# A Brief Rundown We've all heard about Newton's run-in with the apple. It's universally taught in primary schools across the world, his stroke of brilliance a shining light in the formative years of today's youth. They don't tend to teach you gravitation in pre-school though. One day... # Newton's Law of Gravitation Newton himself realised gravity obeyed a certain **inverse-square law**, where the strength of gravity decreased incrementally with the relationship: $F = \frac{k}{r^2}$ Where $r$ is the radius from a point source while $k$ is a constant. Below is a diagram that visualises this. The number of square panes, used to visualise the gravitational strength per unit area, increases constantly with distance, as the total gravitational force remains the same. ![[inverse square law (light-brightness).png|720]] %%replace this diagram%% For example, the gravitational force on an asteroid two times closer to the sun than a comet would be four times the force on the comet. So, Newton figured out that gravity obeyed a similar relationship, which allowed him to create his equation. $F_{g} = G\left( \frac{M_{1}M_{2}}{r^2} \right)$ Where $G$ is the Universal Gravitation Constant, $6.67 \times 10^{-11}$ >But why isn't it an inverse-cube law? Objects are three dimensional, surely the strength should decrease cubically! Well, we can imagine the strength of gravity as depreciating as spheres, yes. **However**, we're actually taking the strength of gravity as a measure of the **surface area** of the sphere, not the volume. That's why it's an inverse-square law! # Superposition - Gravitation Edition For a system that is gravitationally bound to each other, the total gravitational force on each **individual** particle can be expressed using the *principle of superposition*: $F_{1,net} = F_{2} + F_{3} + F_{4} + F_{5} \dots F_{n}$ Let's generalise this into a summation for our own eyes: $F_{1,net} = \sum_{i=2}^n F_{1i}$ Note that force is a vector, meaning the net gravitational force $F = G \frac{m_{1}{m_{2}}}{r^2}$ is a slow and steady process of addition. As a result, the gravitational influence on a point can be generalised for all points $F_i$. ## A Case Study - The Gravitational Force of a Ring We can take a uniform (infinitely small) mass element on the ring called $dm$. %%diagram! you know what i'm talking about. %% ## Useful Example - The Gravitational Force of a Disk # Gravitation on a Surface We humans typically live on the surface of planets. Advances in technology have given us the opportunity to transcend this boundary, but only for so long. So, gravity acts on bodies even just a few metres away from its' centre of mass, meaning that there's a gravitational acceleration on a body on the surface. So, how do we model this? Let's pretend that there's a human standing in a field on the Earth. Like any good physics question, this human is now a sphere, a phenomenon that is growing ever more common in developed countries. So, let's just visualise this with Newton's law of Gravitation. $F = \frac{G(M_{1}m)}{r^2}$ Where $m$ is the mass of the human. This formula actually gives the **resultant force** on the object thanks to gravity. Therefore, we can use Newton's second law $F=ma$ to find the gravitational acceleration of **all** bodies on a surface, as $m$ can vary: $a = \frac{G(M_{1}m)}{r^2m}$ $a = \frac{GM}{r^2}$ >[!Abstract]- Further Constraints >Note that these equations are given for a perfectly spherical body with a constant value of $g$ across it. Note that there are three conditions we need to take into account when creating an accurate model for surface gravitational acceleration: > >1. **The Earth's mass is not evenly distributed**. In fact, the density of Earth varies radially, with the inner core having a much higher density than the outer core. >2. **The Earth is not a perfect sphere**. The Earth is actually an ellipse. As a result of its rotation, it is flattened towards the centre (see [[Uniform Circular Motion (Mechanics)|centrifugal force]]). This is why measurements of $g$ increase when on sea level on the equator compared to sea level on the North Pole. >3. **The Earth is rotating**. There is a centripetal acceleration directing every object to the centre of the earth, corresponding to a centripetal net force pointing to the centre of the Earth. > >This means there is a difference in the gravitational accelerations that we must account for. Note that for every object on the Earth there is a centripetal acceleration component, which we denote as $\omega^2r$. Using Newton's Second Law for gravity $F = ma_{g}$ we can create an equation: >$F - ma_{g} = -m\omega^2r$ >Letting $F$ be the normal force on the object. This force can therefore be rewritten as the weight of the object $mg$, giving us: >$mg - ma_{g} = -m\omega^2r$ >Dividing by $m$ on both sides gives us a value for the difference between the nominal gravitational acceleration and the actual value we get: >$g - a_{g} = -\omega^2r$ ## Gravitation within an object Objects still fall towards the centre of mass regardless of if they lie on the surface of a sphere or any object. Using $\rho = \frac{M}{V}$, we can get the equation for $\rho$, assuming a sphere of uniform density: $\rho = \frac{M_{ins}}{\frac{4}{3}\pi r^3} = \frac{M}{\frac{4}{3}\pi R^3}$ %%add steps%% $\rho = M_{ins} = \frac{M\left( \frac{4}{3}\pi r^3 \right)}{\frac{4}{3}\pi R^3}$ We can therefore find the mass of the internal sphere as follows: $M_{ins} = \frac{4}{3}\pi r^3\rho = \frac{M}{R^3}r^3$ Where: - $\rho$ is the average density of the sphere - $r$ is the radius of the sphere within the larger sphere - $R$ is the radius of the larger body (sphere), where $R^3$ is the volume $V$. This gives us a gravitational force (by substituting $m_1$ for the equation above) of: $F = \frac{GMm}{R^3}r^3$ >[!Danger]- Generalising for more shapes >We can generalise this formula for other shapes! Take the ellipsoid (3D ellipse) for example: >$\rho = \frac{M_{ins_{el}}}{4\pi a_{i}b_{i}c_{i}} = \frac{M_{el}}{4\pi abc}$ >$M_{ins_{el}} = \frac{M_{el}}{ abc}a_{i}b_{i}c_{i}$ >Where $a$ is the semi-major axis of the flat ellipse, $b$ is the semi-major axis of the flat ellipse, $c$ is the value of the vertical radius of the ellipsoid, $M_{ins_{el}}$ is the mass of the ellipsoid embedded within the larger body and $M_{el}$ is the mass of the larger ellipsoidal body. By expressing our force in vector form in terms of the position vector $\overrightarrow{r}$, we can get one last expression: $\overrightarrow{F} = -K\overrightarrow{r}$ Where $K$ is a constant of simple harmonic motion. This shows that orbits are actually [[Simple Harmonic and Pendulum Motion (Waves)|simple harmonic oscillations]] - graphing their position-velocity diagrams them will give you a sinusoidal! >[!Danger]- More on Orbits and Oscillations As a result, given a point mass within a surface a general expression for the gravitational acceleration is given by: $a = \frac{4}{3}\pi r\rho G = \frac{GM}{R^3}r^3$ >[!Success]- Finding the Gravitational Force within the Earth >Let's assume that the Earth is a perfect sphere. This means that we can create an equation for the gravitational acceleration: >$a = \frac{4}{3}\pi r\rho G$ >$F = mG \frac{4}{3} \pi r \rho $ >These give us models for points within an object - where we can trace out arbitrary spheres with radii $r$ within said object. >[!Danger]- A More Accurate Earth >Note that no rotating body is actually spherical - there is a higher torque caused by angular momentum acting tangential to the rotation of the body, while the torque at the poles is actually lower, leading to polar flattening. As a result, it would be wiser to interpret the earth as an ellipsoid - though for most bodies a sphere is accurate enough as a model. > >One object that we can't model using this is the star Achernar - a star that spins so fast that it visibly bulges at the middle. Delighting! # Example Questions >[!Example]- Module 13-2 Question 9 >We want to position a space probe along a line that extends directly toward the Sun in order to monitor solar flares. How far from Earth’s centre is the point on the line where the Sun’s gravitational pull on the probe balances Earth’s gravitational pull? > >**Solution:** > >We're going to need the numbers for this one - the mass of the sun is $1.98 \times 10^{30}kg$ and the mass of the Earth is $5.97 \times 10^{24}kg$. Since the forces on the object must be the same (which is also the definition for a Lagrange point %%add link later%%), we can write the expression as a function of two gravitational forces: >$\frac{G(1.98 \times 10^{30}kg)}{r_{S}^2} = \frac{G(5.97 \times 10^{24}kg)}{r_{E}^2}$ >Where $r_E$ is the distance from the Earth to the object and $r_S$ is the distance from the probe to the sun. >Take the gravitational constant as $G = 6.67 \times 10^{11}$ to further simplify our expression: >$\frac{1.98\times10^{20}}{r_{S}^2} = \frac{1.32\times10^{14}}{r_{E}^2}$ >If we take the sun's distance from the Earth at aphelion at 150 million km, we can express the two distances as ratios of each other: >$1.98\times10^{20}r_{E}^2 = 1.32\times10^{14}r_{S}^2$ >$r_{E}^2 = \frac{1.32\times10^{14}r_{S}^2}{1.98\times 10^{20}} = 1.5 \times 10^{-6}r_{S}^2$ >That's an order of magnitude lower if I've ever seen it! Take the square root of both sides: >$r_{E} = (1.225 \times 10^{-3})r_{S}$ >Since $r_E + r_{S} = 1.5 \times 10^8$, we can rewrite the expression as: >$r_{E} = (1.225 \times 10^{-3})(1.5\times 10^8 - r_{E})$ >$r_{E} = 183700 - 1.225 \times 10^{-3}r_{E}$ >$(1 + 1.225 \times 10^{-3})r_{E} = 187300$ >$r_E \approx 187070 \ km$ # What's Next? For the next instalment in the "Gravitation" series, we'll be taking a look at GPE: [[Gravitational Potential Energy]] Navigate back to the physics landing page here: [[The (Incomplete) Physics Almanac]]