Things fall. Pencils, buildings, people, everything falls.
The Earth is falling towards the Sun, but luckily it is falling in a way that doesn't hit the Sun.
An orbit.
Early Theories of Gravity
The history of gravity tells the story of humanity realizing we aren't the center of the universe.
The Greek philosopher Aristotle (384–322 BCE) 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 to correct for flaws in the geocentric model.
In 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 scientific revolution.
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 the heliocentric model even though it contradicted Catholic scripture.
The leaders of the church asked him not to publish information that supported the heliocentric model, but Galileo continued.
In response, the church forced him to recant his findings, they banned his work, and they sentenced him to house arrest from 1633 until his death in 1642.
Question: What advantages did the heliocentric model have over the geocentric model?
answer
At the time it wasn't clear what model was the best one, but the heliocentric model explained the world in a more elegant and simple way.
Issues like retrograde motion of planets was explained naturally in the heliocentric model without the need for epicycles.
One of the reasons Galileo liked the heliocentric model was that he had observed the Venus went through phases, just like the moon.
In the Ptolemaic model, where Venus orbits the Earth, it wouldn't go through the full cycle of phases.
The heliocentric model was actually less accurate than the geocentric model.
This is because the geocentric model was using a series of epicycles to improve on the circular paths both models used.
Eventually, Kepler's Laws showed that with elliptical paths the heliocentric model was more accurate.
Kepler's Laws of Planetary Motion
In 1609, German astronomer,
Johannes Kepler published 3 laws of planetary motion based on the heliocentric theory.
The orbit of a planet is an ellipse with the Sun at one of the two foci. (F1, F2)
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. (A1 = A2)
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
$$ T^2 = r^3$$
\(T\) = orbital period, total time for one orbit [yr, years]
\(r\) = semi-major axis [AU, astronomical units, 1.5 × 1011 meters]
Kepler's laws mostly agreed with the Copernican model, but with a few improvements.
In Kepler's model orbits moved in an ellipse, not a circle.
The Sun wasn't the center of a planetary orbit, it was a foci of an elliptical orbit.
Also, the planets would speed up and slow down as they orbited.
The semi-major axis of an ellipse is the distance from the center to the longest edge.
For the Earth that distance is 149 597 870 700 m, which is called 1 astronomical unit (AU).
The orbital period squared equals the semi-major axis cubed for units of years (yr) and
astronomical units (AU).
The relationship is more complex with other units.
modified equation for units of meters and seconds
$$T^2 = \frac{4 \pi^2}{GM} r^3$$
\(T\) = orbital period, total time for one orbit [yr, years]
\(r\) = semi-major axis [AU, astronomical units, 1.5 × 1011 meters]
\(M\) = Mass of central body [kg, kilograms]
\(G\) = 6.67408 × 10-11 = universal gravitation constant [N m²/kg²]
planet
semi-major axis
orbital period
Mercury
0.387 AU
0.240 yr
Venus
0.723 AU
0.615 yr
Earth
1.000 AU
1.000 yr
Mars
1.524 AU
Jupiter
5.204 AU
11.86 yr
Saturn
9.582 AU
29.46 yr
Uranus
19.22 AU
84.02 yr
Neptune
30.11 AU
164.8 yr
Example: Calculate the time it takes Mars to orbit the Sun in years.
solution
$$T^2 = r^3$$
$$T^2 = (1.524)^3$$
$$T^2 = 3.540$$
$$T = 1.88 \, \mathrm{yr}$$
Each planet in the solar system lands somewhere on this graph.
So this graph tells us that the farther a planet is from the sun, the longer it takes to orbit.
A Jupiter year is much longer than an Earth year.
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 his new mathematical techniques which are now called calculus.
Universal gravitation states that every particle in the universe is attracted to every other particle.
It connects the heavens and the Earth with one equation.
The same force that makes apples fall also controls the motion of stars, planets, and moons.
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$
\(F\) = 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]
Universal gravitation improved on Kepler's laws of planetary motion because it had a more universal application.
Kepler's laws only applied to orbits where one body was much more massive, like the Sun and it's planets.
Newton's gravitation applied to all matter.
No current theory perfectly describes reality. They all have the potential for improvement.
In 1915 universal gravitation was improved on by Einstein's general relativity.
It is hoped that one day we will improve on general relativity by combining it with quantum mechanics.
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.
Question: The force of gravity on you and the force on the Earth are equal but in opposite directions .
Does that mean when you accelerate down at 9.8 m/s² the Earth accelerates up at 9.8 m/s²?
answer
The Earth and you both have the same force, but not the same acceleration.
$$F=ma$$
$$a = \frac{F}{m}$$
Acceleration equals force divided by mass. The Earth has a large mass, so the acceleration from you pulling on the Earth is small.
name
mass (kg)
radius (km)
Sun
2.00 × 1030
695 700
Mercury
3.301 × 1023
2440
Venus
4.867 × 1024
6052
Earth
5.972 × 1024
6371
Moon
7.346 × 1022
1737
Mars
6.417 × 1023
3390
Ceres
9.384 × 1020
470
Jupiter
1.899 × 1027
70 000
Saturn
5.685 × 1026
58 232
Uranus
8.68 × 1025
25 362
Neptune
1.024 × 1026
24 622
Example: Find the force of gravity between the Earth and the Moon. The distance between them is 384 403 km.
solution
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$
$$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{(3.844 \times 10^{8})^{2}}$$
$$ F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24}) (7.346 \times 10^{22})}{14.78 \times 10^{16}}$$
$$ F = \frac{6.674 \times 5.972 \times 7.346}{14.78} \times \frac{10^{-11}10^{24}10^{22}}{10^{16}}$$
$$ F = 19.798 \times 10^{19} \, \mathrm{N}$$
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 × 1012 N. Calculate its distance from the Sun using the comet's mass of 2.2 × 1014 kg.
solution
$$ F = \frac{GM_{1}M_{2}}{r^{2}} $$
$$ r^{2} = \frac{GM_{1}M_{2}}{F} $$
$$ r^{2} = \frac{(6.674 \times 10^{-11})(2.00 \times 10^{30})(2.2 \times 10^{14})}{3.65 \times 10^{12}} $$
$$ \sqrt{r^{2}} = \sqrt{8.04 \times 10^{21}} $$
$$ r = 8.97 \times 10^{10}\,\mathrm{m} $$
$$G = 6.674 × 10^{-11} $$
Question: Notice that universal gravitation constant is very small. 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?
answer
The force of gravity scales with distance and mass.
The Earth is very close and very massive, so it's gravity is large enough to feel.
The force of gravity is very weak though.
If gravity were stronger we might notice the pull from human sized objects.
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.
masses =
Example: Find the acceleration of gravity for an object on the surface of Earth.
Is it really 9.8 m/s²?
Local Massive Objects Data Table
name
mass (kg)
radius (km)
Sun
2.00 × 1030
695 700
Mercury
3.301 × 1023
2440
Venus
4.867 × 1024
6052
Earth
5.972 × 1024
6371
Moon
7.346 × 1022
1737
Mars
6.417 × 1023
3390
Jupiter
1.899 × 1027
70 000
Saturn
5.685 × 1026
58 232
Uranus
8.68 × 1025
25 362
Neptune
1.024 × 1026
24 622
solution
$$ r = 6\,371\,000 \, \mathrm{m} $$
$$ g = \frac{GM}{r^{2}}$$
$$ g = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{(6.371\times10^{6})^{2}}$$
$$ g = \frac{(6.674\times 10^{-11})(5.972\times 10^{24})}{40.590\times10^{12}} $$
$$ g =\frac{(6.674)(5.972)}{40.590}\times 10^{-11+24-12}$$
$$ g=0.981\,95 \times 10^{1}$$ $$ g=9.8195 \, \mathrm{\frac{m}{s^{2}}}$$
Example: Find how far away from the Earth you need to be to only accelerate at half of 9.8 m/s².
solution
$$ g = \frac{GM}{r^{2}} $$ $$ r^{2} = \frac{GM}{g} $$ $$ r^{2} = \frac{(6.674\times
10^{-11})(5.972\times 10^{24})}{9.8 \times
0.5} $$ $$ r = 9\,160\,226 \,\mathrm{m} = 9160 \,\mathrm{km} $$ $$\text{distance above Earth's
surface}$$
$$ 9160\, \mathrm{km} - 6371\,\mathrm{km} = 2789 \,\mathrm{km}$$
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
$$ g = \frac{GM}{r^{2}} $$
$$ M = \frac{gr^2}{G}$$
$$ M = \frac{(0.0057)(11\,266)^2}{6.674\times 10^{-11}}$$
$$M = 1.08 \times 10^{16}\,\mathrm{kg}$$
Inverse-Square Law (1/r²)
All particles are attracted to each other, but the attraction is divided by the distance squared. Physical laws that diminish at 1/r² are common in nature because of how signals spread out in 3 dimensions. If you threw darts in random directions, the chance of hitting your target would obey this rule.
Question: What distance between masses produces a force of zero?
answer
As the distance approaches infinity, the force approaches zero.
The effects of gravity can be felt at any real distance.
Astronomers see gravitation between our galaxy, the Milky Way, and the Andromeda galaxy at a distance of 2.5 million light years (1022 m)
Question: What happens to the force of gravity as the distance between masses approaches zero?
answer
The force rises towards infinity as the distance approaches zero.
When matter gets very dense, a black hole forms.
Understanding gravity at distances near zero require using general relativity and quantum physics at the same time.
Currently these two theories are not compatible, but we know that some strange things would happen.
Time would slow down, and it would become very difficult for anything to escape.
Question: What else, besides gravity, might follow an inverse square law?
answer
Anything that spreads out in 3 dimensions will follow this law.
Light
Sound
Randomly throwing things, like darts.
Example: If you triple the distance between two massive objects, what happens to their force of gravity? What about ten times the distance?
solution
This is easier to understand with some made up numbers. Set the masses and G to equal 1 since they aren't changing.
$$\text{\underline{triple}}$$
$$r = 3$$
$$ F = \frac{1}{r^2}$$
$$ F = \frac{1}{3^2}$$
$$ F = \frac{1}{9}$$
$$\text{\underline{ten times}}$$
$$r = 10$$ $$ F = \frac{1}{r^2}$$
$$ F = \frac{1}{10^2}$$
$$ F = \frac{1}{100}$$
As the distance between two masses increases, the force of gravity decreases.
name
mass (kg)
radius (km)
g (m/s²)
Mercury
3.301 × 1023
2440
3.7
Venus
4.867 × 1024
6052
8.9
Earth
5.972 × 1024
6371
9.8
Mars
6.417 × 1023
3390
3.7
Question: Which terrestrial planet is being graphed?
strategy
Hover your mouse over the graph. Move your mouse horizontally to check the radius for each planet in the chart above. If the acceleration and the radius match, it is the correct planet.
answer
name
mass (kg)
radius (km)
g (m/s²)
Jupiter
1.899 × 1027
70 000
?
Saturn
5.685 × 1026
58 232
?
Uranus
8.68 × 1025
25 362
?
Neptune
1.024 × 1026
24 622
?
Question: Which gas giant is being graphed? You need to do math for this one.
answer
Questions:
Imagine you could to dig into the center of the Earth.
How would the acceleration of gravity change as you went deeper?
Would it increase, decrease, or stay the same?
answer
As you go deeper into the Earth you feel less gravity.
There is mass both above and below, they pull in opposite directions and cancel out.
You can't feel opposing forces of gravity, because gravity pulls on your whole body evenly.
Using a model of the Earth with constant density the gravity decreases linearly.
At 50% to the center of the Earth the gravity is 50% of 9.8 m/s².
At the center gravity is zero.
Earth's actual gravity according to the Preliminary Reference Earth Model (PREM) is more complex.
This is because the density changes with each layer, and increases towards the center.
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
$$U_{g} = mgh \quad g = \color{blue}{\frac{GM}{r^{2}}} $$
$$U_{g} = m{\color{blue}{\frac{GM}{r^{2}}}}h $$
If we define h to be zero when the two masses are zero distance apart, we can rename h to r.
$$U_{g} = \frac{GM_{1}M_{2}}{r^{2}}r$$
$$U_{g} = \frac{GM_{1}M_{2}}{r} $$
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.
$$U_{g} = -\frac{GM_{1}M_{2}}{r} $$
Here is a more complete derivation using calculus.
$$U_{g} = -\frac{GM_{1}M_{2}}{r} $$
\(U_g\) = 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?
masses =
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.
Example: Calculate the gravitational potential energy between two 100 000 kg masses when there is only 2 m between the center of each mass.
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?
strategy
Calculate the gravitational potential energy at each distance. The change in energy is the work.
Example: A 2000 kg spaceship starts at rest 20 000 km from the surface of Earth. How much kinetic energy will the ship have after it falls down to 10 000 km from the surface of Earth?
strategy
Use conservation of energy. Set the initial kinetic and gravitational potential equal to the final kinetic and gravitational potential.
Use the mouse to move the camera. Double click for full screen.
In the simulation above, gravitational potential energy is represented as warping a 2-D fabric into a 3rd dimension.
This is similar to how gravity is thought of in general relativity.
In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!
Example: An asteroid in the main belt has an average orbital distance of 2.25 AU from the Sun. Using Kepler's simple relation, find its orbital period in years.
solution
Kepler's simple relation works when distance is in AU and period is in years.
The period is longer than Earth's because the asteroid is farther from the Sun.
Example: A comet takes 8 years to orbit the Sun. Using Kepler's simple relation, find its average orbital distance in AU.
solution
$$T^2 = r^3$$
$$8^2 = r^3$$
$$64 = r^3$$
$$r = \sqrt[3]{64}$$
$$r = 4\,\mathrm{AU}$$
This is a rearranged Kepler problem. The distance is not 8 AU because the relationship is not linear.
Example: Planet A orbits at 1 AU and planet B orbits at 4 AU. How many times longer is planet B's orbital period?
solution
Planet B's orbital period is 8 times longer, not 4 times longer.
Example: Two 100 000 kg pieces of mining equipment are parked 20 m apart, measured from center to center. Each machine is much taller than a person, but model them as point masses. What is the gravitational force between them?
solution
$$F = \frac{GM_1M_2}{r^2}$$
$$F = \frac{(6.67 \times 10^{-11})(100\,000)(100\,000)}{(20)^2}$$
$$F = 1.67 \times 10^{-3}\,\mathrm{N}$$
This is tiny, even though the masses are large by everyday standards.
Example: A 1200 kg satellite is 400 km above Earth's surface, about the altitude of some low Earth orbit spacecraft. Find the force of gravity on the satellite. Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
The distance in the gravity equation is measured from Earth's center, not from the surface.
The satellite is above the atmosphere, but gravity is still strong there.
Example: A 75 kg astronaut feels a 300 N gravitational force from Earth. How far is the astronaut from Earth's center? Use Earth's mass as 5.97 × 1024 kg.
solution
$$F = \frac{GM_1M_2}{r^2}$$
$$r^2 = \frac{GM_1M_2}{F}$$
$$r = \sqrt{\frac{GM_1M_2}{F}}$$
$$r = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(75)}{300}}$$
$$r = 9.98 \times 10^6\,\mathrm{m}$$
This is distance from Earth's center. The altitude above the surface would be smaller.
Example: A 75 kg person stands on Earth. Earth pulls on the person, and the person pulls on Earth with the same size force. Find the person's acceleration and Earth's acceleration from this gravitational force. Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
First find the force. Then use Newton's second law from the force unit.
The forces are equal in size, but the accelerations are very different because the masses are very different.
Example: Mars has mass 6.42 × 1023 kg and radius 3.39 × 106 m. Its sky often looks dusty orange, but the surface gravity comes from mass and radius. Find the gravitational acceleration at the surface of Mars.
solution
$$g = \frac{GM}{r^2}$$
$$g = \frac{(6.67 \times 10^{-11})(6.42 \times 10^{23})}{(3.39 \times 10^6)^2}$$
$$g = 3.73\,\mathrm{m/s^2}$$
Mars has less surface gravity than Earth because it has much less mass.
Example: Find the gravitational acceleration 400 km above Earth's surface. Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
$$r = 6.37 \times 10^6 + 400\,000$$
$$r = 6.77 \times 10^6\,\mathrm{m}$$
$$g = \frac{GM}{r^2}$$
$$g = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.77 \times 10^6)^2}$$
$$g = 8.69\,\mathrm{m/s^2}$$
This explains why orbiting astronauts are not away from gravity. They are falling around Earth.
Example: At what altitude above Earth's surface is gravitational acceleration one-fourth of its surface value? Use Earth's radius as 6.37 × 106 m.
solution
Gravity follows an inverse-square pattern. To make gravity one-fourth as large, the distance from Earth's center must double.
The altitude is one Earth radius above the surface, not twice Earth's radius above the surface.
Example: A spacecraft moves from a distance of 3 Earth radii from Earth's center to 6 Earth radii from Earth's center. What happens to the gravitational force from Earth?
solution
The distance doubled because it went from 3 Earth radii to 6 Earth radii.
Example: Two 5.0 × 104 kg asteroids are 30 m apart. What is the gravitational force between them?
answer
This cannot be solved from the information given. Newton's law of gravitation needs the distance between the centers of the objects. The prompt only says they are 30 m apart, which could mean edge-to-edge distance.
Example: A 70 kg rover part is weighed on Mars. Use the surface gravity you found for Mars, 3.73 m/s². What is its weight on Mars?
solution
The mass is still 70 kg. The weight changes because the gravitational acceleration is different.
Example: A rock is dropped from rest near the surface of Mars. Assume the gravitational acceleration stays constant at 3.73 m/s² for the short fall. How fast is the rock moving after 6 s?
solution
This uses the motion equation for constant acceleration. The constant acceleration assumption is reasonable for a short fall near the surface.
The rock speeds up more slowly than it would on Earth.
Example: Two asteroids have masses of 100 000 kg and 200 000 kg. Their centers are 5 m apart. What is their gravitational potential energy?
solution
$$U_g = -\frac{GM_1M_2}{r}$$
$$U_g = -\frac{(6.67 \times 10^{-11})(100\,000)(200\,000)}{5}$$
$$U_g = -0.267\,\mathrm{J}$$
The negative sign means the objects are in a bound gravitational system compared with being infinitely far apart.
Example: The two asteroids from the last problem are slowly pulled from 5 m apart to 20 m apart. How much work must be done to separate them that far?
solution
The work done slowly equals the change in gravitational potential energy.
The work is positive because energy must be added to pull the asteroids farther apart.
Example: A 500 kg probe starts at rest 20 000 km above Earth's surface and falls to 10 000 km above Earth's surface. Ignoring air resistance and engines, how much kinetic energy does it gain? Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
Use conservation of energy. The distance in gravitational potential energy is measured from Earth's center.
Example: A 1000 kg payload is lifted from Earth's surface to a circular-orbit altitude of 400 km. Ignoring the rocket's wasted energy, how much work is needed just to increase the payload's gravitational potential energy? Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
This only counts the change in gravitational potential energy. It does not include the kinetic energy needed for orbit.
This is already a huge amount of energy, and it is still only part of what a launch requires.
Example: Find the minimum escape speed from Earth. Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
$$v = \sqrt{\frac{2GM}{r}}$$
$$v = \sqrt{\frac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}}$$
$$v = 1.12 \times 10^4\,\mathrm{m/s}$$
The mass of the escaping object cancels out, so this speed is the same for a baseball and a spacecraft if air resistance is ignored.
Example: Find the minimum escape speed from the Moon. Use the Moon's mass as 7.35 × 1022 kg and its radius as 1.74 × 106 m.
solution
$$v = \sqrt{\frac{2GM}{r}}$$
$$v = \sqrt{\frac{2(6.67 \times 10^{-11})(7.35 \times 10^{22})}{1.74 \times 10^6}}$$
$$v = 2.37 \times 10^3\,\mathrm{m/s}$$
The Moon's escape speed is much lower than Earth's, mostly because the Moon has much less mass.
Example: A small asteroid has mass 8.0 × 1015 kg and radius 500 m. Find its surface gravity and escape speed.
solution
Use the same distance, the asteroid's radius, for both surface calculations.
The surface gravity is lower than Earth's, but not zero. A fast throw could escape.
Example: A probe starts from rest at a distance of 2 Earth radii from Earth's center and falls to Earth's surface. Ignoring atmosphere and Earth's rotation, how fast is it moving when it reaches the surface? Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
Use conservation of energy. The probe loses gravitational potential energy and gains kinetic energy.
This is a large speed because the probe falls through a large change in gravitational potential energy.
Example: A 900 kg spacecraft is at a distance of 2 Earth radii from Earth's center. Find the gravitational acceleration there and the gravitational force on the spacecraft. Use Earth's mass as 5.97 × 1024 kg and Earth's radius as 6.37 × 106 m.
solution
First find the field strength, then multiply by the spacecraft's mass.
This is one-fourth of surface gravity because the spacecraft is twice as far from Earth's center.
Example: Halley's Comet has an orbital period of about 75 years. If Kepler's simple relation is used as an average model for its orbit around the Sun, what is its average distance from the Sun in AU?
solution
This gives an average orbital distance, not the closest or farthest distance, because Halley's orbit is very stretched out.
The average distance is far beyond Saturn's orbit.
Example: Earth is 1.496 × 1011 m from the Sun. Use the Sun's mass as 2.00 × 1030 kg and Earth's mass as 5.97 × 1024 kg. Find the gravitational force on Earth from the Sun, then find Earth's acceleration toward the Sun.
solution
$$F = \frac{GM_1M_2}{r^2}$$
$$F = \frac{(6.67 \times 10^{-11})(2.00 \times 10^{30})(5.97 \times 10^{24})}{(1.496 \times 10^{11})^2}$$
$$F = 3.56 \times 10^{22}\,\mathrm{N}$$
$$a = \frac{F}{m}$$
$$a = \frac{3.56 \times 10^{22}}{5.97 \times 10^{24}}$$
$$a = 5.96 \times 10^{-3}\,\mathrm{m/s^2}$$
The acceleration is small compared with surface gravity, but it acts continuously and keeps Earth in orbit.
Why is the Moon accelerating even though its speed around Earth is approximately constant?
answer
Its velocity direction changes continuously as it travels in a circle. That change in velocity is an acceleration directed toward Earth.
The Moon is about 60 Earth radii from Earth's center. According to an inverse-square gravitational model, how should its gravitational acceleration compare with acceleration at Earth's surface?
answer
It should be about 1/602, or 1/3600, of the surface value. The article's calculated result is close to that ratio.
What was scientifically important about Newton explaining both a falling apple and the Moon with one gravitational model?
answer
It showed that the same physical rule could describe ordinary motion on Earth and motion in the sky. This unification is why the law is called universal gravitation.
Reading (20 minutes): Read The Relativity of Wrong by Isaac Asimov from Skeptical Inquirer. Then answer these questions.
Why does Asimov say that believing Earth is a sphere is not "just as wrong" as believing Earth is flat?
answer
A spherical model makes much better predictions about Earth's shape and large-scale observations. It is still an approximation, but it is far closer to reality than a flat-Earth model.
How does the example of Earth being slightly flattened at the poles show that scientific models can improve?
answer
New measurements can reveal small limits of an earlier useful model. The new model does not erase the old one; it refines it for situations where greater accuracy matters.
What does the essay suggest students should do when they learn that a scientific model has limitations?
answer
They should ask how accurate the model is, where it works, and what evidence suggests a better version. A limited model can still be extremely useful rather than worthless.
Why was it important to test how antimatter falls instead of assuming that the results for ordinary matter must also apply to antimatter?
answer
Experiments had tested gravity only with ordinary matter. Antimatter could have behaved differently, so scientists needed direct evidence rather than an assumption.
The experiment found that antihydrogen fell between 0.46 g and 1.04 g. What conclusion can scientists make, and what can they not yet claim?
answer
They can conclude that antimatter falls downward rather than upward. The range is still broad, so the experiment does not yet show that antimatter falls at exactly the same acceleration as ordinary matter.
Why does the result make negative-mass explanations for cosmic expansion less plausible?
answer
Those ideas would need antimatter to respond to gravity in a dramatically different way. Observing antihydrogen fall downward removes support for that expectation.
