All about the Gravitational Force (For Beginners)

(This is an excerpt from The Book of Forces)

All objects exert a gravitational pull on all other objects. The Earth pulls you towards its center and you pull the Earth towards your center. Your car pulls you towards its center and you pull your car towards your center (of course in this case the forces involved are much smaller, but they are there). It is this force that invisibly tethers the Moon to Earth, the Earth to the Sun, the Sun to the Milky Way Galaxy and the Milky Way Galaxy to its local galaxy cluster.

Experiments have shown that the magnitude of the gravitational attraction between two bodies depends on their masses. If you double the mass of one of the bodies, the gravitational force doubles as well. The force also depends on the distance between the bodies. More distance means less gravitational pull. To be specific, the gravitational force obeys an inverse-square law. If you double the distance, the pull reduces to 1/2² = 1/4 of its original value. If you triple the distance, it goes down to 1/3² = 1/9 of its original value. And so on. These dependencies can be summarized in this neat formula:

F = G·m·M / r²

With F being the gravitational force in Newtons, m and M the masses of the two bodies in kilograms, r the center-to-center distance between the bodies in meters and G = 6.67·10^(-11) N m² kg^(-2) the (somewhat cumbersome) gravitational constant. With this great formula, that has first been derived at the end of the seventeenth century and has sparked an ugly plagiarism dispute between Newton and Hooke, you can calculate the gravitational pull between two objects for any situation.


(Gravitational attraction between two spherical masses)

If you have trouble applying the formula on your own or just want to play around with it a bit, check out the free web applet Newton’s Law of Gravity Calculator that can be found on the website of the UNL astronomy education group. It allows you to set the required inputs (the masses and the center-to-center distance) using sliders that are marked special values such as Earth’s mass or the distance Earth-Moon and calculates the gravitational force for you.


Example 3:

Calculate the gravitational force a person of mass m = 72 kg experiences at the surface of Earth. The mass of Earth is M = 5.97·10^24 kg (the sign ^ stands for “to the power”) and the distance from the center to the surface r = 6,370,000 m. Using this, show that the acceleration the person experiences in free fall is roughly 10 m/s².


To arrive at the answer, we simply insert all the given inputs into the formula for calculating gravitational force.

F = G·m·M / r²
F = 6.67·10^(-11)·72·5.97·10^24 / 6,370,000² N ≈ 707 N

So the magnitude of the gravitational force experienced by the m = 72 kg person is 707 N. In free fall, he or she is driven by this net force (assuming that we can neglect air resistance). Using Newton’s second law we get the following value for the free fall acceleration:

F = m·a
707 N = 72 kg · a

Divide both sides by 72 kg:

a = 707 / 72 m/s² ≈ 9.82 m/s²

Which is roughly (and more exact than) the 10 m/s² we’ve been using in the introduction. Except for the overly small and large numbers involved, calculating gravitational pull is actually quite straight-forward.

As mentioned before, gravitation is not a one-way street. As the Earth pulls on the person, the person pulls on the Earth with the same force (707 N). However, Earth’s mass is considerably larger and hence the acceleration it experiences much smaller. Using Newton’s second law again and the value M = 5.97·1024 kg for the mass of Earth we get:

F = m·a
707 N = 5.97·10^24 kg · a

Divide both sides by 5.97·10^24 kg:

a = 707 / (5.97·10^24) m/s² ≈ 1.18·10^(-22) m/s²

So indeed the acceleration the Earth experiences as a result of the gravitational attraction to the person is tiny.


Example 4:

By how much does the gravitational pull change when the  person of mass m = 72 kg is in a plane (altitude 10 km = 10,000 m) instead of the surface of Earth? For the mass and radius of Earth, use the values from the previous example.


In this case the center-to-center distance r between the bodies is a bit larger. To be specific, it is the sum of the radius of Earth 6,370,000 m and the height above the surface 10,000 m:

r = 6,370,000 m + 10,000 m = 6,380,000 m

Again we insert everything:

F = G·m·M / r²
F = 6.67·10^(-11)·72·5.97·10^24 / 6,380,000² N ≈ 705 N

So the gravitational force does not change by much (only by 0.3 %) when in a plane. 10 km altitude are not much by gravity’s standards, the height above the surface needs to be much larger for a noticeable difference to occur.


With the gravitational law we can easily show that the gravitational acceleration experienced by an object in free fall does not depend on its mass. All objects are subject to the same 10 m/s² acceleration near the surface of Earth. Suppose we denote the mass of an object by m and the mass of Earth by M. The center-to-center distance between the two is r, the radius of Earth. We can then insert all these values into our formula to find the value of the gravitational force:

F = G·m·M / r²

Once calculated, we can turn to Newton’s second law to find the acceleration a the object experiences in free fall. Using F = m·a and dividing both sides by m we find that:

a = F / m = G·M / r²

So the gravitational acceleration indeed depends only on the mass and radius of Earth, but not the object’s mass. In free fall, a feather is subject to the same 10 m/s² acceleration as a stone. But wait, doesn’t that contradict our experience? Doesn’t a stone fall much faster than a feather? It sure does, but this is only due to the presence of air resistance. Initially, both are accelerated at the same rate. But while the stone hardly feels the effects of air resistance, the feather is almost immediately slowed down by the collisions with air molecules. If you dropped both in a vacuum tube, where no air resistance can build up, the stone and the feather would reach the ground at the same time! Check out an online video that shows this interesting vacuum tube experiment, it is quite enlightening to see a feather literally drop like a stone.


(All bodies are subject to the same gravitational acceleration)

Since all objects experience the same acceleration near the surface of Earth and since this is where the everyday action takes place, it pays to have a simplified equation at hand for this special case. Denoting the gravitational acceleration by g (with g ≈ 10 m/s²) as is commonly done, we can calculate the gravitational force, also called weight, an object of mass m is subject to at the surface of Earth by:

F = m·g

So it’s as simple as multiplying the mass by ten. Depending on the application, you can also use the more accurate factor g ≈ 9.82 m/s² (which I will not do in this book). Up to now we’ve only been dealing with gravitation near the surface of Earth, but of course the formula allows us to compute the gravitational force and acceleration near any other celestial body. I will spare you trouble of looking up the relevant data and do the tedious calculations for you. In the table below you can see what gravitational force and acceleration a person of mass m = 72 kg would experience at the surface of various celestial objects. The acceleration is listed in g’s, with 1 g being equal to the free-fall acceleration experienced near the surface of Earth.


So while jumping on the Moon would feel like slow motion (the free-fall acceleration experienced is comparable to what you feel when stepping on the gas pedal in a common car), you could hardly stand upright on Jupiter as your muscles would have to support more than twice your weight. Imagine that! On the Sun it would be even worse. Assuming you find a way not get instantly killed by the hellish thermonuclear inferno, the enormous gravitational force would feel like having a car on top of you. And unlike temperature or pressure, shielding yourself against gravity is not possible.

What about the final entry? What is a neutron star and why does it have such a mind-blowing gravitational pull? A neutron star is the remnant of a massive star that has burned its fuel and exploded in a supernova, no doubt the most spectacular light-show in the universe. Such remnants are extremely dense – the mass of several suns compressed into an almost perfect sphere of just 20 km radius. With the mass being so large and the distance from the surface to the center so small, the gravitational force on the surface is gigantic and not survivable under any circumstances.

If you approached a neutron star, the gravitational pull would actually kill you long before reaching the surface in a process called spaghettification. This unusual term, made popular by the brilliant physicist Stephen Hawking, refers to the fact that in intense gravitational fields objects are vertically stretched and horizontally compressed. The explanation is rather straight-forward: since the strength of the gravitational force depends on the distance to the source of said force, one side of the approaching object, the side closer to the source, will experience a stronger pull than the opposite side. This leads to a net force stretching the object. If the gravitational force is large enough, this would make any object look like a thin spaghetti. For humans spaghettification would be lethal as the stretching would cause the body to break apart at the weakest spot (which presumably is just above the hips). So my pro-tip is to keep a polite distance from neutron stars.

Antimatter Production – Present and Future

When it comes to using antimatter for propulsion, getting sufficient amounts of the exotic fuel is the biggest challenge. For flights within the solar system, hybrid concepts would require several micrograms of antimatter, while pure antimatter rockets would consume dozens of kilograms per trip. And going beyond the solar system would demand the production of several metric tons and more.

We are very, very far from this. Currently around 10 nanograms of anti-protons are produced in the large particles accelerators each year. At this rate it would take 100 years to produce one measly microgram and 100 billion years to accumulate one kilogram. However, the antimatter production rate has seen exponential growth, going up by sixteen orders of magnitude over the past decades, and this general trend will probably continue for some time.

Even with a noticeable slowdown in this exponential growth, gram amounts of anti-protons could be manufactured each year towards the end of the 21st century, making hybrid antimatter propulsion feasible. With no slowdown, the rate could even reach kilograms per year by then. While most physicists view this as an overly optimistic estimate, it is not impossible considering the current trend in antimatter production and the historic growth of liquid hydrogen and uranium production rates (both considered difficult to manufacture in the past).

There is still much to be optimized in the production of antimatter. The energy efficiency at present is only 10-9, meaning that you have to put in one gigajoule of pricey electric energy to produce a single joule of antimatter energy. The resulting costs are a staggering 62.5 trillion USD per gram of anti-protons, making antimatter the most expensive material known to man. So if you want to tell your wife how precious she is to you (and want to get rid of her at the same time), how about buying her a nice anti-matter scarf?

Establishing facilities solely dedicated to antimatter production, as opposed to the by-product manufacturing in modern particle accelerators, would significantly improve the situation. NASA experts estimate that an investment of around 5 billion USD is sufficient to build such a first generation antimatter factory. This step could bring the costs of anti-protons down to 25 billion USD per gram and increase the production rate to micrograms per year.

While we might not see kilogram amounts of antimatter or antimatter propulsion systems in our lifetime, the production trend over the next few decades will reveal much about the feasibility of antimatter rockets and interstellar travel. If the optimists are correct, and that’s a big if, the grandchildren of our grandchildren’s grandchildren might watch the launch of the first spacecraft capable of reaching neighboring stars. Sci-fi? I’m sure that’s what people said about the Moon landing and close-up pictures from Mars and Jupiter just a lifetime ago.

For more information, check out my ebook Antimatter Propulsion.

New Release for Kindle: Antimatter Propulsion

I’m very excited to announce the release of my latest ebook called “Antimatter Propulsion”. I’ve been working working on it like a madman for the past few months, going through scientific papers and wrestling with the jargon and equations. But I’m quite satisfied with the result. Here’s the blurb, the table of contents and the link to the product page. No prior knowledge is required to enjoy the book.

Many popular science fiction movies and novels feature antimatter propulsion systems, from the classic Star Trek series all the way to Cameron’s hit movie Avatar. But what exactly is antimatter? And how can it be used accelerate rockets? This book is a gentle introduction to the science behind antimatter propulsion. The first section deals with antimatter in general, detailing its discovery, behavior, production and storage. This is followed by an introduction to propulsion, including a look at the most important quantities involved and the propulsion systems in use or in development today. Finally, the most promising antimatter propulsion and rocket concepts are presented and their feasibility discussed, from the solid core concept to antimatter initiated microfusion engines, from the Valkyrie project to Penn State’s AIMStar spacecraft.

Section 1: Antimatter

The Atom
Dirac’s Idea
An Explosive Mix
Proton and Anti-Proton Annihilation
Sources of Antimatter
Storing Antimatter
Getting the Fuel

Section 2: Propulsion Basics

Conservation of Momentum
♪ Throw, Throw, Throw Your Boat ♫
So What’s The Deal?
May The Thrust Be With You
Specific Impulse and Fuel Requirements
Chemical Propulsion
Electric Propulsion
Fusion Propulsion

Section 3: Antimatter Propulsion Concepts

Solid Core Concept
Plasma Core Concept
Beamed Core Concept
Antimatter Catalyzed Micro-Fission / Fusion
Antimatter Initiated Micro-Fusion

Section 4: Antimatter Rocket Concepts

Project Valkyrie
Dust Shields

You can purchase “Antimatter Propulsion” here for $ 2.99.

Motion With Constant Acceleration (Examples, Exercises, Solutions)

An abstraction often used in physics is motion with constant acceleration. This is a good approximation for many different situations: free fall over small distances or in low-density atmospheres, full braking in car traffic, an object sliding down an inclined plane, etc … The mathematics behind this special case is relatively simple. Assume the object that is subject to the constant acceleration a (in m/s²) initially has a velocity v(0) (in m/s). Since the velocity is the integral of the acceleration function, the object’s velocity after time t (in s) is simply:

1) v(t) = v(0) + a · t

For example, if a car initially goes v(0) = 20 m/s and brakes with a constant a = -10 m/s², which is a realistic value for asphalt, its velocity after a time t is:

v(t) = 20 – 10 · t

After t = 1 second, the car’s speed has decreased to v(1) = 20 – 10 · 1 = 10 m/s and after t = 2 seconds the car has come to a halt: v(2) = 20 – 10 · 2 = 0 m/s. As you can see, it’s all pretty straight-forward. Note that the negative acceleration (also called deceleration) has led the velocity to decrease over time. In a similar manner, a positive acceleration will cause the speed to go up. You can read more on acceleration in this blog post.

What about the distance x (in m) the object covers? We have to integrate the velocity function to find the appropriate formula. The covered distance after time t is:

2) x(t) = v(0) · t + 0.5 · a · t²

While that looks a lot more complicated, it is really just as straight-forward. Let’s go back to the car that initially has a speed of v(0) = 20 m/s and brakes with a constant a = -10 m/s². In this case the above formula becomes:

x(t) = 20 · t – 0.5 · 10 · t²

After t = 1 second, the car has traveled x(1) = 20 · 1 – 0.5 · 10 · 1² = 15 meters. By the time it comes to a halt at t = 2 seconds, it moved x(2) = 20 · 2 – 0.5 · 10 · 2² = 20 meters. Note that we don’t have to use the time as a variable. There’s a way to eliminate it. We could solve equation 1) for t and insert the resulting expression into equation 2). This leads to a formula connecting the velocity v and distance x.

3) Constant acceleration_html_b85f3ec

Solved for x it looks like this:

3)’ Constant acceleration_html_m23bb2bb3

It’s a very useful formula that you should keep in mind. Suppose a tram accelerates at a constant a = 1.3 m/s², which is also a realistic value, from rest (v(0) = 0 m/s). What distance does it need to go to full speed v = 10 m/s? Using equation 3)’ we can easily calculate this:

Constant acceleration_html_m11de6604


Here are a few exercises and solutions using the equations 1), 2) and 3).

1. During free fall (air resistance neglected) an object accelerates with about a = 10 m/s. Suppose the object is dropped, that is, it is initially at rest (v(0) = 0 m/s).

a) What is its speed after t = 3 seconds?
b) What distance has it traveled after t = 3 seconds?
c) Suppose we drop the object from a tower that is x = 20 meters tall. At what speed will it impact the ground?
d) How long does the drop take?

Hint: in exercise d) solve equation 1) for t and insert the result from c)

2. During the reentry of space crafts accelerations can be as high as a = -70 m/s². Suppose the space craft initially moves with v(0) = 6000 m/s.

a) What’s the speed and covered distance after t = 10 seconds?
b) How long will it take the space craft to half its initial velocity?
c) What distance will it travel during this time?

3. An investigator arrives at the scene of a car crash. From the skid marks he deduces that it took the car a distance x = 55 meters to come to a halt. Assume full braking (a = -10 m/s²). Was the car initially above the speed limit of 30 m/s?


Solutions to the exercises:

Exercise 1

a) 30 m/s
b) 45 m
c) 20 m/s
d) 2 s

Exercise 2

a) 5,300 m/s and 56,500 m
b) 42.9 s (rounded)
c) 192,860 m (rounded)

Exercise 3

Yes (he was initially going 33.2 m/s)


To learn the basic math you need to succeed in physics, check out the e-book “Algebra – The Very Basics”. For an informal introduction to physics, check out the e-book “Physics! In Quantities and Examples”. Both are available at low prices and exclusively for Kindle.

The Jeans Mass, or: How are stars born?

No, this has nothing to do with pants. The Jeans mass is a concept used in astrophysics and its unlikely name comes from the British physicist Sir James Jeans, who researched the conditions of star formation. The question at the core is: under what circumstances will a dark and lonely gas cloud floating somewhere in the depth of space turn into a shining star? To answer this, we have to understand what forces are at work.

One obvious factor is gravitation. It will always work towards contracting the gas cloud. If no other forces were present, it would lead the cloud to collapse into a single point. The temperature of the cloud however provides an opposite push. It “equips” the molecules of the cloud with kinetic energy (energy of motion) and given a high enough temperature, the kinetic energy would be sufficient for the molecules to simply fly off into space, never to be seen again.

It is clear that no star will form if the cloud expands and falls apart. Only when gravity wins this battle of inward and outward push can a stable star result. Sir James Jeans did the math and found that it all boils down to one parameter, the Jeans mass. If the actual mass of the interstellar cloud is larger than this critical mass, it will contract and stellar formation occurs. If on the other hand the actual mass is smaller, the cloud will simply dissipate.

The Jeans mass depends mainly on the temperature T (in K) and density D (in kg/m³) of the cloud. The higher the temperature, the larger the Jeans mass will be. This is in line with our previous discussion. When the temperature is high, a larger amount of mass is necessary to overcome the thermal outward push. The value of the Jeans mass M (in kg) can be estimated from this equation:

M ≈ 1020 · sqrt(T³ / D)

Typical values for the temperature and density of interstellar clouds are T = 10 K and D = 10-22 kg/m³. This leads to a Jeans mass of M = 1.4 · 1032 kg. Note that the critical mass turns out to be much greater than the mass of a typical star, indicating that stars generally form in clusters. Rather than the cloud contracting into a single star, which is the picture you probably had in your mind during this discussion, it will fragment at some point during the contraction and form multiple stars. So stars always have brothers and sisters.

(This was an excerpt from the Kindle book Physics! In Quantities and Examples)

The Difference Between Mass and Weight

In general, it is acceptable to use weight as a synonym for mass. However, in a very strict physical sense this is incorrect. Weight is the gravitational force experienced by an object and accordingly measured in Newtons and not kilograms. An object of mass m has the weight F:

F = m · g

with the gravitational acceleration g. On Earth the value of the gravitational acceleration at the surface is g = 9.81 m/s². So a typical adult with a mass of m = 75 kg has a weight of:

F = 75 kg · 9.81 m/s² = 735.75 N

On the moon (or any other point of the universe), the mass would remain at m = 75 kg. But since the gravitational acceleration on the moon is much lower (g = 1.62 m/s²), the weight changes to:

F = 75 kg · 1.62 m/s² = 121.5 N

Keep this distinction in mind. Mass is a fundamental property of an object that does not depend on the conditions outside the object, while weight is a variable that changes with the strength of surrounding gravitational field.

(This was an excerpt from Physics! In Quantities and Examples)

What is Mass? A Short and Simple Explanation

Mass is such a fundamental property of matter that it is hard to define without drifting into philosophical realms. Newton’s Second Law provides a great way to understand mass from a physical point of view. The law states that force F (in N) is the product of mass m (in kg) and acceleration a (in m/s²):

F = m · a

So according to this, mass is a measure of an object’s resistance to a change in speed. If the mass is small, a small force is sufficient to produce a noticeable acceleration. However, much more force is necessary to produce the same acceleration for a massive object.

Another way of looking at mass is provided by Newton’s Law of Gravitation. Newton found that the attracting gravitational force between two objects is proportional to the product of their masses m and M:

F ~ m · M

So additionally to creating resistance to changes in state of motion, mass is also the source of gravitational attraction. It seems obvious that in both cases we are talking about the same quantity. But is this actually the case? Is the inertial mass, the mass responsible for opposing changes in velocity, really the same as the gravitational mass, that gives rise to gravity?

This question has led to heated debates among physicist for centuries. All experiments conducted so far, with ever increasing accuracy, have shown that indeed the inertial mass is identical to the gravitational mass. Today, almost all physicists have accepted this equivalence as reality.

The SI unit of mass is kilograms. Ever since 1889, one kilogram has been defined as the mass of the international prototype kilogram (IPK) that is stored in the International Bureau of Weights and Measures in Paris. However, during the 24th General Conference on Weights and Measures that took place in 2011, physicists have agreed to redefine this unit by connecting it to the Planck constant.

Other units that are commonly used for mass are grams (1/1000 of a kilogram), the pound (equal to about 0.45 kilograms) and the tonne (equal to 1000 kilograms). For atoms and molecules scientists use the atomic mass unit u. One u is equivalent to 1.66 · 10-27 kg, which is roughly the mass of a neutron or proton.

(This was an excerpt from Physics! In Quantities and Examples)

Released Today for Kindle: Physics! In Quantities and Examples

I finally finished and released my new ebook … took me longer than usual because I always kept finding new interesting topics while researching. Here’s the blurb, link and TOC:

This book is a concept-focused and informal introduction to the field of physics that can be enjoyed without any prior knowledge. Step by step and using many examples and illustrations, the most important quantities in physics are gently explained. From length and mass, over energy and power, all the way to voltage and magnetic flux. The mathematics in the book is strictly limited to basic high school algebra to allow anyone to get in and to assure that the focus always remains on the core physical concepts.

(Click cover to get to the Amazon Product Page)


Table of Contents:

(Introduction, From the Smallest to the Largest, Wavelength)

(Introduction, Mass versus Weight, From the Smallest to the Largest, Mass Defect and Einstein, Jeans Mass)

Speed / Velocity
(Introduction, From the Smallest to the Largest, Faster than Light, Speed of Sound for all Purposes)

(Introduction, From the Smallest to the Largest, Car Performance, Accident Investigation)

(Introduction, Thrust and the Space Shuttle, Force of Light and Solar Sails, MoND and Dark Matter, Artificial Gravity and Centrifugal Force, Why do Airplanes Fly?)

(Introduction, Surface Area and Heat, Projected Area and Planetary Temperature)

(Introduction, From the Smallest to the Largest, Hydraulic Press, Air Pressure, Magdeburg Hemispheres)

(Introduction, Poisson’s Ratio)

(Introduction, From the Smallest to the Largest, Bulk Density, Water Anomaly, More Densities)

(Introduction, From the Smallest to the Largest, Thermal Expansion, Boiling, Evaporation is Cool, Why Blankets Work, Cricket Temperature)

(Introduction, Impact Speed, Ice Skating, Dear Radioactive Ladies and Gentlemen!, Space Shuttle Reentry, Radiation Exposure)

(Introduction, From the Smallest to the Largest, Space Shuttle Launch and Sound Suppression)

(Introduction, Inverse Square Law, Absorption)

(Introduction, Perfectly Inelastic Collisions, Recoil, Hollywood and Physics, Force Revisited)

Frequency / Period
(Introduction, Heart Beat, Neutron Stars, Gravitational Redshift)

Rotational Motion
(Extended Introduction, Moment of Inertia – The Concept, Moment of Inertia – The Computation, Conservation of Angular Momentum)

(Extended Introduction, Stewart-Tolman Effect, Piezoelectricity, Lightning)

(Extended Introduction, Lorentz Force, Mass Spectrometers, MHD Generators, Earth’s Magnetic Field)

Scalar and Vector Quantities
Measuring Quantities
Unit Conversion
Unit Prefixes
Copyright and Disclaimer

As always, I discounted the book in countries with a low GDP because I think that education should be accessible for all people. Enjoy!

Car Dynamics – Sliding and Overturning

In this post we will take a look at car performance in curves. Of central importance for our considerations is the centrifugal force. Whenever a body is moving in a curved path, this force comes into play. You probably felt it many times in your car. It is the force that tries to push you out of a curve as you go through it.

The centrifugal force C (in N) depends on three factors: the velocity v (in m/s) of the car, its mass m (in kg) and the radius r (in m) of the curve. Given these quantities, we can easily compute the centrifugal force using this formula:

C = m · v² / r

Note the quadratic dependence on speed. If you double the car’s speed, the centrifugal force quadruples. With this force acting, there must be a counter-force to cancel it for the car not to slide. This force is provided by the sideways friction of the tires. The frictional force F (in N) can be calculated from the so called coefficient of friction μ (dimensionless), the car mass m and the gravitational acceleration g (in m/s²).

F = μ · m · g

The coefficient of friction depends mainly on the road type and condition. On dry asphalt we can set μ ≈ 0.8, on wet asphalt μ ≈ 0.6, on snow μ ≈ 0.2 and on ice μ ≈ 0.1. At low speeds the frictional force exceeds the centrifugal force and the car will be able to go through the curve without any problems. However, as we increase the velocity, so does the centrifugal force and at a certain critical velocity the forces cancel each other out. Any increase in speed from this point on will result in the car sliding.

We can compute the critical speed s (in m/s) by equating the expressions for the forces:

m · s² / r = μ · m · g

s = sqrt (μ · r · g)

This is the speed at which the car begins to slide. Note that there’s no dependence on mass anymore. Since both the centrifugal as well as the frictional force grow proportionally to the car’s mass, it doesn’t play a role in determining the critical speed for sliding. All that’s left in terms of variables is the coefficient of friction (lower friction, lower critical speed) and the radius of the curve (smaller radius, more narrow curve, smaller critical speed).

However, sliding is not the only problem that can occur in curves. Under certain circumstances a car can also overturn. Again the centrifugal force is the culprit. Assuming the center of gravity (in short: CG) of the car is at a height of h (in m), the centrifugal force will produce a torque T acting to overturn the car:

T = h · C = m · v² · h / r

On the other hand, there’s the weight of the car giving rise to an opposing torque T’ that grows with the width w (in m) and mass m of the car:

T’ = 0.5 · m · g · w

At low speeds, the torque caused by the centrifugal force will be lower than the one caused by the gravitational pull. But at a certain critical speed o (in m/s), the torques will cancel each other and any further increase in speed will result in the car overturning. Equating the above expressions, we get:

m · o² · h / r = 0.5 · m · g · w

o = sqrt (0.5 · r · g · w / h)

Aside from the coefficient of friction, the determining factor here is the ratio of width to height. The larger it is, the harder it will be for the centrifugal force to overturn the car. This is why lowering a car when intending to go fast makes sense. If you lower the CG while keeping the width the same, the ratio w / h, and thus the critical speed for overturning, will increase.

Let’s look at some examples before drawing a final conclusion from these truly great formulas.


According to the center of gravity of a 2014 BMW 435i is h = 0.5 m above the ground. The width of the car is about w = 1.8 m. Calculate the critical speed for sliding and overturning in a curve of radius r = 300 m on a dry asphalt road (μ ≈ 0.8).

Nothing to do but to apply the formulas:

s = sqrt (0.8 · 300 m · 9.81 m/s²)

s ≈ 49 m/s ≈ 175 km/h ≈ 108 mph

So with normal driving behavior you certainly won’t get anywhere near sliding. But note that sudden steering in a curve can cause the radius of the your car’s path to be considerably lower than the actual curve radius.

Onto the critical overturning speed:

o = sqrt (0.5 · 300 m · 9.81 m/s² · 3.6)

o ≈ 73 m/s ≈ 262 km/h ≈ 162 mph

Not even Michael Schumacher could bring this car to overturn.


How would the critical speeds change if we drove the 2014 BMW 435i through the same curve on an icy road? In this case the coefficient is considerably lower (μ ≈ 0.1). For the critical sliding speed we get:

s = sqrt (0.1 · 300 m · 9.81 m/s²)

s ≈ 17 m/s ≈ 62 km/h ≈ 38 mph

So even this sweet sport car is in danger of sliding relatively quickly under these conditions. What about the overturning speed? Well, it has nothing to do with the friction of the tires, so it will still be at 73 m/s.


This was an excerpt from More Great Formulas Explained. Interested in more car dynamics? Take a look at my post on How to Compute Maximum Car Speed. For other interesting physics articles, check out my BEST OF. I hope you enjoyed and drive safe!