Gravity simulation with particles. Poke the particles with a click.
Kepler's Laws of Planetary Motion
In 1609 Johannes Kepler suggested 3 laws of planetary motion.
1. The orbit of a planet is an ellipse with the sun at one of the two foci.
2. A line segment joining a planet and the sun sweeps out equal areas during equal intervals of time.
3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
Kepler's Laws are pretty accurate, but limited in scope. They only apply to objects in orbit around an approximately stationary central body, like the planets orbiting the sun.
Newton's Law of Universal Gravitation
Universal gravitation states that every particle in the universe is attracted to every other particle.
In 1687 Isaac Newton discovered the equation by watching the motion of planets and making empirical observations. Amazingly, Newton's universal gravitation managed to connect the heavens and the earth with one equation. Imagine the impact of that discovery in those superstitious times.
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$
\(F\) = force of gravity [N, Newtons, kg m/s²]
vector
\(G\) = 6.67408 × 10
^{-11} = universal gravitation constant [N m²/kg²]
\(M\) = mass [kg, kilograms]
\(r\) = distance between the center of each mass [m, meters]
In 1916 universal gravitation was improved on by Einstein's general relativity, but universal gravitation is still used for low mass and low speed situations, like the earth and its satellites.
Each mass feels an equal but opposite force as predicted by Newton's 3rd law. This means that the same force of gravity you feel towards the Earth the Earth feels towards you. So why don't we notice the Earth accelerate towards you?
Notice how small G is (6.6x10 ^{-11}). Out of the four fundamental forces gravity is by far the weakest. If gravity is so weak why do we notice its effects so easily?
Local Massive Objects Data Table
name | mass (kg) | radius (km) | density (g/cm ^{3}) |
---|---|---|---|
Sun | 2.00×10 ^{30} | 695,700 | 1.408 |
Mercury | 3.301×10 ^{23} | 2,440 | 5.427 |
Venus | 4.867×10 ^{24} | 6,052 | 5.243 |
Earth | 5.972×10 ^{24} | 6,371 | 5.515 |
Moon | 7.346×10 ^{22} | 1,737 | 3.344 |
Mars | 6.417×10 ^{23} | 3,390 | 3.933 |
Jupiter | 1.899×10 ^{27} | 70,000 | 1.326 |
Saturn | 5.685×10 ^{26} | 58,232 | 0.687 |
Uranus | 8.682×10 ^{25} | 25,362 | 1.270 |
Neptune | 1.024×10 ^{26} | 24,622 | 1.638 |
solution
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{(3.844 \times 10^{8})^{2}}$$ $$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{14.78 \times 10^{16}}$$ $$ F = \frac{6.674 \times 5.972 \times 7.346}{14.78} \times \frac{10^{-11}10^{24}10^{22}}{10^{16}}$$ $$ F = 19.798 \times 10^{19} N$$solution
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (2.00 \times 10^{30})}{(149.6 \times 10^{9})^{2}}$$ $$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (2.00 \times 10^{30})}{22380 \times 10^{18}}$$ $$ F = 3.561 \times 10^{22} N$$solution
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(100)}{ 10,000,000^{2}}$$ $$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(100)}{ 10^{14}}$$ $$ F = 398.57 N$$solution
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$ M_{1} = \frac{Fr^{2}}{GM_{2}}$$ $$ M_{1} = \frac{(100)(10,000,000)^{2}}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}$$ $$ M_{1} = \frac{(100)(10^{14})}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}$$ $$ M_{1} = 25.01kg$$solution
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$ r^{2} = \frac{GM_{1}M_{2}}{F} $$ $$ r^{2} = \frac{(6.674 \times 10^{-11})(2.00 \times 10^{30})(2.2 \times 10^{14})}{3.65 \times 10^{12}} $$ $$ \sqrt{r^{2}} = \sqrt{8.04 \times 10^{21}} $$ $$ r = 8.97 \times 10^{10} m $$Gravitational Acceleration
It is useful to adapt the universal gravitation equation to predict acceleration. To find acceleration we just need to divide an object's gravitational force by its mass.
derivation of universal gravitational acceleration
$$ F = mg $$ $$ \frac{F}{m} = g $$Newton's second law tells us we can replace F/M with acceleration.
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$ $$ \frac{F}{M_{2}} = \frac{GM_{1}}{r^{2}} $$ $$ g = \frac{GM_{1}}{r^{2}} $$$$ g = \frac{GM}{r^{2}} $$
\(g\) = acceleration of gravity [m/s²]
vector
\(G\) = 6.67408 × 10
^{-11} = universal gravitation constant [N(m/kg)²]
\(M\) = mass of the body pulling [kg, kilograms]
(not the body experiencing the acceleration)
\(r\) = distance between the center of each mass [m, meters]
The acceleration vector is pointed towards the center of the mass producing the acceleration.
The mass of the body being accelerated isn't used in this equation. Use the mass of the body producing the acceleration. To find the acceleration of objects on Earth use Earth's mass.
The simulation below shows a vector
field. Each vector shows the gravitational acceleration potentially felt at that location. These diagrams
are helpful for predicting how a particle will accelerate.
Click the mouse to push the masses around. Number of particles =
Is it really 9.8m/s²?
solution
$$ r = 6,371,000m $$ $$ g = \frac{GM}{r^{2}} $$ $$ g = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{(6.371\times10^{6})^{2}} $$ $$ g = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{40.590\times10^{12}} $$ $$ g = \frac{(6.674)(5.972)}{40.590} \times 10^{-11+24-12}$$ $$ g=0.98195 \times 10^{1}$$ $$ g=9.8195 \frac{m}{s^{2}}$$solution
$$ g = \frac{GM}{r^{2}} $$ $$ r^{2} = \frac{GM}{g} $$ $$ r^{2} = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{9.8 \times 0.5} $$ $$ r = 9,160,226 m = 9,160 km $$Let's find the distance above Earth's surface.
$$ 9,160km - 6,371km = 2789 km$$solution
$$ g = \frac{GM}{r^{2}} $$ $$ M = \frac{gr^2}{G}$$ $$ M = \frac{(0.0057)(11266)^2}{6.674\times 10^{-11}}$$ $$M = 1.08 \times 10^{16}kg$$Inverse-Square Law
All particles are attracted to each other, but the attraction is divided by the distance squared (1/r²). Physical laws that diminish at 1/r² are common in nature because of how signals spread out in 3 dimensions. Two of the four fundamental forces follow an inverse square law.
If you threw darts in random directions, the chance of hitting your target would obey this rule. Notice how the density of lines decreases with distance.
As you can see in the graph below, a 1/r ^{2} function has some interesting results at r=0 and f=0.
What distance between masses produces a force of zero?
What happens to the force of gravity as the distance between masses approaches zero?
What else, besides gravity, might follow an inverse square law?
solution
This is easier to understand with some made up numbers. Set the masses and G to equal 1 since they aren't changing.
The force gets smaller by a factor of 1/9 and 1/100.
Gravitational Potential Energy
Our old gravitational potential energy equation (U=mgh) can be made more accurate if we replace g = 9.8 with a calculated gravitational acceleration. This version of gravitational potential energy now works beyond Earth's surface.
derivation of universal gravitational potential energy
$$U_{g} = mgh \quad g = \color{blue}{\frac{GM}{r^{2}}} $$ $$U_{g} = m\color{blue}{\frac{GM}{r^{2}}}h $$ $$U_{g} = \frac{GM_{1}M_{2}}{r^{2}}r $$ $$U_{g} = \frac{GM_{1}M_{2}}{r} $$When distance is very big the energy goes to zero, so it makes sense to choose the zero of this gravitational potential energy at an infinite distance away. This means that as we bring a mass closer to another mass we are doing negative work. $$U_{g} = -\frac{GM_{1}M_{2}}{r} $$
$$U_{g} = -\frac{GM_{1}M_{2}}{r} $$
\(U_g\) = gravitational potential energy [J, Joules]
\(G\) = 6.67408 × 10
^{-11} = universal gravitation constant [N(m/kg)²]
\(M\) = mass [kg, kilograms]
\(r\) = distance between the center of each mass [m, meters]
What is the gravitational potential energy when two masses are at zero distance away?
What is the gravitational potential energy when two masses are at an infinite distance away?
Energy is a scalar, not a vector. This means that when we calculate gravitational potential energy it has no direction.
You can see the potential energy in the simulation below as a scalar field. Think of the yellow regions as being
deeper into the
gravity well.
Poke around with your mouse. Number of particles =
solution
$$U_{g} = -\frac{GM_{1}M_{2}}{r}$$ $$U_{g} = -\frac{(6.674 \times 10^{-11}) (100,000)(100,000)} {2} $$ $$U_{g} = -0.33J$$The energy is negative because it would take positive work to separate the masses.
solution
$$U_{i} = -\frac{GM_{1}M_{2}}{r}$$ $$U_{i} = -\frac{(6.674 \times 10^{-11})(100,000)(100,000)}{2}$$ $$U_{i} = -0.33J$$$$U_{f} = -\frac{(6.674 \times 10^{-11})(100,000)(100,000)}{10}$$ $$U_{f} = -0.068J$$
$$E_i = E_f$$ $$U_{i} + W = U_{f} $$ $$-0.33 + W = -0.068$$ $$W = 0.262J$$
solution
$$E_{i} = E_{f}$$ $$U_{gi} = U_{gf} + K$$ $$ K = U_{gi} - U_{gf}$$ $$K = -\frac{GM_{1}M_{2}}{r} +\frac{GM_{1}M_{2}}{r}$$ $$K = -\frac{(6.674\times 10^{-11})(5.972\times 10^{24})(2000)}{6,371,000 + 20,000,000}$$ $$+ \frac{(6.674\times 10^{-11})(5.972\times 10^{24})(2000)}{6,371,000 + 10,000,000}$$ $$K = -3.02 \times 10^{10}+4.86 \times 10^{10} $$ $$K = 1.85 \times 10^{10} J$$(hint: Use conservation of energy to find the energy needed to bring both the kinetic and potential energy to zero.)