Calculating exoplanet properties - President's Dream Colloquium - Simon Fraser University
m mass of a planet or satellite. M mass of the sun. G. Newton's gravitational constant. The period of revolution. This calculation is for the case of a circular orbit. which is the amount of work done on the object by the planet when the object is initially "infinitely" far out in space but then proceeds near the. However, at large distances from the Earth, or around other planets or moons, the acceleration is different. The acceleration due to gravity depends on the mass.
An example of using the mass-luminosity relationship to determine stellar mass is illustrated in the image below. From the image, it can be seen that stars which are twice as massive as the sun, such as Sirius, are more than eight times as bright.
That is, we can say that there is a linear relationship between the logarithm of the actual luminosity of a star and the logarithm of its mass.
Acceleration Due to Gravity Formula
The period'P', is the orbital period of the exoplanet, and comes directly from the measured period using, for example, the transit or radial velocity detection methods Detection Methods page. The mass of the star, 'M', was calculated above using the mass-luminosity relationship of stars. Finally, the mass of the exoplanet, 'm', in the equation can be ignored, since it is much smaller than the mass of the parent star. As an example, since the Sun is about three hundred thousand times heavier than the Earth, ignoring the mass of the Earth in this calculation woud introduce an error of less than 0.
The equation can be solved for the only remaining variable which is the orbital radius, 'a'. The force of gravity can be determined from the Doppler shift measured using the radial velocity method.
The equation can be solved for the final remaining variable, 'm2', which is the mass of the exoplanet. To determine the planet radius, the brightness drop of the parent star that occurs during a planetary transit is measured.
This brightness drop is directly related to the ratio of the planet radius to the radius of its parent star, as shown in the image below. Drawing Conclusions from Exoplanet Properties What sort of conclusions can be made about an exoplanet given these properties?Physics - Mechanics: Gravity (11 of 20) Eccentricity Of A Planet's Orbits
First, the orbital radius plays a key role in determining whether or not the planet can support life. If the planet is too close to its parent star, the planet will be tremendously hot; the planet's molecules will be travelling so quickly, and with so much energy, that most of the chemistry that is seen on Earth will not be possible.
Indeed, it is unlikely that any molecules capable of supporting life will form, and certainly not liquid water. Planets that are sufficiently close to their star will also likely be stripped of any atmosphere that they might have had.
Many scientists assume that liquid water is necessary for life, and if this is true, then there are lower and upper bounds on the orbital radius which would allow the exoplanet to harbour life. Since we are defining everything in terms of the weight or pressure per unit of area, the area of the layer can be taken as one unit, and we can ignore that.
As a result the extra weight dw would be equal to dr times the local value of the planet's internal gravity, gr, times the thickness of the layer, dr. Note that the letter d is used here in two ways -- as the density, and as an indicator that we are using a small amount of some unit. With these definitions and statements in mind, we can now calculate the pressure and weight at various depths inside a planet simply by adding up all the values of dw and dp, starting at the surface of the planet and going downward to its center.
Up to this point, although the mathematics appears difficult all we have been doing is defining things and rearranging them, and no actual arithmetic has been done. In other words, at least in the case of planets with uniform density, the pressure at various points inside the planets is directly proportional to the square of the surface gravity, exactly the same result we obtained when presuming that the gravity inside was everywhere the same as at the surface.
This does not, of course, mean that things work out so nicely when the density changes from place to place, and the way in which the pressure changes as you go into the planet would not be the same if calculated using the incorrect assumption of a uniform gravity as it is when calculated correctly, but it does show that if we want to compare the pressures inside various planets, squaring their surface gravities isn't a bad first approximation.
Finally, having calculated the actual value for pressure at various points inside a uniform planet, we can calculate how the pressure changes from place to place, as shown in the diagram below. Near the surface, where the gravity is close to the surface gravity, pressure increases more or less uniformly, but as you near the center of the planet, where the gravity approaches zero, the pressure levels off and becomes nearly constant in the central parts of the planet.
The pressure inside a uniform planet.