Ocean Planets on the Brink of Detection

ZonkerWilliam writes "It seems, at least theoretically, that there may be 'ocean planets' out there in the galaxy. If there are, we are closer than ever to detecting them. The formation of such planets is fairly likely, reports the PhysOrg article, despite the lack of an obvious example in our own solar system. We may have a former ocean planetoid in the neighborhood, orbiting the planet Jupiter: the moon Europa. These water worlds are the result of system formation castoffs, gas giant wannabes that never grew large enough. If any of these intriguing object exist nearby, the recently launched CoRoT satellite will be the device we use to see it. The article explains some of the science behind 'ocean worlds', as well as the new technology we'll use to find them."

Comments

Time to get my geek on.

[Lendrick]

If you know how far away you are from an object and how quickly you're orbiting it (assuming your orbit is roughly circular) you can use simple algebra to get a rough idea of its mass.

Acceleration due to gravity is calculated as follows:

a = G * (m / r^2) ...where a is the accelelration, G is the gravitational constant, and r is the distance between your two objects. Note that we're ignoring the acceleration of the sun toward the earth, which isn't technically correct, but this answer will be close enough.

Since we're looking for the Sun's mass, we solve this equation for m.

m = (a * r^2) / G

The first thing we need to figure out is the value of a, or how fast things accelerate toward the sun. The earth is 1.5e11 meters from the sun, and travels in a (roughly) circular orbit once every 365.25 days (or 3.16e7 seconds). If you calculate the circumferance of the earth's orbit given the radius, you get 9.42e11 meters. The earth is moving at roughly 2.98e4 meters per second.

The next step is to figure out how far the earth falls toward the sun every second. We can do this (again, roughly) without using calculus. Let's say that, for one second, the earth continues to travel in a straight line instead of a circle. If you subtract the earth's real orbital radius from this hypothetical one, you end up with the number of meters that earth falls every second, or a. Note that this isn't an exact calculation -- I would need to use calculus to do that -- but it's still "close enough". I'm an engineer, not a scientist, so be happy I used 3.14 for pi, as opposed to "about 3." :)

The earth's new distance from the sun, if it travelled at a tangent for sone second, would be calculated using the Pythagorean Theorum, as follows:

d = sqrt(1.5e11 ^ 2 + 2.98e4 ^ 2) = sqrt(2.25e22 + 8.88e8) = 150000000000.00296

Subtracting the original distance from the sun, the earth has fallen about 2.96 millimeters in one second, which means that the earth is accelerating toward the sun at .00592 m/s. That's a. Now we just plug all that into the original equation:

m = 0.00592 * 1.5e11^2 / G

According to Google calculator:
((0.00592 (m / (s^2))) * (1.5e11^2) (m^2)) / gravitational constant = 1.9961037 × 10e30 kilograms

Now, looking up the mass of the sun:
mass of the sun = 1.98892 × 10e30 kilograms

Voila, I've just calculated the mass of the sun with less than 1% error, and I didn't even need to remember any calculus. :)