Tag Archives: physics

Orbital mechanics and flying to the Sun

When I watch a science fiction television show or movie, I tend to divide the physics errors into two categories.  There are the ones that I ascribe to narrative or stylistic necessity, and there are the unnecessary mistakes.

In the first group I put things like sound effects in space scenes.  Of course, we all know that sound is not transmitted in a vacuum, but it is often an artistic choice to put sound into these scenes.  Certainly, there are cases where it is not done, the first example that springs to mind being “2001: A Space Odyssey“.  There, the sudden dead silence in some scenes was both realistic and dramatic.

In the second group, though, are things that generally don’t matter to the plot, that could be depicted realistically, but are not.  That brings us to flying into the Sun.

If one were to posit a spacecraft, starting at the Earth, with a trajectory intended to bring it close to the Sun in a flight time of weeks to months, how do you direct the thrust of your engines?  The common depiction treats a flight to the Sun as essentially driving a truck on an invisible highway.  You point your direction of thrust toward the centre of the Sun, turn on the engine, and fly straight there.  This is incorrect, such a trajectory is extremely inefficient, if achievable at all.

While there are many ways we can examine this problem, the simplest one I can think of uses very little math and fairly basic physics.  We will consider only angular momentum.

Angular momentum is a conserved quantity.  If I choose a point in space, the magnitude of my angular momentum relative to that point is equal to the product of my mass, speed and the distance of closest approach between my straight-line extended trajectory and the point of interest.  In mathematical terms, it is the cross product of my momentum and the vector connecting me with the point in space.

Just as momentum is conserved in the absence of a force, angular momentum is conserved in the absence of a torque.  Torque is, similarly, the cross product of the force applied and the vector connecting me with the point in space.

So, let’s consider our angular momentum relative to the centre of the Sun.  Now, when we begin our journey, we are traveling at the same speed as the Earth (approximately 30 km/s), around the Sun.  Our speed is roughly perpendicular to the line connecting the Earth to the Sun, as the Earth’s orbit is quite close to circular.  The magnitude of our angular momentum is, from basic trigonometry, about equal to mass times speed times distance.  That amounts to 4.5E+15 kg m^2/s per kilogram of spacecraft.

Now, we’ve pointed out engine straight away from the Sun, so that our applied force is exactly on the line connecting the spacecraft to the centre of the Sun.  That means that the torque is zero, because the cross product of parallel vectors is zero.  So, in this configuration, the engine cannot change the angular momentum of the spacecraft.  Yes, it can push us toward the Sun, but our initial sideways deflection keeps trying to push us out of the way.  Imagine that you just kept adjusting your engine so the thrust was always directed straight into the Sun, and you managed to just barely touch the edge of the Sun while swinging by.  The Sun’s radius is about 700000 km.  If, for simplicity, we ignore any change in the mass of the spacecraft, that means that it must have a speed along the surface of the Sun of 6430 km/s.  Now, it’s deeper in the Sun’s gravity.  An object starting at the Earth’s orbit and ending at the surface of the Sun would be expected to gain no more than 620 km/s in speed due to gravitational forces.  That leaves a deficit of about 5800 km/s that must come from the spacecraft’s engines.  That is a stupendous delta-V.  Holding onto this number, we now look at a better option.

If, instead, the spacecraft were to fire its engines so as to oppose its motion around the Sun, a much different result is seen.  We already mentioned that the Earth is moving at 30 km/s around the Sun.  So, we now set our engines to drive us backwards along the Earth’s orbit around the Sun.  We run the engines long enough to apply a delta-V of 30 km/s.  Our spacecraft is now sitting still in space, with zero angular momentum relative to the Sun.  But it can’t stay there.  The Sun’s gravity is pulling, and the spacecraft will fall into the Sun.  First slowly, then faster and faster, it will hit the Sun in just under 6 months.  If the trip is intended to take less time, then after the first burn has completed, the engines can now be directed to thrust toward the Sun, and thrust applied as needed.  Earlier thrust is more efficient than later thrust.

So, pushing toward the Sun, about 5800 km/s.  Pushing at 90 degrees to the Sun, about 30 km/s.  And yet you rarely see this handled correctly.

Musings on the "Impact" miniseries

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I watched that 4-hour television miniseries, “Impact”, yesterday. I’m now going to set down some of my observations, from a physics and astronomy viewpoint. Even a broadcast like that can teach you something, if it is used as a starting point to explain the things the writers got wrong. And there is a lot of teaching available here, even in the first ten minutes.

OK, we start off with people observing this “biggest meteor shower in 50000 years”. It is seen starting up, so we know the observations were simultaneous. There were groups in New Mexico, the East coast of the US, and in Germany, all watching the meteor shower begin. Sunset times between those locations are as much as 9 hours apart, and in the summer time (when this movie appears to have been set) there aren’t 9 hours of full darkness. This is a common mistake, movies and television shows will often show two participants in a phone call sitting half a world apart, both in full daylight.

Then, that meteor shower was a disappointment. There are recent records of much more intense meteor showers. The Leonid showers of 1833 and 1966 were, from their descriptions, much more spectacular than the shower shown in this movie.

Two astronomers are observing the meteor swarm through telescopes, before it reaches the Earth. We see a field of rocks large enough to be seen through telescopes, and so densely packed as to block sight lines so that astronomers couldn’t see another object at the back of the swarm. This isn’t a meteor swarm, a meteor swarm is rocks smaller than pebbles, separated from one another by kilometres of empty space. This is an avalanche in space.

Next, we find out that an object, visible while it’s still moving in space, was traveling with the cloud of meteors and is going to strike the moon. Seen from the ground, this object had a visibly different track across the sky, which doesn’t make sense if the objects were all traveling together. But if it were traveling in the same direction as the visible meteors, it wouldn’t stand out and so would be less desirable from a dramatic standpoint, so we’ll let that one pass.

So, this mystery object. Let’s forget about the “brown dwarf” babble, and just describe it as a super-dense, magnetized object with, as they say in the movie, a mass twice that of the Earth’s. It hit the moon, and bad things happened.

Now, the science used to explain the effects on Earth is all nonsense, of course. The “levitating frog” experiment did not produce anti-gravity. It exerted a force on a frog. A string tied to the frog’s leg would also exert a force. This was just like that, but it used a magnetic field to apply the force. Gravity was still affecting the frog, but the frog was being supported against the force of gravity by a force of magnetic origin, one related to the gradient of the magnetic field (how much the field changes over a short distance). So, starting from a misunderstanding of an old news release, the writers created weird fantasy effects where objects that are not too small and not too large levitate in spooky ways in random places on the Earth, then crash to the ground. Whatever, we’re not going to talk about that anymore.

OK, back to “small, very heavy object hits the moon”. Our astronomers mention that the moon has 1/6 the mass of the Earth. No, it doesn’t. It has about 1/6 the surface gravity, but only about 1/80 of the mass of the Earth. This is a common mistake, believing that gravity is a function solely of the mass of the object, and ignoring the different sizes. To take a dramatic example, Saturn has almost 100 times the mass of the Earth, but the force of gravity exerted at the cloud tops is not much higher than the force of gravity at the surface of the Earth, because the cloud tops are over 9 times as far from the centre of Saturn as the surface of the Earth is from its centre.

The moon gets hit by something very small that weighs two Earth masses, and is traveling very fast. And they stick together. 160 times the mass of the moon smacks into it with a speed of, let’s say, several kilometres per second. This object wouldn’t stop. It would barely even notice the moon. If the entire moon got in its way, it would sweep it up and continue on its path practically unaware that it was now carrying a moon with it. Since the thing is small, only a bit of the moon gets in its way. It’s a very small and extremely fast bullet striking a very large soap bubble. You don’t expect the soap bubble to be carried away by the bullet, you expect to find a punctured bubble. You certainly don’t expect the bullet to stop dead in the bubble.

160 times the mass. Imagine you’re driving down a highway, and a raccoon is crossing the road. Just as your car is about to hit it, the raccoon jumps straight up and hits the front of your car. Your car stops dead as if it had struck a concrete wall, and the mid-air raccoon is barely pushed at all. Even cartoons don’t try to get you to believe that.

That much mass, stopping all at once within the moon. Just the kinetic energy released is about the same as the total output of the sun over the space of 48 hours. Not the light hitting the Earth, the light leaving the entire solar sphere. The moon would vanish in a puff of gas. The Earth would vanish in a larger puff of gas.

OK, so suddenly the moon weighs twice what the Earth does. This would have some fairly obvious effects. For one thing, the tides on the surface of the Earth would go from a few metres to a few hundred metres in amplitude. That would have a serious effect on the coastal regions (and with those tides, Missouri is a coastal region).

Increase the mass of the Earth-moon system, and the rotational period will decrease. A month would go from about 30 days to about 10 days. But in the movie, the moon was making complete orbits around the Earth on plot-driven timescales. Sometimes the orbital period was a few days, and toward the end of the movie the orbital period seems to have become about 90 days, because they had deduced that the moon would hit the Earth on this orbit, but they still had 40 days left to try to find a solution. And these weird, sudden “orbital shifts” don’t make sense. Yes, an uneven mass distribution can result in smooth and gradual changes to orbits, but the moon didn’t have an uneven mass distribution. It was a big mass travelling in orbit, with a light, insignificant, moon stuck to it like a bug on a windshield. The pre-impact mass of the moon isn’t even an important perturbation on the mass distribution.

Good news, everybody! We just happen to have a lunar expedition fueled up and ready to go, prepared before the impact. Our heroes can fly to the moon and use some special technology to push the big new mass out of the moon. Well, the mass weighs 160 times what the moon does, so Newton’s laws tell us that you’re not going to push the mass out of the moon, you’re going to push the moon away. The mass won’t be appreciably disturbed. The plan is to push the mass out of the moon so it flies toward the sun, but really all you’d do is send the moon away at high speed while the big dense mass stays firmly in its orbit around the Earth.

Now, about this lunar mission. The good news is that you don’t need as much fuel to cross over, because of the changes to the shape of the gravitational potential fields in the vicinity. The bad news is that the moon’s surface gravity is at least 25 times that of the Earth. That assumes that the colliding mass is at the centre of the moon. In the movie, it’s actually only partway down, and our heroes have to land near it, so they’ll feel a gravitational force much higher than that. OK, they’ve been working out, they can walk and work in 25 gravities of force. But their lander was designed to land on rockets in 1/6 normal gravity. It would be like designing a parachute to land you safely, and then you decide that you’ll change the parameters, the parachute will be used to land a bit more weight. You plus 159 of your friends, all hanging on the one parachute. You might reasonably conclude that the parachute was not designed for that kind of treatment. A similar argument can be made for the lunar lander and that little rocket jumper vehicle, whose engines certainly cannot supply the thrust to land under the new conditions. The lander would probably crumple under its own weight just trying to sit still on the moon, and taking off from the surface would be similarly difficult because of the changed conditions.

OK. Gravitation, tides, astronomy, orbital motion. If you learned something new from this movie, it’s almost certainly wrong.