Newton's law of gravity
Almost everyone has heard the story of Newton sitting under an apple tree and being inspired towards understanding gravity thanks to a falling apple. You may have heard the version of that story where the apple falls on Newton's head but I recently read that it actually landed next to him. I am more inclined to believe this version of the story because I know that if an apple fell on my head I'd be too busy cursing the tree to have a flash of inspiration.
What was the revelation Newton had about gravity? Well basically, he gave us a simple, universal description of gravity. The same force that causes the apple to fall towards the ground also keeps the moon orbiting the Earth and the Earth orbiting the sun. The mathematical expression for the force of gravity between two objects that Newton left us with is:
F is the force of gravity between two objects of masses M and m kilograms, with their centres separated by distance r meters. G is the gravitational constant and is equal to 6.67 × 10-11 km3 kg-1 s-2 .
However, in our day-to-day experiences, it is not forces, per se, that we are most aware of but accelerations. For example, if you are on a train moving at a constant speed (not accelerating, that is; not slowing down or speeding up), you can't tell how fast you are going solely from its motion. The only thing that would give it away is the bumping up and down thanks to uneven tracks. However, it's easy to tell when the train is slowing down or speeding up because you are either pushed backwards or forwards in your seat (depending on which way you are facing).
Although gravity is always pulling us towards the centre of the Earth, what we actually feel is the ground (or floor or chair or whatever) holding us up and preventing us from falling towards the centre of the Earth. It's gravity accelerating us into the floor that we experience as weight. If there was no floor and we were just falling, we would actually feel weightless (air resistance notwithstanding). The International Space Station is well within Earth's gravitational field but it is constantly falling, which is what makes the astronauts inside feel weightless. Luckily it's not just falling straight down; it also has a horizontal velocity which means that in the time it falls downwards a certain amount, the curvature of the Earth results in the ground also falling away by the same amount. This is called an orbit, and I will talk more about them in the next section.
Back to the equation above. How do we turn this into what we experience every day on Earth and then into what we would experience if we were living on another planet. The important quantity to calculate is g, the acceleration due to gravity. Every force (or sum of forces) can be described as an acceleration acting on a mass. F = ma is the famous equation and is the mathematical representation of Newton's second law (side fact: Newton's law of gravitation didn't get a number, his three laws refer to more basic laws of mechanics). To find acceleration due to gravity on the surface of a planet, we equate F = mg (rather than F = ma, since g is the symbol we use for acceleration due to gravity) with Newton's law of gravity:
To find g at the surface of the Earth, you need to substitute in the mass of the Earth for M and the radius of the Earth for r. On Earth, g = 9.8 meters per second per second, which means that, if we ignore air resistance, something that is falling will gain 9.8 m/s of speed each second. It also means that the force with which we are constantly pushed into the ground/chair/bed is equal to our mass times 9.8.
On planets other than Earth which are smaller, larger, heavier or lighter, there are different accelerations due to gravity which we can find by throwing the right numbers into the equation above. Some examples from rocky† planets and moons in our solar system (where g is acceleration due to gravity on Earth's surface):
- Moon: 1.7 m/s2 = 0.17 g (about a sixth of Earth's gravity, so you would feel a sixth as heavy.)
- Mars: 3.7 m/s2 = 0.38 g (between a third and two fifths of Earth's gravity)
- Mercury: 3.8 m/s2 = 0.39 g (coincidentally very close to Mars)
- Ganymede: 1.5 m/s2 = 0.15 g (also about a sixth of Earth's gravity)
And because this is easily applicable to extrasolar planets — that is, planets outside of our solar system — I have also calculated some surface accelerations due to gravity for a few known exoplanets (links below are to Wiki, but I got my values from the Exoplanet iOS app; see below).
- Gliese 1214 b: 8.6 m/s2 = 0.88 g (Just under nine tenths that of Earth. However, it's just inside its host star's habitable zone (the star is called Gliese 1214) which means it'll probably be too hot for human habitation. It could even have a runaway greenhouse effect like Venus. It's also possible that this planet has a very thick atmosphere making it more similar to a small gas giant like Neptune than to Earth.)
- CoRoT 7 b: 18.4 m/s2 = 1.9 g (Just under twice Earth gravity so you would feel almost twice as heavy and, more vitally, your organs would all press down on each other twice as strongly. According to NASA (pdf, sorry), this isn't terribly sustainable in the long term for humans as we are now. Personally, I don't think it would take an awful lot of genetic engineering to fix this for us (we are talking science fiction, after all). The bigger problem with this planet is that it's much to close to its sun for our comfort or survival.)
- Kepler 11f: 3.4 m/s2 = 0.35 g (About a third of Earth's gravity. Compare with Mars or Mercury. Unfortunately, it's also slightly too close to its star to be habitable. Incidentally, the whole Kepler 11 system is quite interesting with six confirmed planets so far.)
* Technically, the distance from the centre of mass, but for round or roundish things like planets the centre of mass is generally the centre of the planet.
† Rocky because you can't stand on the surface of the gas giants. It's possible to calculate the acceleration due to gravity experienced by a hovering platform or similar, however.
As we've established, gravity is what keeps things in orbit around other things. As such, we need to use what we know about gravity to work out how fast something has to orbit for different distances and masses of objects. In fact, Kepler had worked this out to some degree before Newton came along, but Kepler's third law was slightly less specific than could be calculated using Newton's law. Kepler realised that for orbits, the ratio between the cube of the semi-major axis and the square of the period was constant. The semi-major axis is the same as the distance to the larger body from the smaller (like r in the equations earlier) for circular orbits and half the length of the longest side of an ellipse (oval). the period, T, is the time taken to complete one orbit, so if we're talking about planets, then it's the length of a year. We can derive the constant part of Kepler's third law using Newton's law of gravity and the equation that describes centripetal force.
As it happens, I talked about a lot of the ingredients for working out how fast a planet should orbit around its star in my recent post about space elevators. The set of equations below starts by equating the centripetal force (the force required to keep something of mass m moving around in a circle with radius r and at speed v) with the gravitational force (centripetal on the left, gravitational on the right of the equals sign), then shows the derivation of Kepler's third law (the second last line) and finally gives the period, T, of a planet orbiting around a star of mass M at a distance r. Feel free to let your eyes glaze over if maths isn't your thing. You have been warned.
The final line gives us the period in seconds, which for most things isn't terribly helpful. To find out what the length of your planet's year is in days or Earth years you will need to divide your answer for the period by 86400 or 3.15 × 107 respectively.
The reason I chose to rearrange Kepler's law to solve for the period rather than for the semi-major axis is because for science fictional purposes, the distance from the star is more likely to be fixed for plot purposes. For example, an human-inhabited planet has to be in the habitable zone. Exactly what the habitable zone is will be, I think, the subject of a future blog post.
I should also point out that these equations are completely applicable to man-made satellites or moons as well, you just need set the planet's mass to be M instead of the star's mass. Just remember that if your comms satellite is x km above the surface of the Earth/whatever planet, you have to add on the radius of the planet to find r to throw into the equations I've talked about today.