s gravitational constant, M is the object’s mass and c is the speed of light.  Old Reliable says … about 3 kilometers.  Question is, what happens when we throw a rock in there?  To keep things simple, I’m going to model dropping the rock gentle-like, dead-center and with negligible velocity relative to the hole, OK?”

<crickets>

“Say the rock has the mass of the Earth, almost exactly 3×10-6 the Sun’s mass.  The gravitational potential energy released when the rock hits the event horizon from far, far away would be E=G·M·m/rs, which works out to be … 2.6874×1041 joules.  What happens to that energy?”

rs depends on mass, Mr Moire, so the object will expand.  Won’t that push on what’s around it?”

“You’re thinking it’d act like a spherical piston, Jeremy, pushing out in all directions?”

“Yeah, sorta.”

“After we throw in a rock with mass m, the radius expands from rs to rp=2G·(M+m)/c².  I set m to Earth’s mass and Old Reliable says the new radius is … 3.000009 kilometers.  Granted the event horizon is only an abstract math construct, but suppose it’s a solid membrane like a balloon’s skin.  When it expands by that 9 millimeters, what’s there to push against?  The accretion disk?  Those rings might look solid but they’re probably like Saturn’s rings — a collection of independent chunks of stuff with an occasional gas molecule in-between.  Their chaotic orbits don’t have a hard-edged boundary and wouldn’t notice the 9-millimeter difference.  Inward of the disk you’ve got vacuum.  A piston pushing on vacuum expends zero energy.  With no pressure-volume work getting done that can’t be where the infall energy goes.”

“How about lift-a-weight work against the hole’s own gravity?”

“That’s a possibility, Vinnie.  Some physicists maintain that a black hole’s mass is concentrated in a shell right at the event horizon.  Old Reliable here can figure how much energy it would take to expand the shell that extra 9 millimeters.  Imagine that simple Newtonian physics applies — no relativistic weirdness.  Newton proved that a uniform spherical shell’s gravitational attraction is the same as what you’d get from having the same mass sitting at the shell’s geometric center.  The gravitational pull the shell exerts on itself originally was E=G·M²/rs.  Lifting the new mass from rs to rp will cost ΔE=G·(M+m)²/r– G·M²/rs.  When I plug in the numbers…  That’s interesting.”

Vinnie’s known me long enough to realize “That’s interesting” meant “Whoa, I certainly didn’t expect THAT!

“So what didja expect and whatcha got?”

“What I expected was that lift-it-up work would also be just a small fraction of the infall energy and the rest would go to heat.  What I got for ΔE here was 2.6874×1041 joules, exactly 100% of the input.  I wonder what happens if I use a bigger planet.  Gimme a second … OK, let’s plot a range …  How ’bout that, it’s linear!”

“Alright, show us!”

All the infall energy goes to move the shell’s combined mass outward to match the expanded size of the event horizon.  I’m amazed that such a simple classical model produces a reasonable result.”

“Like Miss Plenum says, Mr Moire, sometimes the best science comes from surprises.”

“I wouldn’t show it around, Jeremy, except that it’s consistent with Hawking’s quantum-physics result.”

“How’s that?”

“Remember, he showed that a black hole’s temperature varies as 1/M.  We know that temperature is ΔE/ΔS, where the entropy change ΔS varies as .  We’ve just found that ΔE varies as M.  The ΔE/ΔS ratio varies as M/M²=1/M, just like Hawking said.”

Then Jennie got into the conversation.

~~ Rich Olcott

# Weight And Wait, Two Aspects of Time

I was deep in the library stacks, hunting down a journal article so old it hadn’t been digitized yet.  As I rounded the corner of Aisle 5 Section 2, there he was, leaning against a post and holding a clipboard.

“Vinnie?  What are you doing here?”

“Waiting for you.  You weren’t in your office.”

“But how…?  Never mind.  What can I do for you?”

“It’s the time-dilation thing.  You said that there’s two kinds, a potential energy kind and a kinetic energy kind, but you only told me about the first one.”

“Hey, Ramona broke up that conversation, don’t blame me.  You got blank paper on that clipboard?”

“Sure.  Here.”

“Quick review — we said that potential energy only depends on where you are.  Suppose you and a clock are at some distance r away from a massive object like that Gargantua black hole, and my clock is way far away.  I see your clock ticking slower than mine.  The ratio of their ticking rates, tslow/tfast = √[1-(2G·M/r·c²)], only depends on the slow clock’s position.  Suppose you move even closer to the massive object.  That r-value gets smaller, the fraction inside the parentheses gets closer to 1, the square root gets smaller and I see your clock slow down even more.  Sound familiar?”

“Yeah, but what about the kinetic thing?”

“I’m getting there.  You know Einstein’s famous EEinstein=m·c² equation.  See?  The formula contains neither a velocity nor a position.  That means EEinstein is the energy content of a particle that’s not moving and not under the influence of any gravitational or other force fields.  Under those conditions the object is isolated from the Universe and we call m its rest mass.  We good?”

“Yeah, yeah.”

“OK, remember the equation for gravitational potential energy?”

E=G·M·m/r.

“Let’s call that Egravity.  Now what’s the ratio between gravitational potential energy and the rest-mass energy?”

“Uh … Egravity/EEinstein = G·M·m/r·m·c² = G·M/r·c². Hey, that’s exactly half the fraction inside the square root up there. tslow/tfast = √[1-(2 Egravity/EEinstein)].  Cool.”

“Glad you like it.  Now, with that under our belts we’re ready for the kinetic thing.  What’s Newton’s equation for the kinetic energy of an object that has velocity v?”

E=½·m·v².

“I thought you’d know that.  Let’s call it Ekinetic.  Care to take a stab at the equation for kinetic time dilation?”

“As a guess, tslow/tfast = √[1-(2 Ekinetic/EEinstein)]. Hey, if I plug in the formulas for each of the energies, the halves and the mass cancel out and I get tslow/tfast = √[1-2(½m·v²/m·c²)] = √[1-(v²/c²)].  Is that it?”

“Close.  In Einstein’s math the kinetic energy expression is more complicated, but it leads to the same formula as yours.  If the velocity’s zero, the square root is 1.0 and there’s no time-slowing.  If the object’s moving at light-speed (v=c), the square root is zero and the slow clock is infinitely slow.  What’s interesting is that an object’s rest energy acts like a universal energy yardstick — both flavors of time-slowing are governed by how the current energy quantity compares to EEinstein.”

“Wait — kinetic energy depends on velocity, right, which means that it’ll look different from different inertial frames.  Does that mean that the kinetic time-slowing depends on the frames, too?”

“Sure it does.  Best case is if we’re both in the same frame, which means I see you in straight-line motion.  Each of us would get the same number if we measure the other’s velocity.  Plug that into the equation and each of us would see the same tslow for the other’s clock.  If we’re not doing uniform straight lines then we’re in different frames and our two dilation measurements won’t agree.”

“… Ramona doesn’t dance in straight lines, does she, Sy?”

“That reminds me of Einstein’s quote — ‘Put your hand on a hot stove for a minute, and it seems like an hour. Sit with a pretty girl for an hour, and it seems like a minute. That’s relativity.‘  You’re thinking curves now, eh?”

“Are you boys discussing me?”

<unison> “Oh, hi, Ramona.”

~~ Rich Olcott

# Gravity’s Real Rainbow

Some people are born to scones, some have scones thrust upon them.  As I stepped into his coffee shop this morning, Al was loading a fresh batch onto the rack.  “Hey, Sy, try one of these.”

“Uhh … not really my taste.  You got any cinnamon ones ready?”

“Not much for cheddar-habañero, huh?  I’m doing them for the hipster trade,” waving towards all the fedoras on the room.  “Here ya go.  Oh, Vinnie’s waiting for you.”

I navigated to the table bearing a pile of crumpled yellow paper, pulled up a chair.  “Morning, Vinnie, how’s the yellow writing tablet working out for you?”

“Better’n the paper napkins, but it’s nearly used up.”

“What problem are you working on now?”

“OK, I’m still on LIGO and still on that energy question I posed way back — how do I figure the energy of a photon when a gravitational wave hits it in a LIGO?  You had me flying that space shuttle to explain frames and such, but kept putting off photons.”

“Can’t argue with that, Vinnie, but there’s a reason.  Photons are different from atoms and such because they’ve got zero mass.  Not just nearly massless like neutrinos, but exactly zero.  So — do you remember Newton’s formula for momentum?”

“Yeah, momentum is mass times the velocity.”

“Right, so what’s the momentum of a photon?”

“Uhh, zero times speed-of-light.  But that’s still zero.”

“Yup.  But there’s lots of experimental data to show that photons do carry non-zero momentum.  Among other things, light shining on an an electrode in a vacuum tube knocks electrons out of it and lets an electric current flow through the tube.  Compton got his Nobel prize for that 1923 demonstration of the photoelectric effect, and Einstein got his for explaining it.”

“So then where’s the momentum come from and how do you figure it?”

“Where it comes from is a long heavy-math story, but calculating it is simple.  Remember those Greek letters for calculating waves?”

(starts a fresh sheet of note paper) “Uhh… this (writes λ) is lambda is wavelength and this (writes ν) is nu is cycles per second.”

“Vinnie, you never cease to impress.  OK, a photon’s momentum is proportional to its frequency.  Here’s the formula: p=h·ν/c.  If we plug in the E=h·ν equation we played with last week we get another equation for momentum, this one with no Greek in it:  p=E/c.  Would you suppose that E represents total energy, kinetic energy or potential energy?”

“Momentum’s all about movement, right, so I vote for kinetic energy.”

“That’s potential energy ’cause it depends on where you’re comparing it to.”

“OK, back when we started this whole conversation you began by telling me how you trade off gravitational potential energy for increased kinetic energy when you dive your airplane.  Walk us through how that’d work for a photon, OK?  Start with the photon’s inertial frame.”

“That’s easy.  The photon’s feeling no forces, not even gravitational, ’cause it’s just following the curves in space, right, so there’s no change in momentum so its kinetic energy is constant.  Your equation there says that it won’t see a change in frequency.  Wavelength, either, from the λ=c/ν equation ’cause in its frame there’s no space compression so the speed of light’s always the same.”

“Bravo!  Now, for our Earth-bound inertial frame…?”

“Lessee… OK, we see the photon dropping into a gravity well so it’s got to be losing gravitational potential energy.  That means its kinetic energy has to increase ’cause it’s not giving up energy to anything else.  Only way it can do that is to increase its momentum.  Your equation there says that means its frequency will increase.  Umm, or the local speed of light gets squinched which means the wavelength gets shorter.  Or both.  Anyway, that means we see the light get bluer?”

“Vinnie, we’ll make a physicist of you yet.  You’re absolutely right — looking from the outside at that beam of photons encountering a more intense gravity field we’d see a gravitational blue-shift.  When they leave the field, it’s a red-shift.”

“Keeping track of frames does make a difference.”

Al yelled over, “Like using tablet paper instead of paper napkins.”

~~ Rich Olcott

# Ya got potential, kid, but how much?

Dusk at the end of January, not my favorite time of day or year.  I was just closing up the office when I heard a familiar footstep behind me.  “Hi, Vinnie.  What’s up?”

“Energy, Sy.”

“Energy?”

“Energy and LIGO.  Back in flight school we learned all about trading off kinetic energy and potential energy.  When I climb I use up the fuel’s chemical energy to gain gravitational potential energy.  When I dive I convert gravitational potential energy into  kinetic energy ’cause I speed up.  Simple.”

“So how do you think that ties in with LIGO?”

“OK, back when we pretended we was in those two space shuttles (which you sneaky-like used to represent photons in a LIGO) and I got caught in that high-gravity area where space is compressed, we said that in my inertial frame I’m still flying at the same speed but in your inertial frame I’ve slowed down.”

“Yeah, that’s what we worked out.”

“Well, if I’m flying into higher gravity, that’s like diving, right, ’cause I’m going where gravity is stronger like closer to the Earth, so I’m losing gravitational potential energy.  But if I’m slowing down I’ve gotta be losing kinetic energy, too, right?  So how can they both happen?  And how’s it work with photons?”

“Interesting questions, Vinnie, but I’m hungry.  How about some dinner?”

We took the elevator down to Eddie’s pizza joint on the second floor.  I felt heavier already.  We ordered, ate and got down to business.

“OK, Vinnie.  Energy with photons is different than with objects that have mass, so let’s start with the flying-objects case.  How do you calculate gravitational potential energy?”

“Like they taught us in high school, Sy, ‘little g’ times mass times the height, and ‘little g’ is some number I forget.”

“Not a problem, we’ll just suppose that ‘little g’ times your plane’s mass is some convenient number, like 1,000.  So your gravitational potential energy is 1000×height, where the height’s in feet and the unit of energy is … call it a fidget.  OK?”

“Saves having to look up that number.”

“Uhh… twenty million fidgets.”

“Great.  You maintain level flight to Denver.  As you pass over the Rockies you notice your altimeter now reads 6,000 feet because of that 14,000-foot mountain you’re flying over.  What’s your gravitational potential energy?”

“Six million fidgets.  Or is it still twenty?”

“Well, if God forbid you were to drop out of the sky, would you hit the ground harder in California or Colorado?”

“California, of course.  I’d fall more than three times as far.”

“So what you really care about isn’t some absolute amount of potential energy, it’s the relative amount of smash you experience if you fall down this far or that far.  ‘Height’ in the formula isn’t some absolute height, it’s height above wherever your floor is.  Make sense?”

“Mm-hm.”

“That’s an essential characteristic of potential energy — electric, gravitational, chemical, you name it.   It’s only potential.  You can’t assign a value without stating the specific transition you’re interested in.  You don’t know voltages in a circuit until you put a resistance between two specific points and meter the current through it.  You don’t know gravitational potential energy until you decide what location you want to compare it with.”

“And I suppose a uranium atom’s nuclear energy is only potential until a nuke or something sets it off.”

“You got the idea.  So, when you flew into that high-gravity compressed-space sector, what happened to your gravitational potential energy?”

“Like I said, it’s like I’m in a dive so I got less, right?”

“Depends on what you’re going to fall onto, doesn’t it?”

“No, wait, it’s definitely less ’cause I gotta use energy to fly back out to flat space.”

“OK, you’re comparing here to far away.  That’s legit.  But where’s that energy go?”

“Ahh, you’re finally getting to the kinetic energy side of my question –”

“Whoa, look at the time!  Got a plane to catch.  We’ll pick this up next week.  Bye.”

“Hey, Sy, your tab! …  Phooey, stuck for it again.”

~~ Rich Olcott

# What’s that funnel about, really?

If you’ve ever watched or read a space opera (oh yes, you have), you know about the gravity well that a spacecraft has to climb out of when leaving a planet.  Every time I see the Museum’s gravity well model (photo below), I’m reminded of all the answers the guy gave to, “Johnny, what can you make of this?

The model’s a great visitor-attracter with those “planets” whizzing around the “Sun,” but this one exhibit really represents several distinct concepts.   For some of them it’s not quite the right shape.

The simplest concept is geometrical.  “Down” is the direction you move when gravity’s pulling on you.

A gravity well model for that concept would be just a straight line between you and the neighborhood’s most intense gravity source.

You learned the second concept in high school physics class.  Any object has gravitational potential energy that measures the amount of energy it would give up on falling.  Your teacher probably showed you the equation GPE = m·g·h, where m is the mass of the object, h is its height above ground level, and g is a constant you may have determined in a lab experiment.

If the width of the gravity well model at a given height represents GPE at that level, the model is a simple straight-sided cone.

But of course it’s not that simple.  Newton’s Law of Gravity says that the potential energy at any height r away from the planet’s center is proportional to 1/r.

Hmm… that looks different from the “proportional to h” equation.  Which is right?

Both equations are valid, but over different distance scales.  The HS teachers didn’t quite lie to you, but they didn’t give you the complete picture either.  Your classroom was about 4000 miles (21,120,000 feet) from Earth’s center, whereas the usual experiments involve height differences of at most a dozen feet.  Even the 20-foot drop from a second-story window is less than a millionth of the way down to Earth’s center.

Check my numbers:

Height h 1/(r+h)
× 108
Difference in 1/(r+h)
× 1014
0 4.734,848,484 0
20 4.734,844,001 4.48
40 4.734,839,517 8.97
60 4.734,835,033 13.45
80 4.734,830,549 17.93
100 4.734,826,066 22.42

Sure enough, that’s a straight line (see the chart).  Reminds me of how Newton’s Law of Gravity is valid except at very short distances.  The HS Law of Gravity works fine for small spans but when the distances get big we have to use Newton’s equation.

We’re not done yet. That curvy funnel-shaped gravity well model could represent the force of gravity rather than its potential energy.  Newton told us that the force goes as 1/r2 so it decreases much more rapidly than the potential energy does as you get further away.  The gravity force well has a correspondingly sharper curve to it than the gravity energy well.

The funnel model could also represent the total energy required to get a real spacecraft off the surface and up into space.  Depending on which sci-fi gimmickry is in play, the energy may come from a chemical or ion rocket, an electromagnetic railgun, or even a tractor beam from some mothership way up there.

No matter the technology, the theoretical energy requirement to get to a given height is the same.  In practice, however, each technology is optimal for some situations but forbiddingly inefficient in others.  Thus, each technology’s funnel  has its own shape and that shape will change depending on the setting.

In modern physics, the funnel model could also represent Einstein’s theory of how a mass “bends” the space around it.  (Take a look at this post, which is about how mass curves space by changing the local distance scale.)  Cosmologists describe the resulting “shapes” with embedding diagrams that are essentially 2D pictures of 3D (or 4D) contour plots.  The contours are closest together where space is most compressed, just as lines showing a steep hillside on a landscape contour map are close together.

The ED around a non-spinning object looks just like the force model picture above.  No surprise — gravitational force is how we we perceive spatial curvature.

~~ Rich Olcott