Living Close to a Star

The following is an excerpt from the book, Exoplanets and Alien Solar Systems, by Tahir Yaqoob, Ph.D. Please note that the excerpt is from the eBook so the links to the references are not functional on this webpage.

Excerpt

In this section we are going to do a thought experiment and imagine what it would be like to live on a planet that was orbiting very close to its host star, when the star-planet separation is only a few percent of the Earth-Sun separation.

However, we first need to look at the effects of gravity on the surface of a planet and how gravity determines orbital motion. One of the very first things we learn about as babies is falling. Everyone is familiar with Aristotle's realization that the acceleration of a falling object near the surface of the Earth is independent of the mass of the falling object. If the distance traveled by the falling object is not too great compared to the size of the Earth, then the acceleration is constant, making an object increase its speed of fall by the same amount for every equal interval of time. This happens to be about 21.8 miles per hour per second (or 35.3 kilometers per hour per second) near the surface of the Earth. We are of course neglecting the effects of air resistance in all of this discussion. In other words, when you fall, your speed relative to Earth will increase by 21.8 miles (35.3 kilometers) per hour every second so that it takes a mere 5 seconds of falling to reach a speed of approximately 100 miles per hour (or about 160 kilometers per hour). The actual value of the acceleration near the surface of a planet is determined by how much mass of the planet is packed into a given size, although it is not quite the same as density. Nevertheless, the acceleration is determined by the mass of the planet and the distance between the center of mass of the falling object and the center of mass of the planet. Therefore, on the surface of a planet, the acceleration, or surface gravity as it is otherwise known, is determined by the mass and radius of the planet.

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The first thing to appreciate is that what you feel as your weight is the force of resistance to your tendency to fall. So when you are stationary on the ground and not falling, the back reaction of the Earth in response to you trying to fall into the ground is your weight. When you are in free fall and there is no ground beneath your feet to stop your free fall, there is no back reaction, and under these circumstances you are weightless. Everything in your body is falling at essentially the same rate so there is no back reaction between different parts of your body either. The second thing to appreciate is that an object that is in orbit only under the influence of gravity is also free falling. There is no fixed reference point in space if an object is in orbit around another, and they are really both falling towards each other. Thus the Moon is falling towards Earth and the Earth is falling towards the Moon. The reason why they don't crash into each other (i.e., why the Moon never hits the ground) is that they have a component of motion that is not along the line joining their centers, so they are in effect always trying to fall to hit each other but never do because their direction of motion is changing at every instant so they continuously miss each other. In the special case of a circular orbit, the physics works out such that the size and speed of the orbit take on values such that the amount that two objects fall towards each other is exactly compensated by their motion in a direction perpendicular to the fall so that the two objects remain at a fixed distance apart. Of course, if there are any energy losses (for example in the case of a satellite around Earth suffering from atmospheric drag), the two objects do get closer to each other, resulting in a decaying orbit, which ultimately does result in a crash. In general, orbits are elliptical, and the parameters of the ellipse depend on the initial conditions, such as how the two objects ended up in orbit, including the initial energy of the system, and the relative initial direction of motion. Whether an orbit is elliptical or circular, the objects partaking in the orbital motion are in free fall and therefore weightless.

Now let's consider the fact that Earth is also in orbit around the Sun. Aren't the Earth and the Sun falling towards each other as well? Yes, they are. Moreover, you are falling towards the Sun as well because you orbit the Sun as well. Associated with every acceleration there is a force, so what is the equivalent force that is pulling you towards the Sun? On the surface of the Earth the force from the ground that is stopping your fall is your weight. We can express the force between you and the Sun as some percentage of your regular Earth weight to get a sense of how small it is. The size of that force is a competition between the mass of the Sun (which is about a third of a million times that of Earth), and the distance between you and the Sun. It turns out that the force of the Sun pulling on something on the Earth's surface (e.g., you) is about 1,600 times smaller than the weight of that object on the surface of the Earth due to Earth's gravity. That is, about 0.06% of an object's weight is a good estimate of the force between that object and the Sun. For example, if you weigh 160 pounds (72.6 kilograms), you are being pulled towards the Sun by an equivalent weight of approximately 0.1 pounds (about 45 grams). You might think that is small, but it is not negligible: that is probably the weight of your ear! More importantly, if you were not in orbit around the Sun and were allowed to fall towards it head-on, you would reach a speed of 100 miles per hour (160 kilometers per hour) in just 2 hours and 2 minutes due to that “tiny” force of 1.6 ounces (45 grams).

There's more, because now we realize some interesting consequences. Suppose you are standing at a position and time on Earth such that the Sun is directly overhead and you weigh yourself and find that you weigh 160 pounds. The Earth is pulling you straight down but the Sun is pulling you straight up. Your net rate of falling (acceleration) towards Earth is slightly less because of the small opposing tendency to fall towards the Sun. Therefore your net weight is slightly less than it would be without the Sun. What happens when the Earth has spun half a revolution so that the Sun is on the other side of Earth? Well, you will want to fall towards the Earth and towards the Sun as before, but now these forces are in the same direction and not in opposition so the net result will be that you will weigh slightly more than you would without the Sun's gravity. So we have the curious situation that you would weigh less during the daytime and more during the night! To be sure, the spin of the Earth is associated with centripetal forces as well so the situation is much more complicated but the general point of this discussion is that the Sun's gravity can work both in opposition and in the same direction as an object's gravitational attraction towards its home planet. On Earth we don't notice this effect but imagine living on a planet that was so close to its host star that the star's gravitational effect at the surface of the planet was comparable to the planet's surface gravity due to its own mass. Then we get some very interesting effects. Let's ignore for the moment that you would not be able to survive on such a planet because it would be so hot that you would be evaporated. (You could imagine that your government has built heat-resistant, cocooned dwelling areas for you to live in.)

In the morning, at “star-rise” you will feel yourself being pulled towards the horizon. The details depend on how your planet's spin axis is oriented, but for the sake of argument, let's suppose that you are on the equator and the planet's spin axis is perpendicular to its orbital plane. As well as wanting to fall towards the horizon, you will still want to fall to the center of your planet so your net weight will be in a diagonal direction into the ground. If you climbed onto a fence and jumped off, your “down” direction will be diagonal. As the morning wears on you feel the direction of “down” changing to become more and more horizontal. As the morning coffee break approaches you feel the urge to be dragged along the ground towards the horizon. The feeling passes and then you start to feel lighter and lighter as the star's gravitational pull becomes more and more “vertical.” As lunchtime approaches and your star approaches the overhead position you continue to feel lighter, until finally, at noon, you could actually be weightless and float up into the air (depending on the relative strengths of gravity from your planet and your star, at your position).

You can imagine how complicated the infrastructure of a city on such a planet would be in order to accommodate everyone, with everything starting to float around at lunchtime. Lunchtime cafes could be situated at the top of large pillars, for example (so that you could just “float” in). Parachutes could be available for people who have lingered too long after lunchtime. That's because after the star has passed overhead you start to feel heavier and heavier again as the planet's gravity once again dominates. At “star-set” you feel pulled towards the horizon, in the opposite direction that was “downwards” in the morning. At night your star is on the other side of your planet and as the night progresses you feel heavier and heavier. When the star is directly opposite you (with the planet in between you and the star), you feel incredibly weighed down and barely able to move. A stark contrast to the floating fun you were having at lunchtime.

What if your star's gravity were so strong that at high noon your star's gravity doesn't just balance the surface gravity on your planet, but is so strong that it dominates? Unfortunately, there would then be a possibility that you could be whisked off into space and end up escaping from your planet, possibly going into an eternal orbit around your star (again, ignoring evaporation). Presumably, your government will have built precautionary structures in heavily populated areas, but being caught in an unprotected area at lunchtime would carry a high risk. These are not frivolous observations. On the contrary they have direct relevance to the close-in exoplanets that are mysteriously underdense. You see, what if the planet is not made of rock but is gaseous or liquid? If the planet is so close to its host star that the starlit side of the planet experiences a gravitational force from the star that is comparable to, or larger than, the local surface gravity, what is going to happen? The planet is held together by gravity, and the gravity usually maintains the spherical shape because in equilibrium the inward collapse due to gravity is exactly balanced by the back reaction due to the compressibility limit. However, if the star's gravity in the part of the planet that is facing the star is so strong that it overcomes the local surface gravity, that material there will become weightless and may even become unbound from the planet. In other words, the planet could undergo continuous loss of mass if the material is gaseous or liquid (the liquid may of course vaporize in the process). Certainly, there will be a distortion in the shape of the planet, as the gravitational competition between the star and the planet will vary over different parts of the planet. The planet will be elongated along the line joining the planet and the star. The material on the dark side of the planet will experience reinforced acceleration in the same direction (towards the star) and that material will therefore be heavier, and the least likely to escape. The effects are similar to gravitational tides, but not quite. Tidal stress usually refers to the stress in the structure of the planet caused by the star's gravity having a different strength over different regions of the planet, causing it to be pulled apart and broken. Instead, what I have talked about here is simply the strength of the local surface gravity of the planet at a given point on the planet, compared with the acceleration towards the star at the same point. If the two are comparable, material will be simply lifted off from the surface layers of the planet. The process is dynamic because as the mass loss proceeds, the rate of loss will change because the surface gravity will change. The star-planet distance is also likely to change as the planet becomes less massive. The details will depend on the exact structure of the planet, amongst other things.

So, what about the new, close-in, anomalously underdense exoplanets that have been discovered? At the surface of these planets, which is stronger, the surface gravity due the planet itself, or the host star gravity?

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