The Sun is a massive object. To put it into perspective, about one million Earths could fit inside it… Another fun fact is that if space could carry sound waves, the Sun would be as loud as a jackhammer! Even from the distance of 93 million miles!
Hello all and welcome back to another post in my astrophysics series! Today’s topic will be focused on using the equation for gravitational force to calculate the mass of the Sun. Following this, the concept of gravitational potential energy will be investigated. Without further adieu let’s get into it!
To start off today’s calculations, we’re going to begin with two equation of force and set them equal to each other. One of the equations you learned in the previous post. The equations are:
$$\mathcal{F} = \frac{G m M}{r^2}$$
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, $\frac{V^2}{r_E}$ 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,
$$\frac{G m M}{r^2} = \frac{M_E V^2}{r_E}$$
Now substitute the correct variables in for the gravitational force equation:
$$\frac{G M_E M_S}{r_E^2} = \frac{M_E V^2}{r_E}$$
Where:
Rearranging the problem algebraically and cancelling where necessary so that we solve for the mass of the Sun, we have:
$$M_S = \frac{r_E V^2}{G}$$
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.
Here, we saw that we can manipulate equations and set them equal to each in order to solve for a specific property, in this case, mass.
In future posts, we may derive the gravitational potential energy equation, but for now it’s sufficient to just state it:
$$\mathcal{U} = -\frac{G m M}{r}$$
From the outset, this looks remarkably similar to the gravitational force equation. However, there is one important difference. In the denominator there is only an r and not an $r^2$. In addition, a crucial difference is this. the whole value of the force is preceeded by a negative sign. What does this mean?
Simply put, when you are solving for energy, but the equations dictates that that energy is negative, it means that the object in question is bounded by something. In atoms, electrons are bound to the nucleus, and in scenarios involving gravity, small objects are bound to very large objects. Also, this means that as the absolute value of the energy decreases, the negative sign means that the energy is really becoming less negative. Less negative really means that the energy is increasing overall so that when the energy reaches a value of precisely 0, it has just enough energy to break away from whatever it was bound to. This concept will become very important once escape velocity is presented.
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:
$$\vert \mathcal{U}\vert = (K + heat)$$
This is the virial theorem in more specific language: Exactly half of the change in potential gravitational energy is lost as heat.
The significance of this theorem is displayed when examining stars, but that’s not till later. For now though, I hope you enjoyed this post and perhaps learned something new. This is still relatively simple and basic, but I hope you enjoyed it all the same.
In my next post, we’ll take our first look at electromagnetic radiation. For all that and more, I’ll see you in the next post!