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.
____________________________________

Posts Tagged ‘astronomical unit’
Earth’s Orbital Velocity
Sunday, July 19th, 2015Earth’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.
____________________________________

The Astronomical Unit — It’s So Relative
Thursday, May 7th, 2015
Last time we learned that early scientists used the Earth itself as an optical rangefinder to gauge its distance to the moon, and we posed the question: Can Earth be used in the same way to gauge distance to the sun? Unfortunately not. As we learned earlier in this blog series, the more distant the object, the larger the rangefinder that’s required, and the Earth just plain isn’t big enough to be used to compute that distance. The sun is 390 times farther away than the moon is, and that presents a real problem. Today we’ll learn about an alternate method to gauge this great distance. Johannes Kepler made a great contribution towards finding Earth’s distance to the sun when he developed his Third Law of Planetary Motion. Through his observation of planetary movements he was able to determine each planet’s relative distance from the sun in terms of what he dubbed the astronomical unit (AU), a yardstick by which the distance between all planets in our solar system and our shared sun could be judged. Kepler established the distance between Earth and the sun to be that astronomical unit, depicted in the illustration as r. The reason Kepler developed the AU was because in his day there was no known way to measure the distance to the sun. The AU, an abstract term with no real numerical value in terms of the distance being measured became a sort of placeholder term until r could actually be measured. The best the AU could do was allow him to determine how far a planet was from the sun relative to Earth’s distance from it. For example, his Third Law states that Venus’ distance to Earth is approximately onequarter the distance between Earth and the sun, or 0.28 AU. That meant that Venus’ distance to the sun was threequarters the distance of Earth to the sun, or 0.72 AU. Together, these two distances equaled one AU, the as yet unquantifiable distance r between Earth and the sun. Why all the fuss over Venus? Because Edmund Halley, a scientist who came after Kepler and shared in his fascination with interplanetary distances, would use Venus’ proximity to Earth to set up an optical rangefinder relationship between them and the object of his fixation, the sun. Crucial to this accomplishment would be making use of Venus’ orbital movement and the moment it would come into a direct line between Earth and the sun. We’ll explore that further next time.
____________________________________
