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 = (F_{g} × r^{2}) ÷ (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 • second^{2} 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, F_{g} , are respectively, m = 5.96 × 10^{24} kilograms r = 149,000,000,000 meters F_{g} = 3.52 × 10^{22} Newtons Inserting these values into Newton’s equation to determine the mass, M, of the sun we get: M = [(3.52 × 10^{22}) × (149,000,000,000)^{2}] ÷ [(5.96 × 10^{24}) × (6.67 × 10^{11})] M = 1.96 × 10^{30} kilograms So how big is 1.96 × 10^{30} kilograms? To get a better idea, let’s divide the sun’s mass, M, by the Earth’s mass, m, (1.96 × 10^{30} kilograms) ÷ (5.96 × 10^{24} 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. 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.
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Posts Tagged ‘Universal Gravitational Constant’
How Big is the Sun?
Monday, August 10th, 2015Centripetal 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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What is Earth’s Mass?
Friday, November 7th, 2014
Last time we learned how Henry Cavendish used Christiaan Huygens’ work with pendulums to determine the value of g, the acceleration of gravity factor for Earth, to be 32.3 ft/sec^{2}, or 9.8 m/sec^{2}. From there Cavendish was able to go on and arrive at values for other factors in Isaac Newton’s gravity formula, namely G, the universal gravitational constant, and M, Earth’s mass. Today, we’ll discuss how Cavendish was able to calculate the Earth’s mass. Newton’s formula for gravity, once again, is: M = (g × R^{2}) ÷ G where M stands for the mass of the heavenly body being quantified. For our case today M will represent the mass of Earth, which was originally quantified in slugs, a British unit of measurement. Today the measurement unit of choice in most parts of the world is the kilogram, which is the metric equivalent of a slug. With regard to the other variables in Newton’s gravity formula, namely, R and G, their values had previously been determined. Eratosthenes’ measurement of shadows cast by the sun on Earth’s surface had revealed Earth’s radius, R, to be 6,371 kilometers, or 6,371,000 meters. And Cavendish’s experiments led him to conclude that the universal gravitational constant, G, was 6.67 × 10^{11} cubic meters per kilogramsecond squared. Plugging these values into Newton’s equation, we calculate Earth’s mass to be: M = ((9.8 m/sec^{2}) × (6,371,000 m)^{2}) ÷ (6.67 × 10^{11} m^{3}/kgsec^{2}) M = 5.96 × 10^{24} kilograms Incidentally, 5.96 × 10^{24 } is scientific notation, or mathematical shorthand, for the number 5,960,000,000,000,000,000,000,000. That’s a whole lot of zeros! Calculating the mass of Earth was an impressive accomplishment. Now that its value was known, scientists would be able to calculate the mass and acceleration of gravity for any heavenly body in the universe. We’ll see how that’s done next time.
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How Big is the Earth?
Wednesday, October 8th, 2014
Last time we discussed Isaac Newton’s Law of Gravitation and how he used it to arrive at conclusions concerning gravity. He theorized the existence of a universal gravitational constant, G, a set numerical value for all heavenly bodies in our universe, and he developed a formula to determine the acceleration of gravity, g. Newton felt sure that the gravity at play on the surface of any heavenly body, such as stars and planets, could be determined if one knew the value of G, along with the object’s mass, M, and radius, R, and he developed this equation to do so, g = (G × M) ÷ R^{2} At this point you may be thinking, Finding the mass and radius of a heavenly body is hard enough, but what is this universal gravitational constant?? Good point. Back in Newton’s time, the existence of G was purely speculative. He conceived it to be a numerical value which would act as a fudge factor, enabling his equation for determining g to work. As a matter of fact, Newton had no clue of how to determine G and was convinced that it would be beyond anyone’s ability to do so. The mysterious G factor and its numerical value were not actually determined until more than a hundred years later by Henry Cavendish. In 1796 Cavendish was focused on determining the Earth’s mass, M, by using Newton’s equation. To arrive at a value for G, Cavendish conducted experiments which measured the gravitational attraction between two lead spheres attached by way of a torsion balance. After much testing he eventually concluded that he had computed G to a reasonable degree of accuracy and that its value was equal to 3.439 x 10^{8} cubic feet per slug per second squared. In this case a slug is not a slimy creature living in the garden, but rather a unit of measurement used to quantify the mass of an object. For the full story, see this article on Cavendish’s experiment by The Physics Classroom. But even after determining G, Cavendish still had to obtain values for g and R in order to calculate M. This was made possible thanks to the work of two men who came before him. One of these was the Greek mathematician Eratosthenes, who way back in 230 B.C. discovered that the radius of the earth, R, could be calculated by simply measuring the shadows of objects cast on Earth’s surface. All he needed was a measuring stick and geometry. For the full story see this fascinating article on the subject from Bucknell University. As for the value of g, the acceleration of gravity on Earth, Cavendish was aided by the previous efforts of a Dutch mathematician from Newton’s time, Christiaan Huygens. You may recall that Huygens was first introduced in a previous blog series on spur gear geometry, where we learned that he studied the motion of clock pendulums. Through observation, Huygens was able to arrive at a mathematical formula capable of predicting the pendulums’ often erratic motion on ships at sea. Next time we’ll see how Huygens’ insights gained by watching pendulums ultimately made it possible for him to arrive at a numerical value for Earth’s acceleration of gravity, g. _______________________________________

Newton’s Law of Gravitation and the Universal Gravitational Constant
Monday, September 29th, 2014
Last time we introduced the term acceleration of gravity, a physical phenomenon posited by Sir Isaac Newton in his book Philosophia Naturalis Principia Mathematica. Newton’s Law of Gravitation is also presented in this book. It provides the basis for his mathematical formula to calculate the acceleration of gravity, g, for any heavenly body in the universe. Newton’s formula to compute the acceleration of gravity is, g = (G × M) ÷ R^{2} where, g is the acceleration of gravity, M the mass of the heavenly body, R the radius, and G the universal gravitational constant. As for the values of the variables in his equation, Newton theorized that G would be a constant, holding the same numerical value throughout the universe. This universal gravitational constant would be the glue that bound together M, the mass of the object being measured, and R, its radius, and render Newton’s formula a workable equation. Without these three values, scientists would be unable to determine the acceleration of gravity rate, g, for the heavenly body under study, and Newton’s equation would be useless, relegated to the depths of pure mathematical theory. In fact, the value for G wasn’t determined until 1796. At that time Henry Cavendish derived its value as an adjunct to calculating the mass of Earth. In the end he was able to arrive at values for Earth’s mass, M, as well as its radius, R. He also came up with the much needed value for G, the universal gravitational constant. He was able to accomplish so much by building upon the work of other scientists before him. We’ll see who those earlier scientists were and how they contributed to the world’s discoveries concerning gravity next time.
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