gravitational constant Archives

March 19, 2015February 27, 2017 by Fraser Cain

Could the Death Star Destroy a Planet?

In the movie Star Wars, the Darth Vader’s Death Star destroyed a planet. Could this really happen?

You’ve watched Star Wars right? Is that still a thing? With the Starring and the Warring? Anyway, there’s this classic scene where the “Death Star” sidles up to Alderaan, and it is all like “Hey Planetoid, you lookin’ fine tonight” and then it fires up the superlaser and destroys the entire orb in a single blast. “BOOM”. Shortly followed by some collective group screaming on the interstellar forceway radio.

This is generally described as “science fiction”. And when you’re making up stories, anything you like can happen in them. George Lucas’ hunger for your childhood toy money wasn’t hampered by the pesky constraints of physics in any meaningful way.

Here at the Guide to Space, we get to take our own flights of fancy and pointlessly speculate for your amusement. That’s our job. Well, that and snark. Let’s consider what it would actually take to destroy a planet with a ‘pew pew’ style laser beam, and what kinds of energy would need to be harnessed in a fully armed and operational battle station.

Let’s go back and carefully review our “evidence”. The Death Star drifts in, charges up all its lasers into a superlaser blast focused on Alderaan. The planet then detonates and chunks fly off in every direction just like the pie eating contest in “Stand By Me”.

What we saw was every part of Alderaan given enough of a kick so that it was traveling at escape velocity from every other part of the planet. If the Death Star hadn’t delivered enough explosive energy, the planet might have fluffed up for a moment, but then the collective gravity would suck it all back in together, and then the slightly re-arranged, and likely now uninhabited planet would continue orbiting its star.

You can imagine doing this the slow way. Take each continent on Alderaan, load it up into a rocket and blast that rocket off into space as though it was on escape trajectory from the planet. Sure, you’d would need an incomprehensible number of rocket launches to get that material off the planet. But hey, midichlorians, blue finger lightning and ESP.

Fortunately, as you carted away more and more of the busted up rock, it would have less mutual gravity, and so the rocket launches would require less and less energy to get the job done. Eventually, you’d just be left with one last chunk of rock that you could just force ninja kick into the neighboring star.

So how much energy is that going to take? Well, there’s an “easy” calculation you can make. The energy you’d need is equal to 3 times the gravitational constant (6.673 x 10^-11) times the mass of the planet squared divided by 5 times the planet’s radius. Do this math for an Earth-sized/mass world, and let’s see that’s, two and one, carry the 5… and you get 2 x 10^36 joules. That’s a two followed by 36 zeros in joules. Is that a lot? That sounds like a lot.

Well, our own Sun puts out 3 x 10^26 joules per second. So, if you poured all the energy from the Sun into the task of tearing apart the Earth, it wouldn’t have enough energy to do it. In fact, you’d need to focus the light of the Sun for a full week to get that level of planet destruction done.

According to ancient Star Warsian dork scholars, the Death Star (SOLUS MORTIS) is powered by a hyperreactor with the output of multiple main sequence stars. So there you go, problem solved. It’s the size of a small moon, but it’s more powerful than many stars. Of course it can destroy a planet.

The Death Star clearly destroyed Alderaan. We watched it explode. I saw it, you saw it. We heard the screams of millions of souls cry out. It happened. But what if it wasn’t a beam thingy?

Our math is good, but clearly we’re not enlightened enough to comprehend the true wisdom hidden within the Lucasian scriptures. Perhaps the Death Star’s superlaser was just a targeting laser. Directing the placement of gigantic antimatter bomb. According to Ethan Siegel, from “Starts With a Bang,” you’d only need 1.24 trillion tonnes of antimatter.

Imagine you made a bomb out of that much antimatter iron – if that’s even a thing – you’d only need a sphere about 3 km across. If the Death Star is 150 km across or so, they could carry a bunch of these. Very carefully. Like super carefully. Okay, maybe it’d be a good idea if everyone took off their boots, and make sure they only talked with their inside voices.

Obviously, Star Wars is a story, so anything, ANYTHING can happen. The future is unknown, and we might discover all kinds of weirdo physics and harness them into all kinds of powerful weapons. I’m only suggesting, that a space station capable of deploying a week’s worth of solar energy in a single second might be a stretch. And maybe, George, if you just done a little back of the napkin math, we wouldn’t be talking about this right now. Also, maybe no Ewoks. I’m just saying.

Where do you stand on the feasibility of imaginary space station weaponry? How big a planet can your imagination destroy?

October 26, 2009December 24, 2015 by Jean Tate

Gravity Constant

Anaglyph images created from an ESA video animation of global gravity gradients. A more accurate global map will be generated by ESA's GOCE craft. Credit: ESA and Nathaniel Burton Bradford.

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The constant of gravity, or gravity constant, has two meanings: the constant in Newton’s universal law of gravitation (so is commonly called the gravitational constant, it also occurs in Einstein’s general theory of relativity); and the acceleration due to gravity at the Earth’s surface. The symbol for the first is G (big G), and the second g (little g).

Newton’s universal law of gravitation in words is something like “the gravitational force between two objects is proportional to the mass of each and inversely proportional to the square of the distance between them“. Or something like F (the gravitational force between two objects) is m₁ (the mass of one of the objects) times m₂ (the mass of one of the other object) divided by r² (the square of the distance between them). The “is proportional to” means all you need to make an equation is a constant … which is G.

In other words: F = Gm₁m₂/r²

The equation for little g is simpler; from Newton we have F = ma (a force F acting on a mass m produces an acceleration a), so the force F on a mass m at the surface of the Earth, due to the gravitational attraction between the m and the Earth is F = mg.

Little g has been known from at least the time of Galileo, and is approximately 9.8 m/s² – meters per second squared – it varies somewhat, depending on how high you are (altitude) and where on Earth you are (principally latitude).

Obviously, big G and little g are closely related; the force on a mass m at the surface of the Earth is both mg and GmM/r², where M is the mass of the Earth and r is its radius (in Newton’s law of universal gravitation, the distance is measured between the centers of mass of each object) … so g is just GM/r².

The radius of the Earth has been known for a very long time – the ancient Greeks had worked it out (albeit not very accurately!) – but the mass of the Earth was essentially unknown until Newton described gravity … and even afterwards too, because neither G nor M could be estimated independently! And that didn’t change until well after Newton’s death (in 1727), when Cavendish ‘weighed the Earth’ using a torsion balance and two pairs of lead spheres, in 1798.

Big G is extremely hard to measure accurately (to 1 part in a thousand, say); today’s best estimate is 6.674 28 (+/- 0.000 67) x 10^-11 m³ kg^-1 s ^-2.

The Constant Pull of Gravity: How Does It Work? is a good NASA webpage for students, on gravity; and the ESA’s GOCE mission webpage describes how satellites are being used to measure variations in little g (GOCE stands for Gravity field and steady-state Ocean Circulation Explorer).

The Pioneer Anomaly: A Deviation from Einstein’s Gravity? is a Universe Today story related to big G, as is Is the Kuiper Belt Slowing the Pioneer Spacecraft?; GOCE Satellite Begins Mapping Earth’s Gravity in Lower Orbit Than Expected is one about little g.

No surprise that the Astronomy Cast episode Gravity covers both big G and little g!

July 19, 2009October 18, 2016 by John Carl Villanueva

What is the Gravitational Constant?

Visualization of a massive body generating gravitational waves (UWM)

The gravitational constant is the proportionality constant used in Newton’s Law of Universal Gravitation, and is commonly denoted by G. This is different from g, which denotes the acceleration due to gravity. In most texts, we see it expressed as:

G = 6.673×10^-11 N m² kg^-2

It is typically used in the equation:

F = (G x m₁ x m₂) / r² , wherein

F = force of gravity

G = gravitational constant

m₁ = mass of the first object (lets assume it’s of the massive one)

m₂ = mass of the second object (lets assume it’s of the smaller one)

r = the separation between the two masses

As with all constants in Physics, the gravitational constant is an empirical value. That is to say, it is proven through a series of experiments and subsequent observations.

Although the gravitational constant was first introduced by Isaac Newton as part of his popular publication in 1687, the Philosophiae Naturalis Principia Mathematica, it was not until 1798 that the constant was observed in an actual experiment. Don’t be surprised. It’s mostly like this in physics. The mathematical predictions normally precede the experimental proofs.

Anyway, the first person who successfully measured it was the English physicist, Henry Cavendish, who measured the very tiny force between two lead masses by using a very sensitive torsion balance. It should be noted that, after Cavendish, although there have been more accurate measurements, the improvements on the values (i.e., being able to obtain values closer to Newton’s G) have not been really substantial.

Looking at the value of G, we see that when we multiply it with the other quantities, it results in a rather small force. Let’s expand that value to give you a better idea on how small it really is: 0.00000000006673 N m² kg^-2

Alright, let’s now see what force would two 1-kg objects exert on one another when their geometrical centers are spaced 1 meter apart. So, how much do we get?

F = 0.00000000006673 N. It really doesn’t matter much if we increase both masses substantially.

For example, let’s try the heaviest recorded mass of an elephant, 12,000 kg. Assuming we have two of these, spaced 1 meter apart from their centers. I know it’s difficult to imagine that since elephants are rather stout, but let’s just proceed this way because I want to put emphasis on the significance of G.

So, how much did we get? Even if we rounded that off, we’d still obtain only 0.01 N. For comparison, the force exerted by the earth on an apple is roughly 1 N. No wonder we don’t feel any force of attraction when we sit beside someone… unless of course you’re a male and that person is Megan Fox (still, it’d be safe to assume that the attraction would only be one way).

Therefore, the force of gravity is only noticeable when we consider at least one mass to be very massive, e.g. a planet’s.

Allow me to end this discussion with one more mathematical exercise. Assuming you know both your mass and your weight, and you know the radius of the earth. Plug those into the equation above and solve for the other mass. Voila! Wonder of wonders, you’ve just obtained the mass of the Earth.

You can read more about the gravitational constant here in Universe Today. Want to learn more about a new study that finds fundamental force hasn’t changed over time? There’s also some insights you can find among the comments in this article: Record Breaking “Dark Matter Web” Structures Observed Spanning 270 Million Light Years Across

There’s more about it at NASA. Here are a couple of sources there:

Here are two episodes at Astronomy Cast that you might want to check out as well:

Sources: