# Universal Gravitation

Gravity simulation with particles. Poke the particles with a click.

Kepler's Laws of Planetary Motion

In 1609, just before Newton's universal gravitation in 1687, 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 Issac 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?

All particles are attracted to each other, but the attraction is divided by the distance squared (1/r²). Physical laws that diminish with a 1/r² rule are common in physics because of how signals spread out in 3 dimensions.

What happens to the force of gravity as the distance between masses approaches zero?

### Local Massive Objects Data Table

name mass (kg) radius (km) density (g/cm3) rotation axis tilt (°)
Sun 2.00×1030 695,700 1.408 ~30 d 0
Mercury 3.301×1023 2,440 5.427 58.785 d ~0
Venus 4.867×1024 6,052 5.243 243.686 d 177.36
Earth 5.972×1024 6,371 5.515 23.9345 h 23.45
Moon 7.346×1022 1,737 3.344 27.322 d 1.5
Mars 6.417×1023 3,390 3.933 24.6229 h 25.19
Jupiter 1.899×1027 70,000 1.326 9.9250 h 3.13
Saturn 5.685×1026 58,232 0.687 10.656 h 26.73
Uranus 8.682×1025 25,362 1.270 17.24 h 97.77
Neptune 1.024×1026 24,622 1.638 16.11 h 28.32
exoplanets data table
Example: Find the force of gravity between the Earth and the Moon. The distance between them is 384,403 km.
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{(3.844 \times 10^{8})^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{14.78 \times 10^{16}}$$ $$F = \frac{6.674 \times 5.972 \times 7.346}{14.78} \times \frac{10^{-11}10^{24}10^{22}}{10^{16}}$$ $$F = 19.798 \times 10^{19} N$$
Example: Find the force of gravity between the Earth and the Sun. The distance between them is 149.6 billion m.
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (2.00 \times 10^{30})}{(149.6 \times 10^{9})^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (2.00 \times 10^{30})}{22380 \times 10^{18}}$$ $$F = 3.561 \times 10^{22} N$$
Example: Find the force of gravity from the Earth and a 100kg person 10,000,000m from the center of Earth.
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(100)}{ 10,000,000^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(100)}{ 10^{14}}$$ $$F = 398.57 N$$
Example: How massive must an object be in order to feel a force of 100N at a distance of 10,000,000m from the center of the Earth?
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$M_{1} = \frac{Fr^{2}}{GM_{2}}$$ $$M_{1} = \frac{(100)(10,000,000)^{2}}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}$$ $$M_{1} = \frac{(100)(10^{14})}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}$$ $$M_{1} = 25.01kg$$
Example: When it is closest to the sun on its 75 year orbit Halley's Comet feels a force of gravity from the Sun of 3.65×1012 N. Calculate its distance from the sun using the comet's mass of 2.2×1014 kg.
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 $$F = M_{2}a$$ $$\frac{F}{M_{2}} = a$$

Newton's second law tells us we can replace F/M with acceleration.

$$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$\frac{F}{M_{2}} = \frac{GM_{1}}{r^{2}}$$ $$a = \frac{GM_{1}}{r^{2}}$$

## $$a = \frac{GM}{r^{2}}$$

$$a$$ = 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]
$$r$$ = 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 =

Example: Find the acceleration of gravity for an object on the surface of Earth.
Is it really 9.8m/s²?
solution $$r = 6,371,000m$$ $$a = \frac{GM}{r^{2}}$$ $$a = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{(6.371\times10^{6})^{2}}$$ $$a = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{40.590\times10^{12}}$$ $$a = \frac{(6.674)(5.972)}{40.590} \times 10^{-11+24-12}$$ $$a=0.98195 \times 10^{1}$$ $$a=9.8195 \frac{m}{s^{2}}$$
Example: Find how far away from the earth you need to be to only accelerate at half of 9.8m/s².
solution $$a = \frac{GM}{r^{2}}$$ $$r^{2} = \frac{GM}{a}$$ $$r^{2} = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{9.8 \times 0.5}$$ $$r = 9,160,226 m = 9,160 km$$

Let's find the distance above Earth's surface.

$$9,160km - 6,371km = 2789 km$$

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 $$U_{g} = mgh \quad a = \color{blue}{\frac{GM}{r^{2}}}$$ $$U_{g} = m\color{blue}{\frac{GM}{r^{2}}}h$$ $$U_{g} = \frac{GM_{1}M_{2}}{r^{2}}r$$ $$U_{g} = \frac{GM_{1}M_{2}}{r}$$
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 =

Example: A 2000kg space ship starts at rest 20,000km from the surface of Earth. How much kinetic energy will the ship have after it falls down to 10,000km from the surface of Earth?
solution $$E_{i} = E_{f}$$ $$U_{gi} = U_{gf} + K$$ $$U_{gi} - U_{gf} = K$$ $$-\frac{GM_{1}M_{2}}{r} +\frac{GM_{1}M_{2}}{r} = K$$ $$-\frac{(6.674\times 10^{-11})(5.972\times 10^{24})(2000)}{6,371,000 + 20,000,000} +\frac{(6.674\times 10^{-11})(5.972\times 10^{24})(2000)}{6,371,000 + 10,000,000} = K$$ $$-3.02 \times 10^{10}+4.86 \times 10^{10} = K$$ $$1.85 \times 10^{10} J = K$$
Example: Find the minimum escape velocity for a 100kg person on Earth.
solution $$E_{i} = E_{f}$$ $$U_{gi} + K = U_{gf} + K$$
kinetic and potential energy will go to zero once the person escapes the earth's gravity $$U_{gi} + K = 0$$ $$-U_{gi} = K$$ $$\frac{GM_{1}M_{2}}{r} = (1/2)M_{2}v^2$$ $$\frac{GM_{1}}{r} = (1/2)v^2$$ $$\sqrt{\frac{2GM_{1}}{r}} = v$$ $$\sqrt{\frac{2(6.674\times 10^{-11}) (5.972 \times 10 ^{24})}{6,371,000}} = v$$ $$11,183 \small \frac{m}{s} = \normalsize v$$