# Universal Gravitation

particles

Things fall. Pencils, buildings, people, everything falls. The Earth is falling towards the Sun. Luckily, it is falling in a way that consistently misses the Sun.

Early Theories of Gravity

Physics often contradicts our feelings for how it should work. The history of gravity tells the story of humanity realizing we aren't the center of the universe.

The Greek philosopher Aristotle (384–322 BC) believed that the natural place for the element of earth and water was down. The natural place for the elements of air and fire was up. He also believed that heavier objects fell faster than light objects.

Like many early philosophers, Aristotle organized the sky into a geocentric model. Moving planetary spheres surrounded a stationary and spherical Earth.

The geocentric models showed flaws. Planets would inexplicably change in speed and direction. To correct these flaws several small circles called epicycles were added to the path of the planets. As telescopes and data collection methods improved, the number of epicycles grew.

Around 1543, Polish astronomer, Nicolaus Copernicus published a book on the heliocentric theory. The idea that the planets revolved around the Sun, not the Earth. His heliocentric theory marks the beginning of the the scientific revolution.

In 1609, German astronomer, Johannes Kepler published 3 laws of planetary motion based on the heliocentric theory. In his model orbits moved in an ellipse, not circles.

Italian natural philosopher, Galileo Galilei, discovered that falling objects all accelerate at the same rate as long as air resistance isn't a significant factor.

Galileo also championed Copernican heliocentrism. This blasphemous idea upset the Catholic church so much they held a trail and found him guilty of heresy. They forced him to recant his findings, they banned his work, and they sentenced him to house arrest from 1633 until his death in 1642.

## Universal Gravitation

In 1687 Isaac Newton published his book, Mathematical Principles of Natural Philosophy. The book contained his laws of motion and his law of universal gravitation. His reasoning was based on geometric proofs and new mathematical techniques which are now called calculus.

Universal gravitation states that every particle in the universe is attracted to every other particle. Amazingly, universal gravitation connects the earth and sky with one equation. The same force that makes apples fall also makes stars, planets, and moons orbit each other!

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

$$F$$ = 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 (6.674 x 10 -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
exoplanets data table
Example: Use the data table above to find the force of gravity between the Earth and the Moon. The distance between them is 384,403 km.
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{(3.844 \times 10^{8})^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{14.78 \times 10^{16}}$$ $$F = \frac{6.674 \times 5.972 \times 7.346}{14.78} \times \frac{10^{-11}10^{24}10^{22}}{10^{16}}$$ $$F = 19.798 \times 10^{19} \, \mathrm{N}$$
Example: Find the force of gravity between the Earth and the Sun. The distance between them is 149.6 billion m.
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (2.00 \times 10^{30})}{(149.6 \times 10^{9})^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (2.00 \times 10^{30})}{22380 \times 10^{18}}$$ $$F = 3.561 \times 10^{22} \, \mathrm{N}$$
Example: Find the force of gravity between the Earth and a 100 kg person that is 10,000,000 m from the center of the Earth.
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(100)}{ 10,000,000^{2}}$$ $$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(100)}{ 10^{14}}$$ $$F = 398.57 \, \mathrm{N}$$
Example: How massive must an object be in order to feel a force of 100 N at a distance of 10,000,000 m from the center of the Earth?
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$M_{1} = \frac{Fr^{2}}{GM_{2}}$$ $$M_{1} = \frac{(100)(10,000,000)^{2}}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}$$ $$M_{1} = \frac{(100)(10^{14})}{(6.674 \times 10^{-11})(5.972 \times 10^{24})}$$ $$M_{1} = 25.01 \, \mathrm{kg}$$
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 3.65×10 12 N. Calculate its distance from the Sun using the comet's mass of 2.2×10 14 kg.
solution $$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$r^{2} = \frac{GM_{1}M_{2}}{F}$$ $$r^{2} = \frac{(6.674 \times 10^{-11})(2.00 \times 10^{30})(2.2 \times 10^{14})}{3.65 \times 10^{12}}$$ $$\sqrt{r^{2}} = \sqrt{8.04 \times 10^{21}}$$ $$r = 8.97 \times 10^{10}\,\mathrm{m}$$

## 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
$$G = \small 6.674 \times 10^{-11}$$ = universal gravitation constant [m³/kg/s²]
$$M$$ = 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 =

Example: Find the acceleration of gravity for an object on the surface of Earth.
Is it really 9.8 m/s²?
solution $$r = 6,371,000 \, \mathrm{m}$$ $$g = \frac{GM}{r^{2}}$$ $$g = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{(6.371\times10^{6})^{2}}$$ $$g = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{40.590\times10^{12}}$$ $$g = \frac{(6.674)(5.972)}{40.590} \times 10^{-11+24-12}$$ $$g=0.98195 \times 10^{1}$$ $$g=9.8195 \, \mathrm{\frac{m}{s^{2}}}$$
Example: Find how far away from the Earth you need to be to only accelerate at half of 9.8 m/s².
solution $$g = \frac{GM}{r^{2}}$$ $$r^{2} = \frac{GM}{g}$$ $$r^{2} = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{9.8 \times 0.5}$$ $$r = 9,160,226 \,\mathrm{m} = 9,160 \,\mathrm{km}$$ $$\text{distance above Earth's surface}$$ $$9,160\, \mathrm{km} - 6,371\,\mathrm{km} = 2789 \,\mathrm{km}$$
Example: Mars has two very small moons, Phobos and Deimos. Phobos has a surface gravity of 0.0057 m/s² and a surface radius of 11,266 m. Calculate the mass of Phobos? (Does your answer agree with wikipedia?)
solution $$g = \frac{GM}{r^{2}}$$ $$M = \frac{gr^2}{G}$$ $$M = \frac{(0.0057)(11266)^2}{6.674\times 10^{-11}}$$ $$M = 1.08 \times 10^{16}\,\mathrm{kg}$$

## 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?

Example: If you triple the distance between two massive objects, what happens to their force of gravity? What about ten times the radius?
solution

This is easier to understand with some made up numbers. Set the masses and G to equal 1 since they aren't changing.

$$\text{triple}$$ $$r = 3$$ $$F = \frac{1}{r^2}$$ $$F = \frac{1}{3^2}$$ $$F = \frac{1}{9}$$
$$\text{ten times}$$ $$r = 10$$ $$F = \frac{1}{r^2}$$ $$F = \frac{1}{10^2}$$ $$F = \frac{1}{100}$$

As the distance between two masses increases, the force of gravity decreases.

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 have negative potential energy. $$U_{g} = -\frac{GM_{1}M_{2}}{r}$$

# $$U_{g} = -\frac{GM_{1}M_{2}}{r}$$

$$U_g$$ = 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 reddish regions as being deeper into the gravity well.

Poke around with your mouse. Number of particles =

Example: How much gravitational potential energy exists between two 100,000 kg masses at a distance of 2 meters?
solution $$U_{g} = -\frac{GM_{1}M_{2}}{r}$$ $$U_{g} = -\frac{(6.674 \times 10^{-11}) (100,000)(100,000)} {2}$$ $$U_{g} = -0.33\,\mathrm{J}$$

The energy is negative because it would take positive work to separate the masses.

Example: Two 100,000 kg masses are 2 meters apart. How much work would it take to bring them to 10 m apart?
solution $$U_{i} = -\frac{GM_{1}M_{2}}{r}$$ $$U_{i} = -\frac{(6.674 \times 10^{-11})(100,000)(100,000)}{2}$$ $$U_{i} = -0.33\,\mathrm{J}$$
$$U_{f} = -\frac{(6.674 \times 10^{-11})(100,000)(100,000)}{10}$$ $$U_{f} = -0.068\,\mathrm{J}$$
$$E_i = E_f$$ $$U_{i} + W = U_{f}$$ $$-0.33 + W = -0.068$$ $$W = 0.262 \, \mathrm{J}$$
Example: A 2000 kg space ship starts at rest 20,000 km 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?
hint
Use conservation of energy. Set the initial kinetic and gravitational potential equal to the final kinetic and gravitational potential.

solution $$E_{i} = E_{f}$$ $$U_{gi} = U_{gf} + K$$ $$K = U_{gi} - U_{gf}$$ $$K = -\frac{GM_{1}M_{2}}{r} +\frac{GM_{1}M_{2}}{r}$$ $$K = -\frac{(6.674\times 10^{-11})(5.972\times 10^{24})(2000)}{6,371,000 + 20,000,000}$$ $$+ \frac{(6.674\times 10^{-11})(5.972\times 10^{24})(2000)}{6,371,000 + 10,000,000}$$ $$K = -3.02 \times 10^{10}+4.86 \times 10^{10}$$ $$K = 1.85 \times 10^{10} \, \mathrm{J}$$
Example: Find the minimum escape velocity for a 100kg person on Earth.
hint
Use conservation of energy to find the energy needed to bring both the kinetic and potential energy to zero.

solution $$E_{i} = E_{f}$$ $$U_{gi} + K = U_{gf} + K$$ $$U_{gi} + K = 0$$ $$-U_{gi} = K$$ $$\frac{GM_{1}M_{2}}{r} = (1/2)M_{2}v^2$$ $$\frac{GM_{1}}{r} = (1/2)v^2$$ $$\sqrt{\frac{2GM_{1}}{r}} = v$$ $$\sqrt{\frac{2(6.674\times 10^{-11}) (5.972 \times 10 ^{24})}{6,371,000}} = v$$ $$11,183 \, \mathrm{\tfrac{m}{s}} =v$$