Archive for August, 2015

Computing Potential Energy

Tuesday, August 25th, 2015

      Last week we discovered that objects acquire potential energy as it relates to gravity based on the height those objects are elevated above the ground.   We also introduced an equation to calculate the potential energy of a coffee mug perched on a shelf.   We’ll work with that equation today and compute the latent energy that’s hidden within that mug.

      Here again is the equation to determine potential energy, put in terms relating to gravity,

PEgravitational = m × g × h

where m is the mass of the mug, h is the height it’s been elevated above the floor, and g is the Earth’s acceleration of gravity factor, as explained in my previous blog entitled, Sir Isaac Newton and the Acceleration of Gravity.

      The equation above can be solved using either English or metric units.   In the US it’s generally standard practice to perform calculations using English units, such as feet and pounds.   But when measuring mass a less familiar English unit, the slug, comes into play.   If you’re interested in learning more about this unit, go to a previous blog article entitled, The Force of Gravity.

      The kilogram is the metric equivalent of a slug.   Since it’s the unit of mass most commonly used throughout the world, we’ll use it to perform our calculation.

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      Let’s say our mug has a mass of 0.25 kilograms, the shelf it’s resting on is 2 meters above the floor, and g is 9.8 meters/second2.   The mug’s gravitational potential energy would then be expressed as,

PEgravitational = (0.25 kg) × (9.8 meters/second2) × (2 meters)

PEgravitational = 4.9 kg • meter2/second2

      Next time we’ll expand on our discussion of potential energy and discuss the Law behind the phenomenon and the fact that energy can only be converted from one form to another.

Copyright 2015 – Philip J. O’Keefe, PE

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Gravitational Potential Energy on Earth

Monday, August 17th, 2015

     Last week we concluded our discussion on the force of gravity within our solar system.   Today we’ll turn our attention to the subject of gravity on Earth and exploring the physics behind falling objects.   We’ll start off by discussing potential energy as it relates to gravity, or the latent energy acquired by an object when it’s been elevated above ground level.

      Potential energy was the term adopted by 19th Century Scottish scientist William Rankine to represent the latent, or masked, energy hidden within objects.   As an example, let’s say you’ve placed your favorite coffee mug in its designated spot on the kitchen shelf.   Sitting there so still you wouldn’t dream it was brimming with gravitational potential energy, but if your cat came along and brushed against it, sending it freefalling to the floor, your mug would quickly become a projectile, gaining speed at a uniform rate as it accelerated towards ground level.

      Where did that once passive little cup acquire its mounting energy?  Simply by virtue of the fact it had been lifted by your arm and placed in an elevated position.   You see, Earth’s gravitational pull is forever exerting its invisible tug on objects.   It was tugging at the mug as you lifted it, and the higher you lifted it, the more gravitational potential energy the mug received.   Once perched on the shelf it bridled with latent energy, only to be set free when the cat caused it to lose its support.

      To illustrate the relationship between the coffee mug, the shelf, and Earth’s gravitational pull, we’ll employ the equation used to compute potential energy, notated in terms of gravity,

PEgravitational = m × g × h

      This equation states that the mug’s gravitational potential energy, PEgravitational, is a factor of its mass, m, Earth’s gravitational pull, g, and the mug’s height above ground level, h.

      Within the scientific community g is referred to as Earth’s acceleration of gravity, a phenomenon commonly accepted to be the uniform accelerating rate at which an object falls on Earth, equal to 9.8 meters per second per second, or meters/second2.   It represents a rate of constant acceleration, which happens to be precisely the same whether the object falling is a brick, feather, or coffee mug.

      Next time we’ll work with the potential energy equation which will enable us to see how the curious orange kitty sets loose the latent power held within that mug.

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Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog


How Big is the Sun?

Monday, August 10th, 2015

      Last time we calculated the sun’s force of gravity acting upon Earth.   It was the final unknown quantity within Newton’s equation to determine the mass of the sun, an equation we’ve been working with for some time now.   Today we’re set to discover just how big the sun is.

      Newton’s formula, introduced in a past blog in this series entitled, Gravity and the Mass of the Sun is again,

M = (Fg × r2) ÷  (m × G)

where G is the universal gravitational constant as determined by Henry Cavendish and discussed in our blog, How Big is the Earth? and is equal to,

G = 6.67 × 10-11 meters per kilogram • second2

      As discussed in last week’s blog, The Sun’s Gravitational Force, Earth’s mass, m, its distance from the sun, r, and the force of the sun’s gravity acting upon Earth, Fg , are respectively,

m = 5.96 × 1024 kilograms

r = 149,000,000,000 meters

Fg = 3.52 × 1022 Newtons

      Inserting these values into Newton’s equation to determine the mass, M, of the sun we get:

M = [(3.52 × 1022) × (149,000,000,000)2] ÷ [(5.96 × 1024) × (6.67 × 10-11)]

M = 1.96 × 1030 kilograms

      So how big is 1.96 × 1030 kilograms?   To get a better idea, let’s divide the sun’s mass, M, by the Earth’s mass, m,

(1.96 × 1030 kilograms) ÷ (5.96 × 1024 kilograms) = 328,859.06

      That’s a big number, and it translates to the sun being over 300,000 times more massive than Earth.   The picture below displays this comparison in stunning visual terms.

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      Once 19th Century scientists had calculated the mass of the sun, they went on to calculate the masses of other heavenly bodies in our solar system and the gravitational forces at play on each of them.   Armed with this information mankind was able to subsequently build exploratory probes capable of extending their reach into the far unknowns of our solar system and beyond.

      This ends our discussion on gravity within our solar system.   Next time we’ll return to Earth and begin exploring the physics behind falling objects.


The Sun’s Gravitational Force

Monday, August 3rd, 2015

      Last time we discovered that Earth zips around the sun at the mind boggling speed of 29,680 meters per second.   This is the final bit of information required to calculate Fg, the gravitational force exerted upon Earth by its sun, as set out in Newton’s equation on the subject and derived from his Second Law of Motion.   We’ll calculate that quantity today.

      Newton’s formula that we’ll be working with is,

Fg = [m ×  v2] ÷  r

where Earth’s speed, or orbital velocity, is the v in the equation.   The other variables, m and r, have previously been determined in this blog series.   For a refresher see Centripital Force Makes the Earth Go Round, What is Earth’s Mass, and Calculating the Distance to the Sun.   Earth’s mass, m, is valued at 5.96 × 1024 kilograms, while r is Johannes Kepler’s astronomical unit, equal to about 149,000,000,000 meters.

      Inserting these numerical values into Newton’s equation to determine the sun’s gravitational force acting upon Earth we arrive at,

Fg = [(5.96 × 1024 kilograms) × (29,680 meters per second)2] ÷ 149,000,000,000 meters

Fg = 3.52 × 1022 kilogram • meter per second2

      This metric unit of force, kilogram • meter per second2, represents kilograms multiplied by meters, and their product divided by seconds squared.   It’s known in scientific circles as the Newton, in honor of Sir Isaac Newton, widely recognized as one of the greatest scientists of all time and a key figure in the scientific revolution that began over three centuries ago.   Therefore the sun’s gravitational force acting upon Earth is typically referred to as,

Fg = 3.52 × 1022 Newtons

      Here in the US where we like to use English units such as feet and pounds, the Newton is said to equal 0.225 pounds of force.   Therefore in English units the sun’s gravitational force is expressed as,

Fg = (3.52 × 1022 Newtons) × (0.225 pounds of force per Newton)

Fg = 7.93 × 1021 pounds

      That’s scientific notation for 7,930,000,000,000,000,000,000 pounds!   That’s the amount of force exerted by the sun’s gravitational pull on Earth.   Seems about right — right?

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      Now that we know Fg, we have everything we need to calculate the mass of the sun, which in turn enables us to determine the mass and gravity of other planets in our solar system.   We’ll calculate the sun’s mass next time.