# Intro - Falling Objects *See [[Power, Work Done and Energy#Potential Energy|potential energy]] for prerequisite knowledge.* Notice that it's considered common knowledge for objects to speed up as they fall. You drop an apple on top of an unsuspecting teachers' head, they'll get a big bruise, you'll get a stern talking to. It's time for us to generalise this phenomenon - known as gravitational potential energy! %%diagram%% For objects falling linearly (that is, parallel to the Earth (the horizontal plane)), the formula for GPE is as simple as: $U = mgh$ Where: - $g$ is the gravitational constant - $h$ is the linear distance above the ground of the object In a perfect vacuum, thanks to [[Power, Work Done and Energy#Conservation of Energy (Principle of)|conservation of energy]] all of this potential energy will be converted into kinetic energy, creating the identity: $mgh = \frac{1}{2}mv^2$ # Continuing with GPE %%diagram queue!!!%% However, in a setting where the objects instead exert **gravitational forces** on each other, we can take the formula for work done $W = \int F(x) \, dx$ and the potential energy equivalence $W = -\Delta U$ to get a generalised expression for GPE: $U = -\frac{GMm}{r}$ >[!Success]- Doing the Maths - Proving the Potential Energy Equation (Detailed) >Since $W = \int F(x) \, dx$, we can find the work done of the object as: >$W = \int F(r) \cdot dr $ >Note that from a radial distance $r$ from the main body's centre that we can model the gravitational force exerted an object with lower bound $r$ and upper bound $\infty$. This is as the gravitational force requires an infinite distance to stop influencing an object. We can model this change with the variable $dr$ as the object moves away from the larger parent body. > >As these two quantities are vectors, it is possible to create a [[Vector Operations (Maths)#The Scalar (Dot Product)|dot product]] between the force vector and the change in displacement between the two objects $r$ (Note that an integral is just multiplying by a ratio $dr$) As a result, we can use the identity $|a \cdot b| = |a||b|\cos \phi$ to get: >$F(r) \cdot \overrightarrow{dr} = F(r)dr\cos \phi$ >Substitute the equation for gravitational force $\frac{GMm}{r^2}$ into F(r) and 180$^\circ$ into $\phi$. This is as the gravitational force decreases as $dr$ is positive, and doing so gives us: >$F(r) \cdot \overrightarrow{dr} = -\frac{GMm}{r^2}dr$ >We can integrate this by substituting back into the original equation: >$U = -GMm \int_{r}^{\infty} \frac{1}{r^2} \, dr = -GMm [-\frac{1}{r}]_{r}^{\infty} = 0 - \frac{GMm}{r}$ >$U = -\frac{GMm}{r}$ >There is another way to describe this using $-W = \Delta U$; note that we cannot bring an object down from a potential energy at the boundary condition infinity - where the GPE is given with 0 as the object is no longer gravitationally influenced. > >Therefore, all objects in the universe actually exert a gravitational influence on us - we just can't feel it most of the time. > >Using this information, we can create the following identity: >$U_{\infty} - U_{r} = -W$ >$-U_{r} = -W$ >$-\frac{GMm}{r} = W$ >This means that the work done by the gravitational force is actually equal to the potential energy of the gravitational force! Notice that we can see the potential energy - work done relation above, where the change in the potential energy is the negative of the change in the work done - so we've integrated our force diagram to get the equation! Mind you, this solution only applies if there are two bodies - so no other tidal effects from who knows where! For a system with multiple particles, remember that we can model the total net force on one arbitrary point with the formula: $F_{net} = \sum_{i=2}^nF_{n}$ Where $i = 2$ denotes that we take the force from the next particle as the starting point for our summation. This is an example of [[Newton's Law of Gravitation - The Intro#Gravitation Superposition|superposition]] - where Since the force function $F$ is directly intertwined with the potential energy equation (by virtue of being in the integral) we can also apply this logic for gravitational potential energy: $U_{net} = \sum_{i=2}^n \ (\frac{Gm_{net}m_{i}}{r_{i}})$ Where $m_{net}$ is the mass of the particle we want to find. Going the other way (since the force is a direct representation and thus has an opposite influence on the object), we can differentiate the potential energy formula to get: $F = -\frac{d}{dr}\left( -\frac{GMm}{r} \right) = -\frac{GMm}{r^2}$ This proves our original mass-energy equivalence - the force acts radially towards the centre of the main body, much like our potential energy does. # Escape Velocities ![[Pasted image 20240729140820.png]] Note that we can also use the principle of escape velocities to take a look at escape velocities - the minimum velocity required to escape the gravitational bounding of a body. Escape velocity is given by the formula: $v = \sqrt{ \frac{2GM}{r} }$ >[!Abstract]- Proving the Formula >Let's use an example - imagine a rocket is launched for a moon mission with a mass $m$, kinetic energy $E_k$ and potential energy $U$. > >As the escape velocity refers to the minimum speed required to overcome gravity, the total range of the object will be infinite. This means kinetic and potential energy can both be taken as 0 when the object reaches an infinitely far point, allowing us to take the sum of potential and kinetic energies of the object to be zero thanks to conservation of energy: >$K = -U$ $\frac{1}{2}mv^2 = \frac{GMm}{r}$ >Isolating the $v$ velocity yields: >$v = \sqrt{ \frac{2GM}{r} }$ >As a result, escape velocity scales with mass - the escape velocity of Ceres, a dwarf planet is 0.64 km/s while Earth's escape velocity is 11.2 km/s. These aren't easy speeds to reach, mind you! # More Gravitational Energies --> kinetic and pot. energy comparisons to make mech. energy! The # Example Questions >[!Example]- Module 13-5 Problem 40 >A projectile is shot directly away from Earth’s surface. Neglect the rotation of Earth. What multiple of Earth’s radius $R_{E}$ gives the radial distance a projectile reaches if: > >*a)* its initial speed is 0.500 of the escape speed from Earth? >*b)* its initial kinetic energy is 0.500 of the kinetic energy required to escape Earth? >*c)* What is the least initial mechanical energy required at launch if the projectile is to escape Earth? > >**Solution a)** > >The mechanical energy of the system is given by: >$E = K + U$ >By the principle of conservation of energy our initial mechanical energy must equal to the final mechanical energy as shown. At the maximum height of the object, it actually has a velocity and kinetic energy of 0, allowing us to create an identity for the final radius $r_f$: >$\frac{1}{2}m v_{i}^2 - \frac{GMm}{r} = -\frac{GMm}{r_{f}}$ >Notice that each term contains the mass of the arbitrary projectile - let's get rid of it by dividing on all sides: >$\frac{1}{2}v_{i}^2 - \frac{GM}{r} = -\frac{GM}{r_{f}}$ >Divide both sides by $GM$ >$\frac{v_{i}^2}{2 GM} - \frac{1}{r} = \frac{1}{r_{f}}$ >Notice that we can actually substitute $v_i$ with our numerical expression [[Gravitational Potential Energy (Mechanics)#Escape Velocities|given above]]. This means that $v_i$ is: >$v_{i} = \frac{1}{2} \sqrt{ \frac{2GM}{r} }$ >$\frac{2GM}{8 GMr} = \frac[](Gravitational%20Potential%20Energy.md#Escape%20Velocities)e velocity the projectile travels to a maximum height of 4 times Earth's radius. > >**Solution b)** >Once again, define the initial and final energies of the system: >$K_{i} + U_{i} = K_{f} + U_{f}$ >Remember that the final kinetic energy is still going to be 0 at the highest point: >$K_{i} - \frac{GMm}{r_{i}} = -\frac{GMm}{r_{f}}$ >We can model the velocity of the object when the kinetic energy of the object is half of the required escape energy as follows: >$\frac{1}{2}mv^2 = mv_{p}^2$ >Where $v_p$ denotes the velocity of the actual object while $v$ is the escape velocity. Further expansion and simplification of the expression yields an expression for $v_{p}$: >$\frac{1}{2} \frac{2GM}{r} = v_{p}^2$ >$\frac{GM}{r} = v_{p}^2$ >$v_{p} = \sqrt{ \frac{GM}{r} } $ >Substitute this expression into our original equation and divide by $m$ on both sides: >$\frac{1}{2}\left( \frac{GM}{r} \right) - \frac{GM}{r} = -\frac{GM}{r_{f}}$ >Divide both sides by $GM$: >$-\frac{1}{2r} = -\frac{1}{r_{f}}$ >Therefore the object will have travelled a distance of 2 Earth radii by the time it is suspended (not moving). > >**Solution c)** >Initial mechanical energy means that we find the mechanical energy when the gravitational potential energy of the system is zero (or negligible). This means that we just need to find an expression for kinetic energy of the projectile at escape velocity: >$E_{mec} = \frac{1}{2}m \sqrt{ \frac{{ 2GM }}{r} } $ >$E_{mec} = \frac{GMm}{r}$ # What's Next? For the final instalment in the gravitation series, go here: [[Kepler's Laws (with Derivations)]] To navigate back to the physics page, go here: [[The (Incomplete) Physics Almanac]]