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 astronomical unit’
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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The Transit of Venus
Monday, May 18th, 2015
Last time we learned of Johannes Kepler’s Third Law of Planetary Motion and his development of the astronomical unit (AU) and how these contributed to bringing ancient scientists a step closer to calculating Earth’s distance to the sun. Today we’ll see why Kepler’s focus on Venus, specifically its travel through space in relation to Earth and the sun — the socalled transit of Venus — would become the crucial element to solving the puzzle. Astronomers had previously used the Earth itself as an optical rangefinder to calculate distance to the moon. But unlike the moon which is relatively close to Earth, the sun is many tens of millions of miles away, too distant to be used in that manner. When it came to finding the distance from Earth to its sun, they were stumped. Then in 1716 Edmund Halley had the insight to combine Kepler’s Third Law and the parallax principle with Venus’ orbital journeys to devise an ingenious solution to the problem. The transit of Venus, first predicted by Kepler in 1627, is a rare astronomical phenomenon which only occurs every 243 years. At this time Venus becomes clearly visible from Earth and appears as a black dot traveling a straight path across the fiery backdrop of the sun’s surface. Figure 1 shows this phenomenon as it would look from Earth. According to Halley’s plan, two observers with telescopes would be positioned on opposite sides of the Earth. Due to the principle of parallax their lines of sight would provide different perspectives of Venus’ path. See Figure 2. From the perspective of Observer B, Venus’s path would appear higher on the sun’s face than the path seen from the perspective of Observer A. As their lines of sight converge on Venus’ center, an angle forms between them, which we’ll name α. The same angle forms as they look past Venus to the sun in its backdrop. Halley theorized that if the angle α could be measured, Kepler’s Third Law could be used together with trigonometry to calculate the distance between Earth and the sun, Kepler’s socalled AU. We’ll review Halley’s methodology next time.
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