diff --git a/content/posts/caclulating_sun.md b/content/posts/calculating_sun.md similarity index 93% rename from content/posts/caclulating_sun.md rename to content/posts/calculating_sun.md index 8f65635..a369757 100644 --- a/content/posts/caclulating_sun.md +++ b/content/posts/calculating_sun.md @@ -21,16 +21,16 @@ The other equation is: $$\mathcal{F} = {m a}$$ -Now, we can use the second equation by substituting the mass for the Mass of the Earth: $5\times10^{24}$ kilograms. Then, we can then substitute the acceleration for velocity squared divided by distance i.e. the velocity of the Earth as it is going around the Sun squared, divided by the Earth's distance from the Sun. Thus, $\big\frac{v^2}{r_E}\big$ -In summary we now have $\mathcal{F} = \frac{M_E v^2}{r_E}$ which relates the mass, velocity, and distance in terms of the Earth. Note that the velocity of the Earth here equals 30,000 m/sec. +Now, we can use the second equation by substituting the mass for the Mass of the Earth: $5\times10^{24}$ kilograms. Then, we can then substitute the acceleration for velocity squared divided by distance i.e. the velocity of the Earth as it is going around the Sun squared, divided by the Earth's distance from the Sun. Thus, $\frac{V^2}{r_E}$ +In summary we now have $\mathcal{F} = \frac{M_E V^2}{r_E}$ which relates the mass, velocity, and distance in terms of the Earth. Note that the velocity of the Earth here equals 30,000 m/sec. Now here's the important part. We are going to take that prevous equation we just made for force and *set it equal* to the equation of the force of gravity, since both equations are equating force. This is something that is done all the time when trying to calculate properties of celestial bodies, whether that be momentum, acceleration, mass, etc. Thus, -$$\frac{G m M}{r^2} = \frac{M_E v^2}{r_E}$$ +$$\frac{G m M}{r^2} = \frac{M_E V^2}{r_E}$$ Now substitute the correct variables in for the gravitational force equation: -$$\frac{G M_E M_S}{r_E^2} = \frac{M_E v^2}{r_E}$$ +$$\frac{G M_E M_S}{r_E^2} = \frac{M_E V^2}{r_E}$$ Where: - $M_E$ is the mass of the Earth @@ -38,7 +38,7 @@ Where: Rearranging the problem algebraically and cancelling where necessary so that we solve for the mass of the Sun, we have: -$$M_S = \frac{r_E v^2}{G}$$ +$$M_S = \frac{r_E V^2}{G}$$ Putting the correct values for the velocity of the Earth, the distance to the Sun, and the gravitational constant, we can solve for the mass of the Sun. Here we see that the value is $1.99\times10^{30}$, a value that is remarkably close to the accepted value for the mass of the Sun. @@ -58,7 +58,7 @@ Simply put, when you are solving for energy, but the equations dictates that tha As we're talking about gravitational potential energy, it may be wise to introduce a topic called the Virial Theorem. This theorem states that the absolute value of the *change* of the potential gravitational energy is equal to kinetic energy plus heat. Mathematically, it is as follows: -$$\vert \mathcal{U} = (K + heat)$$ +$$\vert \mathcal{U}\vert = (K + heat)$$ This is the virial theorem in more specific language: Exactly half of the change in potential gravitational energy is lost as heat. diff --git a/public/categories/astronomy-and-astrophysics/index.html b/public/categories/astronomy-and-astrophysics/index.html index abe1db2..7577273 100644 --- a/public/categories/astronomy-and-astrophysics/index.html +++ b/public/categories/astronomy-and-astrophysics/index.html @@ -36,7 +36,7 @@