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 F_{g}, 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, F_{g} = [m × v^{2}] ÷ 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 × 10^{24} 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, F_{g} = [(5.96 × 10^{24} kilograms) × (29,680 meters per second)^{2}] ÷ 149,000,000,000 meters F_{g} = 3.52 × 10^{22} kilogram • meter per second^{2} This metric unit of force, kilogram • meter per second^{2}, 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, F_{g} = 3.52 × 10^{22} 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, F_{g} = (3.52 × 10^{22} Newtons) × (0.225 pounds of force per Newton) F_{g} = 7.93 × 10^{21 }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? Now that we know F_{g}, 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. ____________________________________

Posts Tagged ‘force of gravity’
The Sun’s Gravitational Force
Monday, August 3rd, 2015Earth’s High Speed Race Around the Sun
Tuesday, July 28th, 2015
Last time we began our discussion on velocity by focusing on one of its aspects, distance, and we calculated Earth’s orbital distance around the sun to be 5,816,023,200 miles. Today we’ll focus on velocity’s other aspect, time. Together, these aspects will allow us to solve for Earth’s orbital velocity, aka speed. When early astronomers monitored Earth’s journeys, they found it took exactly one year for it to complete its orbit around the sun. They combined this fact with Earth’s orbital travel distance of 5,816,023,200, or 9.36 × 10^{11} meters — meters being the unit of measurement most often used in scientific computations — and solved for Earth’s orbital velocity as follows, v = 9.36 × 10^{11} meters ÷ 1 year = 9.36 × 10^{11} meters per year The scientific notation of 9.36 × 10^{11 }equates to 936,000,000,000 meters, a large and unruly number to work with. We can simplify things further by breaking this number down into units of meters per second, which will then allow us to arrive at Earth’s velocity in terms of miles per hour, something most Earthlings can relate to. One meter per second is equal to 2.237 miles per hour, and there are 31,536,000 seconds in one year, so breaking Earth’s orbital velocity down into meters per second we arrive at, v = 9.36 × 10^{11} meters ÷ 31,536,000 seconds = 29,680 meters per second v = (29,680 meters per second) × (2.237 mph/m/sec) = 66,394 miles per hour Yes, it’s true, Earth whips around the sun at warp speed. Our fastest man made rockets only achieve speeds of about 25,000 miles per hour. Earth beats them twotoone! Now that we know Earth’s orbital velocity, we have everything we need to calculate the gravitational force exerted on Earth by the sun. We’ll do that next time.
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Earth’s Orbital Velocity
Sunday, July 19th, 2015
Last time we introduced Newton’s equation to calculate the sun’s gravitational force acting upon Earth, and today we’ll begin solving for the last remaining unsolved variable within that equation, v, Earth’s orbital velocity. Here again is Newton’s equation, F_{g} = [m × v^{2}] ÷ r For a refresher on how we solved for m, Earth’s mass, and r, the distance between Earth and the sun, follow these links to past blogs in this series, What is Earth’s Mass and Calculating the Distance to the Sun. Velocity, or speed, as it’s most commonly referred to, is based on both time and distance. To bear this out we’ll use an object and situation familiar to all of us, traveling in a car. The car’s velocity is a factor of both the distance traveled and the time it takes to get there. A car traveling at a velocity of 30 miles per hour will cover a distance of 30 miles in one hour’s time. This relationship is borne out by the formula, v_{Car} = distance traveled ÷ travel time v_{Car} = 30 miles ÷ 1 hour = 30 miles per hour Similarly, v is the distance Earth travels during its orbital journey around the sun within a specified period of time. It had been observed since ancient times that it takes Earth one year to complete one orbit, so all that remained to be done was calculate the distance Earth traveled during that time. Vital to calculations was the fact that Earth’s orbit is a circle, which allows geometry to be employed and calculations to be thereby simplified. Refer to Figure l. Figure 1 From geometry we know that the circumference of a circle, C, is calculated by, C = 2× π × r where π is a constant, the well known mathematical term pi, which is equal to 3.1416, and r is the radius of Earth’s circular orbit, determined, courtesy of the work of Johannes Kepler and Edmund Halley, to be approximately 93,000,000 miles. Stated in metric units, the unit of measurement most often employed in science, that comes to 149,000,000,000 meters. Inserting these numerical values for π and r into the circumference formula, scientists calculated the distance Earth travels in one orbit around the sun to be, C = 2 × π × 149,000,000,000 meters = 9.36 x 10^{11} meters Next time we’ll introduce the time element into our equations and solve for v, and from there we’ll go on and finally solve for F_{g},_{ }the sun’s gravitational force acting upon Earth.
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Centripetal Force
Friday, July 3rd, 2015
Have you ever wondered how Earth keeps its steady orbit around its life sustaining sun, or what prevents it from breaking away and flying off willynilly into the universe? It’s more than just simple gravity, it’s the physics behind centripetal force, the topic we’ll be exploring today. We’ve been working our way towards a full discussion on gravity in this long blog series, navigating subjects such as the behavior of falling objects, the acceleration of gravity, the masses of Earth and the sun, and the optical measurement of cosmic distances. We’ve now come full circle from my opening blog on the subject, Gravity and the Mass of the Sun. In that blog an equation was introduced as a means to calculate the mass of the sun, and in that equation is the variable we’ll be working towards solving today, F_{g, }the sun’s gravitational force upon the Earth. Here again is that equation, M = (F_{g} × r^{2}) ÷ (m × G) Gravity, mass, and distance all come into play in forming the structure of our universe, and the variables in this equation reflect that: M, the mass of the sun, r the distance between Earth and the sun, m the Earth’s mass, and G the universal gravitational constant. With the exception of F_{g}, all variables in this equation have already been solved for in previous blogs in this series. For a refresher go to, Calculating the Distance to the Sun, What is Earth’s Mass? and Newton’s Law of Gravitation and the Universal Gravitational Constant. As there is no direct means to measure the cosmic quantity, F_{g}, we’re left to an indirect method for its computation. The indirect method is based on the phenomenon of centripetal force, F_{c} something most children become acquainted with when they experience the thrill of twirling an object attached to a string, say a rubber ball, above their heads. See Figure 1. Figure 1 As the ball is twirled, the string becomes taut. The energy exerted upon it by the child’s hand, coupled with the ball’s mass and traveling speed/velocity, v, make the ball want to move off in a straight trajectory into space, like a launched projectile. But the string it’s attached to prevents it from doing so, forcing the ball to instead travel a circular path around the center point of rotation. The taut string and the ball’s circular path are evidence of centripetal force, F_{c}, at work. Next time we’ll employ Isaac Newton’s Second Law of Motion to the centripetal force phenomenon to see how the sun’s gravitational force keeps Earth in a stable circular orbit around the sun.
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Gravity and the Mass of the Sun
Friday, December 12th, 2014
As a young school boy I found it hard to believe that scientists were able to compute the mass of our sun. After all, a galacticsized measuring device does not exist. But where there’s a will, there’s a way, and by the 18th Century scientists had it all figured out, thanks to the work of others before them. Newton’s two formulas concerning gravity were key to later scientific discoveries, and we’ll be working with them again today to derive a third formula, bringing us a step closer to determining our sun’s mass. Newton’s Second Law of Motion allows us to compute the force of gravity, F_{g,} acting upon the Earth, which has a mass of m. It is, F_{g} = m × g (1) Newton’s Universal Law of Gravitation allows us to solve for g, the sun’s acceleration of gravity value, g = (G × M) ÷ r^{2} (2) where, M is the mass of the sun, r is the distance between the sun and Earth, and G is the universal gravitational constant. You will note that g is a common factor between the two equations, and we’ll use that fact to combine them. We’ll do so by substituting the right side of equation (2) for the g in equation (1) to get, F_{g} = m × [(G × M) ÷ r^{2}] then, using algebra to rearrange terms, we’ll set up the combined equation to solve for M, the sun’s mass: M = (F_{g }× r^{2}) ÷ (m × G)^{ } (3) At this point in the process we know some values for factors in equation (3), but not others. Thanks to Henry Cavendish’s work we know the value of m, the Earth’s mass, and G, the universal gravitational constant. What we don’t yet know is Earth’s distance to the sun, r, and the gravitational attractive force, F_{g}, that exists between them. Next time we’ll introduce some key scientists whose work contributed to a method for computing the distance of our planet Earth to its sun. _______________________________________

The Force of Gravity
Thursday, November 20th, 2014
Last time we saw how Henry Cavendish built upon the work of scientists before him to calculate Earth’s mass and its acceleration of gravity factor, as well as the universal gravitational constant. These values, together with the force of gravity value, F_{g}, which we’ll introduce today, moved scientists one step further towards being able to discover the mass and gravity of any heavenly body in the universe. According to Newton’s Second Law of Motion, the force of gravity, F_{g}, acting upon any object is equal to the object’s mass, m, times the acceleration of gravity factor, g, or, F_{g} = m × g So what is F_{g}? It’s a force at play way up there, in the outer reaches of the galaxy, as well as back home. It keeps the moon in orbit around the Earth and the Earth orbiting around the sun. In the same way, F_{g }keeps us anchored to Earth, and if we were to calculate it, it would be calculated as the force of our body’s mass under the influence of Earth’s gravity. It’s common to refer to this force as weight, but it’s not quite so simple. Using the metric system, the unit of measurement most often used for scientific analyses, weight is determined by multiplying our body’s mass in kilograms by the Earth’s acceleration of gravity factor of 9.8 meters per second per second, or 9.8 meters per second squared. For example, suppose your mass is 100 kilograms. Your weight on Earth would be: Weight = F_{g} = m × g = (100 kg) × (9.8 m/sec^{2}) = 980 kg · m/sec^{2} = 980 Newtons Newtons? That’s right. It’s easier than saying kilogram · meter per second per second. It’s also a way to pay homage to the man himself. In the English system of measurement things are perhaps even more confusing. Your weight is found by multiplying the mass of your body measured in slugs by the Earth’s acceleration of gravity factor of 32 feet per second per second. Slugs is British English speak for pounds · second squared per foot. We normally refer to weight in units of pounds, and in engineering circles it’s pounds force. For example, suppose your mass is 6 slugs, or 6 pounds · second squared per foot. Your weight on Earth would be: Weight = F_{g} = m × g = (6 Lbs · sec^{2}/ft) × (32.2 ft/sec^{2})= 193.2 Lbs To avoid any confusion, you could just step on the bathroom scale. Next time we’ll see how the force of gravity is influenced by an inverse proportionality phenomenon. _______________________________________
