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... from both sides to leave you with</p>]]></content:encoded>
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<?xml version="1.0" encoding="UTF-8"?><rss version="2.0" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:sy="http://purl.org/rss/1.0/modules/syndication/" > <channel> <title>RSS Univers</title> <link>https://www.universator.com/</link> <description>Univers</description> <lastBuildDate>Tue, 08 Jul 2025 13:29:58 +0200</lastBuildDate> <language>en</language> <sy:updatePeriod>daily</sy:updatePeriod> <sy:updateFrequency>1</sy:updateFrequency> <item> <title>Gravitational force Problems</title> <description>Using physics, you can calculate the gravitational force that is exerted on one object by another object. For example, given the weight of, and distance between, two objects, you can calculate how large the force of gravity is ...</description> <content:encoded><![CDATA[<img src="/img/important_characteristics_of_gravitational_force_ga.jpg" alt="Important Characteristics of" align="left" /><p>Using physics, you can calculate the gravitational force that is exerted on one object by another object. For example, given the weight of, and distance between, two objects, you can calculate how large the force of gravity is between them. Here are some practice questions that you can try. Practice questions The gravitational force between objects A and B is 4 newtons. If the mass of B were one-half as large as it currently is while A's mass remains the same, how large is the gravitational force? Calculate the force of gravity between two 3-kilogram ball bearings separated by a distance of 10 centimeters. Round your answer to two significant digits. A 9, 000-kilogram starship is pulled toward Planet X, a behemoth with a radius of 65, 000 kilometers. When the starship is 2, 500 kilometers from the planet's surface, what is the starship's acceleration (providing that its engines are turned off)? Round your answer to two significant digits. Answers The following are the answers to the practice questions: 2 N The force of gravity exerted between objects is proportional to each object's mass. If B's mass is halved — with A's mass remaining unchanged — then the gravitational force between A and B is also halved: Before you can substitute all the given values into the law of universal gravitation, you need to convert the distance between the ball bearings into meters to match the units in the gravitational constant, G: With its engines off, the only force that the starship feels is the gravitational force attracting it to Planet X. Therefore, the net force on the starship must be equal to the force of gravity between the ship and the planet, represents the distance between the centers of the two objects: The distance from the center of Planet X to its surface is 65, 000 kilometers, and the distance from the surface to the starship is another 2, 500 kilometers, making the total distance between the planet and the starship 67, 500 kilometers — or, more importantly, given the units situation, 67, 500, 000 meters Substituting all the data into the equations leaves you with: (You can save yourself a little handwriting by noticing that because appears on both sides of the equation in the second line, you can divide it from both sides to leave you with</p>]]></content:encoded> <category><![CDATA[Gravitational Force]]></category> <link>https://www.universator.com/GravitationalForce/gravitational-force-problems</link> <guid isPermaLink="true">https://www.universator.com/GravitationalForce/gravitational-force-problems</guid> <pubDate>Tue, 08 Jul 2025 09:29:00 +0000</pubDate> </item> <item> <title>Theoretical energy</title> <description>Group Overview The last decades have seen exciting progress in high energy physics with the establishment of the Standard Model of the quark and lepton constituents of matter and their interactions. A unified gauge theory of weak ...</description> <content:encoded><![CDATA[<img src="/img/100_renewable_energy_isnt_theoretical_its.jpg" alt="100% renewable energy is a" align="left" /><p>Group Overview The last decades have seen exciting progress in high energy physics with the establishment of the Standard Model of the quark and lepton constituents of matter and their interactions. A unified gauge theory of weak and electromagnetic interactions received experimental confirmation through the discovery of the W and Z bosons and a host of other experiments. The strong interactions of quarks and gluons - the constituents of the proton, neutron and other hadrons - are described by a similar gauge theory, Quantum Chromodynamics. Our group at Carnegie Mellon works to extract predictions from the Standard Model and on the physics that lies beyond it, such as theories with more complicated Higgs sectors and grand unified theories. We're particularly interested in the nature and origin of the symmetries that characterize the interactions and applications that are of astrophysical or cosmological significance. Member Research Thrusts Fred Gilman's research is in theoretical particle physics, particularly in understanding the nature of CP violation. He looks to the LHC to provide us with additional sources of CP violation that would explain the dominance of matter over antimatter in the universe and to produce in the laboratory the particles that make up the dark matter. Gilman is leading efforts to create and expand the McWilliams Center for Cosmology. He is a member of the Board of Directors of the Large Synoptic Survey Telescope Corporation. Richard Holman's interests center mainly on the interface between cosmology and particle physics with strong interests in the quantum mechanics/field theory involved in inflationary cosmologies. He is developing a formalism to describe quantum fields in non-equilibrium environments, such as occur during and immediately after an inflationary phase. He is also involved in trying to understand how current and future measurements of the Cosmic Microwave Background Radiation could detect effects coming from Planck-Scale physics, as well as constructing models that describe the so-called Dark Energy component of the Universe. Mike Levine is co-director of the Pittsburgh Supercomputer Center. He has developed computational hardware and numerical and algebraic algorithms to perform high order perturbative calculations in quantum electrodynamics. Ira Rothstein is concerned with using the data from the LHC to explain the origin of mass and the nature of the dark matter. He has worked on various topics in this field ranging from theories of extra dimensions to calculating Higgs boson production rates. He has also using quantum field theory to calculate classical gravity wave profiles for inspiralling black holes. He also works on effective field theory techniques to find systematic ways of calculating strong interaction observables at high energies.</p>]]></content:encoded> <category><![CDATA[Dark Energy]]></category> <link>https://www.universator.com/DarkEnergy/theoretical-energy</link> <guid isPermaLink="true">https://www.universator.com/DarkEnergy/theoretical-energy</guid> <pubDate>Sun, 29 Jun 2025 09:20:00 +0000</pubDate> </item> <item> <title>Laws of Gravitation by Isaac Newton</title> <description>Born in England, Isaac Newton was a highly influential physicist, astronomer, mathematician, philosopher, alchemist and theologian. In 1687, Newton published Philosophae Naturalis Principia Mathematica, what is widely regarded to ...</description> <content:encoded><![CDATA[<img src="/img/sir_isaac_newtons_handwritten_notes_about.jpg" alt="Sir Isaac Newton's handwritten" align="left" /><p>Born in England, Isaac Newton was a highly influential physicist, astronomer, mathematician, philosopher, alchemist and theologian. In 1687, Newton published Philosophae Naturalis Principia Mathematica, what is widely regarded to be one of the important books in the history of science. In it he describes universal gravitation and the three laws of motion, concepts that remained at the forefront of science for centuries after. Newton’s law of universal gravitation describes the gravitational attraction between bodies with mass, the earth and moon for example. Newton’s three laws of motion relate the forces acting on a body to its motion. The first is the law of inertia, it states that ‘every object in motion will stay in motion until acted upon by an outside force’. The second is commonly stated as ‘force equals mass times acceleration’, or F = ma. The third and final law is commonly known as ‘to every action there is an equal and opposite reaction’. Other significant work by Newton includes the principles of conservation related to momentum and angular momentum, the refraction of light, an empirical law of cooling, the building of the first practical telescope and much more. Newton moved to London in 1696 and took up a role as the Warden of the Royal Mint, overseeing the production of the Pound Sterling. Newton was known to have said that his work on formulating a theory of gravitation was inspired by watching an apple fall from a tree. A story well publicized to this very day. Famous Isaac Newton quotes include: "Plato is my friend - Aristotle is my friend - but my greatest friend is truth." "If I have seen further it is only by standing on the shoulders of Giants." "I can calculate the motions of the heavenly bodies, but not the madness of people." "I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the sea-shore, and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me."</p>]]></content:encoded> <category><![CDATA[Newton Universal Law]]></category> <link>https://www.universator.com/NewtonUniversalLaw/laws-of-gravitation-by-isaac-newton</link> <guid isPermaLink="true">https://www.universator.com/NewtonUniversalLaw/laws-of-gravitation-by-isaac-newton</guid> <pubDate>Fri, 20 Jun 2025 09:02:00 +0000</pubDate> </item> <item> <title>Formula of gravitational constant</title> <description>You have no doubt seen gravity work in your life. After all, it is the force that keeps your feet on the ground! This lesson explores the acceleration a body experiences due to gravity and provides details of the mechanics behind ...</description> <content:encoded><![CDATA[<img src="/img/formula_for_gravitational.jpg" alt="Formula for Gravitational" align="left" /><p>You have no doubt seen gravity work in your life. After all, it is the force that keeps your feet on the ground! This lesson explores the acceleration a body experiences due to gravity and provides details of the mechanics behind it. Click "next lesson" whenever you finish a lesson and quiz. Got It You now have full access to our lessons and courses. Watch the lesson now or keep exploring. Got It You're 25% of the way through this course! Keep going at this rate, and you'll be done before you know it. Way to go! If you watch at least 30 minutes of lessons each day you'll master your goals before you know it. Go to Next Lesson Take Quiz Congratulations on earning a badge for watching 10 videos but you've only scratched the surface. Keep it up! Go to Next Lesson Take Quiz You've just earned a badge for watching 50 different lessons. Keep it up, you're making great progress! Go to Next Lesson Take Quiz You have earned a badge for watching 20 minutes of lessons. You have earned a badge for watching 50 minutes of lessons. You have earned a badge for watching 100 minutes of lessons. You have earned a badge for watching 250 minutes of lessons. You have earned a badge for watching 500 minutes of lessons.</p>]]></content:encoded> <category><![CDATA[Universal Gravitation Constant]]></category> <link>https://www.universator.com/UniversalGravitationConstant/formula-of-gravitational-constant</link> <guid isPermaLink="true">https://www.universator.com/UniversalGravitationConstant/formula-of-gravitational-constant</guid> <pubDate>Wed, 11 Jun 2025 09:00:00 +0000</pubDate> </item> <item> <title>Newtonian gravitational constant</title> <description>Figures Figure 1: Sketch of the experiment. The Rb atom interferometer operates as a gravity gradiometer and the W masses are used as the source of the gravitational field. For the measurement of , the position of the source ...</description> <content:encoded><![CDATA[<img src="/img/presentation_relativity_supplementary_chapters_s2_and.jpg" alt="Newtonian gravitational" align="left" /><p>Figures Figure 1: Sketch of the experiment. The Rb atom interferometer operates as a gravity gradiometer and the W masses are used as the source of the gravitational field. For the measurement of , the position of the source masses is alternated between configurations F and C. Plots of gravitational acceleration ( az ) produced along the symmetry axis by the source masses are also shown for each configuration; a constant value for Earth’s gravity was subtracted. The spatial regions of the upper and lower atom interferometers are indicated by the thick lines. The vertical acceleration plots show the effect of source mass in cancelling the local gravity gradient at the positions of the atomic apogees. Figure 2: Experimental data. , Typical Lissajous figures obtained by plotting the output signal of the upper atom interferometer versus that of the lower one for the two configurations of the source masses. , Modulation of the differential phase shift for the two configurations of source masses for a given direction of the Raman beams’ vector. Each phase measurement is obtained by fitting a 360-point scan of the atom interference fringes to an ellipse. The error bars, not visible on this scale, are given by the standard error of the least-squares fit to the ellipse. c, Results of the measurements to determine . Each point is obtained by averaging the signals recorded for the two directions of the Raman vector (Methods). Data acquisition for each point took about one hour. These data were recorded on different days during one week in July 2013. The error bars are given by the combined errors in the phase angles of four ellipses. , Histogram of the data in c. Figure 3: Comparison with previous results. Result of this experiment for compared with the values obtained in previous experiments and with the recent CODATA adjustments. Only the experiments considered for the current CODATA 2010 value, and the subsequent BIPM-13 result, are included. For details on the experiments and their identification with the abbreviations used in the figure, see ref. 3 and the additional references in Methods.</p>]]></content:encoded> <category><![CDATA[Universal Gravitation Constant]]></category> <link>https://www.universator.com/UniversalGravitationConstant/newtonian-gravitational-constant</link> <guid isPermaLink="true">https://www.universator.com/UniversalGravitationConstant/newtonian-gravitational-constant</guid> <pubDate>Mon, 02 Jun 2025 08:51:00 +0000</pubDate> </item> <item> <title>Higgs boson LHC</title> <description>Today during the 50th session of “Rencontres de Moriond” in La Thuile Italy, the ATLAS and CMS experiments presented for the first time a combination of their results on the mass of the Higgs boson. The combined mass of the ...</description> <content:encoded><![CDATA[<img src="/img/find_a_higgs_boson_in_lhc.jpg" alt="Find a Higgs boson in LHC" align="left" /><p>Today during the 50th session of “Rencontres de Moriond” in La Thuile Italy, the ATLAS and CMS experiments presented for the first time a combination of their results on the mass of the Higgs boson. The combined mass of the Higgs boson is mH = 125.09 ± 0.24 (0.21 stat. ± 0.11 syst.) GeV, which corresponds to a measurement precision of better than 0.2%. The Higgs boson is an essential ingredient of the Standard Model of particle physics, the theory that describes all known elementary particles and their interactions. The Brout-Englert-Higgs mechanism, through which the existence of the Higgs boson was predicted, is believed to give mass to all elementary particles. Today’s result is the most precise measurement of the Higgs boson mass yet and among the most precise measurements performed at the LHC to date. “Collaboration is really part of our organization’s DNA, ” says CERN Director General Rolf Heuer. “I’m delighted to see so many brilliant physicists from ATLAS and CMS joining forces for the very first time to obtain this important measurement at the LHC”. The Higgs boson decays into various different particles. For this measurement, results on the two decay channels that best reveal the mass of the Higgs boson have been combined. These two decay channels are: the Higgs boson decaying to two photons; and the Higgs boson decaying to four leptons – where the leptons are an electron or muon. Candidate Higgs boson event from collisions between protons in the ATLAS detector on the LHC. From the collision at the centre, the particle decays into four muons (red tracks)(Image:ATLAS/CERN) Each experiment has found a few hundred events in the Higgs to photons channel and a few tens of events in the Higgs to leptons channel. The analysis uses the data collected from about 4000 trillion proton-proton collisions at the Large Hadron Collider (LHC) in 2011 and 2012 at centre-of-mass energies of 7 and 8 TeV. “The Higgs Boson was discovered at the LHC in 2012 and the study of its properties has just begun. By sharing efforts between ATLAS and CMS, we are going to understand this fascinating particle in more detail and study its behaviour, ” says CMS spokesperson Tiziano Camporesi. The Standard Model does not predict the mass of the Higgs boson itself, so it must be measured experimentally. But once supplied with a Higgs mass, the Standard Model does make predictions for all the other properties of the Higgs boson, which can then be tested by the experiments. This mass combination is the first step towards a combination of other measurements of the Higgs boson’s properties, which will include its other decays. "While we are just getting ready to restart the LHC, it is admirable to notice the precision already achieved by the two experiments and the compatibility of their results, ” says CERN Director of Research Sergio Bertolucci. “This is very promising for LHC Run 2.” Up to now, increasingly precise measurements from the two experiments have established that all observed properties of the Higgs boson, including its spin, parity and interactions with other particles are consistent with the Standard Model Higgs boson. With the upcoming combination of other Run 1 Higgs results from the two experiments and with higher energy and more collisions to come during LHC Run 2, physicists expect to increase the precision of the Higgs boson mass even more and to explore in more detail the particle’s properties. During Run 2, they will be able to combine their results promptly and thus increase the LHC’s sensitivity to effects that could hint at new physics beyond the Standard Model.</p>]]></content:encoded> <category><![CDATA[Higgs Boson]]></category> <link>https://www.universator.com/HiggsBoson/higgs-boson-lhc</link> <guid isPermaLink="true">https://www.universator.com/HiggsBoson/higgs-boson-lhc</guid> <pubDate>Sat, 24 May 2025 08:33:00 +0000</pubDate> </item> <item> <title>Force of gravity Calculator</title> <description>If you fill in the height, you'll get the time and speed at the end of your fall. If you're kind enough to supply your mass, you'll also get the energy in joules (newton-meters) when you hit the deck. :-) See the to see how ...</description> <content:encoded><![CDATA[<img src="/img/physics_equations_formulas_calculator.jpg" alt="Physics Equations Formulas" align="left" /><p>If you fill in the height, you'll get the time and speed at the end of your fall. If you're kind enough to supply your mass, you'll also get the energy in joules (newton-meters) when you hit the deck. :-) See the to see how impact velocity varies with height. As is probably obvious, the higher you are, the harder you land. The relationship looks like this, in km/h: In other words, falling from 50m high is the equivalent of getting hit by a car going 112 km/h, or 70 miles per hour — what would happen if you ran out into a busy freeway. If that's not a decent argument against free soloing, I'm not sure what is. Free fall / falling speed equations The calculator uses the standard formula from Newtonian physics to figure out how long before the falling object goes splat: The force of gravity, = 9.8 m/s2 Gravity accelerates you at 9.8 meters per second per second . After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on. Time to splat: sqrt ( 2 * height / 9.8 ) It's the square root because you fall faster the longer you fall. The more interesting question is why it's times two: If you accelerate for 1 second, your average speed over that time is increased by only 9.8 / 2 m/s. Velocity at splat time: sqrt( 2 * g * height ) This is why falling from a higher height hurts more. Energy at splat time: 1/2 * mass * velocity2 = mass * g * height Ignoring air friction: Terminal velocity This calculator doesn't take into account air friction. But think about what happens if you stick your hand out of the window while driving down the freeway: The wind pushes pretty hard against you. That's air friction. The faster you're going, the harder it pushes back. In fact, it pushes back with the square of your speed, whereas the acceleration of gravity is constant. This means that at some point, the force of air against you equals the force of gravity, and you stop accelerating. That point is called terminal velocity (see this wikipedia article for more information). It depends a lot on your position — something shaped like a bullet will have a higher terminal velocity than something shaped like a flat pancake parallel to the earth, because the latter has more surface area exposed to air friction. The calculator doesn't take any of this into account. In practice, terminal velocity on earth will prevent you from going more than about 320 km/h, or about 200 miles per hour. If you're lying belly-to-the-earth, you'll only travel about 195 km/h (122 miles per hour). As you can see from the graph above, you'd have to fall from higher than 50 meters above the ground for this to really matter much, and at that point, you'd be in enough trouble to not care much. Skydivers, however, should go read the Wikipedia article. About the calculator This is a javascript-based calculator. For you history buffs, the first version used a 10-iteration implementation of Newton's method to compute the square root needed for some of the equations, because in the days of yore, many browsers didn't support sqrt natively. You can see the original code here: newton_sqrt.js - Square root using Newton's Method in Javascript.</p>]]></content:encoded> <category><![CDATA[Gravitational Force]]></category> <link>https://www.universator.com/GravitationalForce/force-of-gravity-calculator</link> <guid isPermaLink="true">https://www.universator.com/GravitationalForce/force-of-gravity-calculator</guid> <pubDate>Thu, 15 May 2025 08:29:00 +0000</pubDate> </item> <item> <title>Gravitational force Equations</title> <description>Near the surface of the Earth, use = 9.81 m/s2 (meters per second squared; which might be thought of as "meters per second, per second", or 32.2 ft/s2 as "feet per second per second") approximately. For other planets, multiply by ...</description> <content:encoded><![CDATA[<img src="/img/gravitational_forces_equation_images.jpg" alt="Gravitational Forces Equation" align="left" /><p>Near the surface of the Earth, use = 9.81 m/s2 (meters per second squared; which might be thought of as "meters per second, per second", or 32.2 ft/s2 as "feet per second per second") approximately. For other planets, multiply by the appropriate scaling factor. A coherent set of units for , , and is essential. Assuming SI units, is measured in meters per second squared, so must be measured in meters, in seconds and in meters per second. In all cases, the body is assumed to start from rest, and air resistance is neglected. Generally, in Earth's atmosphere, all results below will therefore be quite inaccurate after only 5 seconds of fall (at which time an object's velocity will be a little less than the vacuum value of 49 m/s (9.8 m/s2 × 5 s) due to air resistance). Air resistance induces a drag force on any body that falls through any atmosphere other than a perfect vacuum, and this drag force increases with velocity until it equals the gravitational force, leaving the object to fall at a constant terminal velocity. Atmospheric drag, the coefficient of drag for the object, the (instantaneous) velocity of the object, and the area presented to the airflow determine terminal velocity. Apart from the last formula, these formulas also assume that negligibly varies with height during the fall (that is, they assume constant acceleration). The last equation is more accurate where significant changes in fractional distance from the center of the planet during the fall cause significant changes in g. This equation occurs in many applications of basic physics. The equations [edit] Measured fall time of a small steel sphere falling from various heights. The data is in good agreement with the predicted fall time of , where h is the height and g is the acceleration of gravity. Examples [edit] The first equation shows that, after one second, an object will have fallen a distance of 1/2 × 9.8 × 12 = 4.9 meters. After two seconds it will have fallen 1/2 × 9.8 × 22 = 19.6 meters; and so on. The second to last equation becomes grossly inaccurate at great distances. If an object fell 10, 000 meters to Earth, then the results of both equations differ by only 0.08%; however, if it fell from geosynchronous orbit, which is 42, 164 km, then the difference changes to almost 64%. Based on wind resistance, for example, the terminal velocity of a skydiver in a belly-to-earth (i.e., face down) free-fall position is about 195 km/h (122 mph or 54 m/s). This velocity is the asymptotic limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the terminal velocity is approached. In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on. Higher speeds can be attained if the skydiver pulls in his or her limbs (see also freeflying). In this case, the terminal velocity increases to about 320 km/h (200 mph or 90 m/s), which is almost the terminal velocity of the peregrine falcon diving down on its prey. The same terminal velocity is reached for a typical .30-06 bullet dropping downwards—when it is returning to earth having been fired upwards, or dropped from a tower—according to a 1920 U.S. Army Ordnance study. Competition speed skydivers fly in the head down position and reach even higher speeds. The current world record is 1, 357.6 km/h (843.6 mph/Mach 1.25) by Felix Baumgartner who skydived from 38, 969.4 m (127, 852.4 ft) above earth on 14 October 2012. The record was set due to the high altitude where the lesser density of the atmosphere decreased drag. For astronomical bodies other than Earth, and for short distances of fall at other than "ground" level, in the above equations may be replaced by G(M+m)/r2 where is the gravitational constant, M is the mass of the astronomical body, m is the mass of the falling body, and r is the radius from the falling object to the center of the body. The time taken for an object to fall from a height to a height , measured from the centers of the two bodies, is given by: where is the sum of the standard gravitational parameters of the two bodies. This equation should be used whenever there is a significant difference in the gravitational acceleration during the fall. Acceleration relative to the rotating Earth [edit] Centripetal force causes the acceleration measured on the rotating surface of the Earth to differ from the acceleration that is measured for a free-falling body: the apparent acceleration in the rotating frame of reference is the total gravity vector minus a small vector toward the north-south axis of the Earth, corresponding to staying stationary in that frame of reference.</p>]]></content:encoded> <category><![CDATA[Gravitational Force]]></category> <link>https://www.universator.com/GravitationalForce/gravitational-force-equations</link> <guid isPermaLink="true">https://www.universator.com/GravitationalForce/gravitational-force-equations</guid> <pubDate>Tue, 06 May 2025 08:27:00 +0000</pubDate> </item> <item> <title>Equation for universal Gravitation</title> <description>Figure 1: Portrait of Isaac Newton (by courtesy of SEDS). Newton's Laws of Motion Law of Inertia: A body continues in its state of constant velocity (which may be zero) unless it is acted upon by an external force. Fundamental ...</description> <content:encoded><![CDATA[<img src="/img/presentation_chapters_13_16_13_universal.jpg" alt="G, in the equation for" align="left" /><p>Figure 1: Portrait of Isaac Newton (by courtesy of SEDS). Newton's Laws of Motion Law of Inertia: A body continues in its state of constant velocity (which may be zero) unless it is acted upon by an external force. Fundamental Law of Dynamics: For an unbalanced force acting on a body, the acceleration a produced is proportional to the force impressed; the constant of proportionality is the inertial mass m of the body. F = m . a Law of Action and Reaction: In a system where no external forces are present, every action force is always opposed by an equal and opposite reaction force. Newton's Law of Universal Gravitation Two bodies attract each other with equal and opposite forces; the magnitude of this force is proportional to the product of the two masses and is also proportional to the inverse square of the distance between the centers of mass of the two bodies. F = G . M . m / r2 where m and m are the masses of the two bodies, r is the distance between the two, and G is the gravitational constant, whose value is : G = 6.67 . 10-11 Newton.metre2/kg2 The force with which the Earth attracts bodies situated near to its surface is called body's weight. The weight of a mass m, located on the Earth's surface is : P = m . g This expression is an immediate consequence of the Universal Gravitation Law and of . Under normal conditions, the value of g is approximately equal to 9.8 metres/second 2. Gravitational Acceleration According to, if a body acts on another with a certain force, the latter one acts on the former with an equivalent force in opposite direction. This happens with the Universal Gravitation Law, where the force that a body with mass m exerts on another body of mass m is the same as that exerted by the body of mass m on the body of mass m (although in opposite direction). However, although these forces are equal, the accelerations are not. By applying we will have: The acceleration value of the body with mass m is: am = F / m = G . M / r2 which does not depend on m, but on m. The acceleration value of the body with mass m is: aM = F / M = G . m / r2 That is: The acceleration of a body subject to the action of gravity does not depend on its own mass, but on that of the body acting on it. The Two-body Problem: Differential Equations of Universal Gravitation Considering that acceleration is the second derivative with respect to time, we have that the acceleration of a body subject to the gravitational action of another body of mass m is: d2r/dt2 = G . M / r2 We use a cartesian co-ordinates permanently situated in the center of the body with mass m. For this reason, the position of the body of mass m coincides with its distance r to the origin of co-ordinates (its radius vector). Changing the previous equation to cartesian co-ordinates (x, y) we obtain: d2x/dt2 = G . M . x / r3 d2y/dt2 = G . M . y / r3 where r = square root (x2 + y2) Depending on the masses of the two bodies, and the initial conditions (the initial positions and velocities) the trajectory of the moving body may be: A free fall in a straight line. A circle (an ellipse with no eccentricity). An ellipse with a low eccentricity. An ellipse with a high eccentricity (a very elongated ellipse). A parabole (the limiting case). You can push one of the following six buttons to see a simulation of these cases. Two additional buttons allow you to stop the simulation and to continue it.</p>]]></content:encoded> <category><![CDATA[Universal Gravitation Constant]]></category> <link>https://www.universator.com/UniversalGravitationConstant/equation-for-universal-gravitation</link> <guid isPermaLink="true">https://www.universator.com/UniversalGravitationConstant/equation-for-universal-gravitation</guid> <pubDate>Sun, 27 Apr 2025 08:24:00 +0000</pubDate> </item> <item> <title>Gravitational field Physics</title> <description>Variation of g with distance from the centre of a uniform spherical mass of radius, R Variation of g on a line joining the centres of two point masses If m1 &gt; m2 then The potential at a point in a gravitational field is equal ...</description> <content:encoded><![CDATA[<img src="/img/physics_focus_first_direct_measurement.jpg" alt="Figure caption expand figure" align="left" /><p>Variation of g with distance from the centre of a uniform spherical mass of radius, R Variation of g on a line joining the centres of two point masses If m1 > m2 then The potential at a point in a gravitational field is equal to the work done bringing a 1kg mass from infinity to that point. The units are Jkg-1. To calculate work done by a force we use the equation w = F.s but in this situation it is a little more complicated because the force is not of constant magnitude. However, we do know how the force varies with distance from the body (Newton’s law of universal gravitation) and it can be shown* that the work done, w, bringing a mass, m, from infinity to point p is given by: where, M is the mass of the body (the earth, in this case). *a very useful phrase if you a) don't know how to do something or b) can't be bothered to do it ! A body at infinity, has zero gravitational potential. A body normally falls to its lowest state of potential (energy) so we must arrange that the equation for potential is such that, as r decreases, the potential decreases. We can do this by including a negative sign in the above equation. Then, as r decreases, V decreases (becomes a greater negative quantity). Therefore, to calculate the potential at a point in the gravitational field of a point mass or a uniformly distributed spherical mass: If a body is thrown upwards fast enough, it never comes back down: it has escaped from the planet. The velocity needed to do this is called the escape velocity of the planet. As the body is moving away from the planet, it is losing kinetic energy and gaining potential energy. To completely escape from the gravitational attraction of the planet, the body must be given enough kinetic energy to take it to a position where its potential energy is zero. The potential energy possessed by a body of mass m, in a gravitational field is given by:</p>]]></content:encoded> <category><![CDATA[Gravitational Field]]></category> <link>https://www.universator.com/GravitationalField/gravitational-field-physics</link> <guid isPermaLink="true">https://www.universator.com/GravitationalField/gravitational-field-physics</guid> <pubDate>Fri, 18 Apr 2025 08:22:00 +0000</pubDate> </item> </channel></rss>If you would like to create a banner that links to this page (i.e. this validation result), do the following:
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