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 ‘the AU’
How Big is the Sun?
Monday, August 10th, 2015The 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 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. ____________________________________

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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Earth’s Distance to the Sun — A Roadmap
Thursday, June 18th, 2015
We left off with Edmund Halley’s proposed method to solve the riddle of Earth’s distance to the sun. Halley posited that when Venus’ orbit brought it directly between the Earth and sun, then principles of astronomy, trigonometry, and geometry could be combined to calculate that distance. Instrumental to Halley’s theory were a number of elements discussed previously in this blog series, including the work of Johannes Kepler. We’ll mesh those elements today and chart the course for future discoveries. To begin things, Halley knew that Kepler’s Third Law of Planetary Motion set out the distance between Earth and the sun in theoretical terms as, 1AU = r_{Venus} ÷ 0.28 which meant that if the distance from Earth to Venus, r_{Venus}, could be calculated, then the distance from Earth to the sun was easily deduced, a matter of simple division. Crucial to the calculation of r_{Venus} is to find the value for the angle α which forms between observers’ lines of sight while charting Venus’ travel across the face of the sun, something which only happens during a rare astronomical event known as the Transit of Venus. See Figure 1. Figure 1 Figure 1 shows two observers positioned on opposite sides of the Earth, busily surveying Venus’ movement across the sun’s face. Their lines of sight converge at a vertex point, or point of intersection, on Venus, then move beyond it to the sun. Due to the principle of vertical angles, which stipulates that angles which share the same vertex point also share the same angle measurement, we know that the angle α that’s formed between Observer A and B‘s lines of sight is of the same value between Earth and Venus as it is between Venus and the sun. Once a is determined, its numerical value can be plopped into an equation we’ve been working with for some time now in this blog series. It’s similar to the equation previously used to calculate Earth’s distance to the moon, r = d x tan(θ) Follow this link to Optically Measuring Cosmic Distances for a review. And here is that equation with terms modified to reflect our new quest, the distance from Earth to Venus, r_{Venus} = d ÷ tan(α) As for the variable d, the distance between the two observers, we’ve worked with that before, too. Follow this link to Determining Chord Length on Circle Earth for a refresher. Next time we’ll see how Venus’ travel path is key to determining the angle α, shown in green on the illustration, and how this angle is crucial to our discovery of the distance between Earth and the sun.
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The Transit of Venus from Different Perspectives
Wednesday, June 3rd, 2015
We’ve been working on calculating Earth’s distance to the sun and discussing how past scientists, including Johannes Kepler and Edmund Halley, contributed to the discussion. Today we’ll see how Halley used his theory on the transit of Venus together with Kepler’s Third Law of Planetary Motion and combined them with known mathematical principles to solve the riddle of this enormous distance, known as one astronomical unit, or AU. Halley’s solution began with stationing two observers with telescopes aimed at Venus on opposite sides of the Earth. Their different lines of sight would cause the principle of parallax to come into play, resulting in them seeing Venus from different perspectives. Their sight would converge at Venus’ center and an angle, α, would form between them. Halley posited that if this angle could be measured, it would be an important first step in calculating the distance between Earth and the sun. See Figure 1. Due to their differing perspectives, Observer A would see Venus traveling a path lower on the sun’s face, while Observer B would see it following a path higher up. See Figure 2. The net result was that the length, L_{A}, of Venus’ path as seen by Observer A was significantly shorter than length, L_{B}, of Venus’ path as seen by Observer B. Because of this, Observer B would have seen Venus pass in front of the sun before Observer A. These differing observations meant that even if both observers were to set their telescopic crosshairs on Venus at the exact same moment it became visible to each of them, it would serve no purpose, because they lacked a common point of reference at which to aim in order to take measurements. This fact made measuring the angle α with a physical device such as a protractor impossible. So Halley gave up on the idea of physically measuring α. Instead, he proposed calculating it based on the time it took for Observers A and B to watch Venus traverse the sun’s face from one side to the other along each of their observational paths. Next time we’ll see how Halley put his idea to work to calculate α and used it in conjunction with Kepler’s Third Law to calculate the AU.
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