This lesson defines what a gravitational field is and some practical examples encountered in the real world. We also develop the equations governing the gravitational field and explain what the units of measurement mean. 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.

]]>A shadow cosmos, woven silently into our own, may have its own rich inner life On September 23, 1846, Johann Gottfried Galle, Director of the Berlin Observatory, received a letter that would change the course of astronomical history. It came from a Frenchman, Urbain Le Verrier, who had been studying the motion of Uranus and concluded that its path could not be explained by the known gravitational forces acting on it. Le Verrier suggested the existence of a hitherto unobserved object whose gravitational pull was perturbing Uranus's orbit in precisely the way required to account for the anomalous observations. Following Le Verrier's directions, Galle went to his telescope that night and discovered the planet Neptune. Purchase to read more You've read the preview. Already purchased? Sign in to access the full article. From this issue SA Special Editions Every Issue. Every Year. 1845 - Present Neuroscience. Evolution. Health. Chemistry. Physics. Technology. Subscribe Now!

]]>Top Question: Define force of gravitation. Answer: The force of attraction which exists between any two objects in the universe is known as force of gravitation. Question: State Newton's law of gravitation. Answer: According to this law, "Every particle in this universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them". Question: Why is G called a universal constant? Answer: G is known as universal constant, because its value remains the same throughout the universe. Question: List out the physical quantities on which the gravitational force between objects depends. Answer: The gravitational force between two objects depends on a) the mass and b) the distance between them Question: What do you mean by a freely falling object? Answer: An object which moves towards the earth due to force of gravity is described as a freely falling object. Question: What is acceleration due to gravity? Answer: The acceleration produced in a body due to force of gravity is known as acceleration due to gravity. Question: Two bodies of mass 10 kg and 12 kg are falling freely. What is the acceleration produced in the bodies due to force of gravity? Answer: The acceleration due to gravity produced in both the bodies is the same as it is independent of the mass of the body. Acceleration produced in both the bodies 10 kg and 12 kg is 9.8 m/s2. Question: What will happen to the force of gravitation between two objects A and B if the distance between them is reduced to half? Answer: Let d be the distance between the two objects A and B of mass m1 and m2 respectively, The force between A and B when distance between them is reduced to half F1 = 4 F i.e., the force increases. The force of gravitation between any two objects increases by a factor 4 if the distance between the objects is reduced to half. Question: What would you observe if there are two massive bodies A and B of equal masses which experience only force of gravitation? Answer: The objects A and B would be moving around each other.

]]>This Hubble Space Telescope composite image shows a ghostly "ring" of dark matter in the galaxy cluster Cl 0024+17. Credit: NASA, ESA, M.J. Jee and H. Ford (Johns Hopkins University) Roughly 80 percent of the mass of the universe is made up of material that scientists cannot directly observe. Known as dark matter, this bizarre ingredient does not emit light or energy. So why do scientists think it dominates? Studies of other galaxies in the 1950s first indicated that the universe contained more matter than seen by the naked eye. Support for dark matter has grown, and although no solid direct evidence of dark matter has been detected, there have been strong possibilities in recent years. The familiar material of the universe, known as baryonic matter, is composed of protons, neutrons and electrons. Dark matter may be made of baryonic or non-baryonic matter. To hold the elements of the universe together, dark matter must make up approximately 80 percent of its matter. [Image Gallery: Dark Matter Across the Universe] The missing matter could simply be more challenging to detect, made up of regular, baryonic matter. Potential candidates include dim brown dwarfs, white dwarfs and neutrino stars. Supermassive black holes could also be part of the difference. But these hard-to-spot objects would have to play a more dominant role than scientists have observed to make up the missing mass, while other elements suggest that dark matter is more exotic. These illustrations, taken from computer simulations, show a swarm of dark matter clumps around our Milky Way galaxy. Image released July 10, 2012. Credit: J. Tumlinson (STScI) Most scientists think that dark matter is composed of non-baryonic matter. The lead candidate, WIMPS (weakly interacting massive particles), have ten to a hundred times the mass of a proton, but their weak interactions with "normal" matter make them difficult to detect. Neutralinos, massive hypothetical particles heavier and slower than neutrinos, are the foremost candidate, though they have yet to be spotted. The smaller neutral axion and the uncharched photinos are also potential placeholders for dark matter. A third possibility exists — that the laws of gravity that have thus far successfully described the motion of objects within the solar system require revision. Proving the unseen Scientists calculate the mass of large objects in space by studying their motion. Astronomers examining spiral galaxies in the 1950s expected to see material in the center moving faster than on the outer edges. Instead, they found the stars in both locations traveled at the same velocity, indicating the galaxies contained more mass than could be seen. Studies of the gas within elliptical galaxies also indicated a need for more mass than found in visible objects. Clusters of galaxies would fly apart if the only mass they contained were visible to conventional astronomical measurements. Albert Einstein showed that massive objects in the universe bend and distort light, allowing them to be used as lenses. By studying how light is distorted by galaxy clusters, astronomers have been able to create a map of dark matter in the universe. All of these methods provide a strong indication that the most of the matter in the universe is something yet unseen. Dark matter versus dark energy After the Big Bang, the universe began expanding outward. Scientists once thought that it would eventually run out of the energy, slowing down as gravity pulled the objects inside it together. But studies of distant supernovae revealed that the universe today is expanding faster than it was in the past, not slower, indicating that the expansion is accelerating. This would only be possible if the universe contained enough energy to overcome gravity — dark energy.

]]>When a transmitter is connected to an antenna and radiates power, it's often interesting to know what is the electromagnetic field strength at a given distance. The following diagram summarizes the problem: A transmitter of power Pt is connected to an antenna of gain that radiates in the surrounding space. We are interested to know the intensity of the field , and at the distance from the transmitting antenna. This measurement must be done in the far field region, otherwise the formula used here are not valid. This means that the measurement must be taken at an adequate minimum distance from the transmitting antenna, so that we can neglect its shape and suppose that we have a nice and spherical wave. This calculator is not intended for near field analysis. Some theory The power provided by the transmitter is radiated by the antenna. As the waves leave the antenna, they spread on the surface of spheres of increasing radius as they travel away. In a very similar way, when dropping a stone in a pond, circular ripples leave the point of impact and become larger and larger in diameter as they travel away. Wave-fronts can be approximated with spheres only if we are far enough from the antenna that originated them, so that its shape can be neglected and considered a point source. For this reason, we can only consider the far field region. As the spheres become bigger and bigger, their surface increases with the square of the distance d2 . The total amount of power carried by the wave doesn't change (without losses), so the same power spreads on a larger and larger surface, explaining the 1/d2 dependence of the power density of the wave and finally of the received power. Furthermore, this simple model doesn't take into account any effect of the ground: the transmitting antenna needs to radiate in free space and to have line of sight with the measuring point. Antennas that use the ground as part of them, like vertical monopoles, cannot be calculated reliably. Ground can also reflect waves and reflections are not take into account, only the straight path is considered. The effect of the ground is less important at microwave frequencies, since it tends to absorb energy instead of reflecting it back and since the distance to the ground in terms of wavelength is much larger. In other words, in most cases, the model is ok for wavelengths shorter than a few meters, say frequencies from the VHF band and up. This being said, with our simple spherical waves model, calculating the power density at a given distance is quite straightforward and depends mainly on the geometrical considerations above, as stated by the following equation:

]]>Learn interesting trivia and information about a wide range of science topics with our fun science facts for kids. Gravity Facts Enjoy our range of interesting gravity facts that help explain how gravity relates to both life on Earth and other objects in our solar system. Learn about important concepts such as acceleration, mass, tides and orbits as well as some useful formulas, strange trivia and other fun information relating to the topic of gravity. Sponsored Links Objects with mass are attracted to each other, this is known as gravity. Gravity keeps Earth and the other planets in our solar system in orbit around the Sun. It also keeps the Moon in orbit around Earth. Tides are caused by the rotation of the Earth and the gravitational effects of the Moon and Sun. Acceleration of objects to due to the gravity on Earth is around 9.8 m/s2. If you ignore air resistance (drag) then the speed of an object falling to Earth increases by around 9.8 metres per second every second. The force of gravity 100 kilometres (62 miles) above Earth is just 3% less than at the Earth’s surface. The human body can handle increased g-forces as seen in activities such as dragster races, airplane acrobatics and space training. The highest known acceleration voluntarily experienced by a human is 46.2 g by g-force pioneer John Stapp. While formula one racing drivers may feel around 5 g’s under heavy braking, they can experience over 100 g’s if a crash causes them to decelerate extremely quickly over a very short distance. Some roller coasters have been known to include g-forces of around 4 to 6 g. The higher something is, the greater its gravitational potential energy. Back in the Middle Ages, weapons called trebuchets were used to take advantage of this principle, using mechanical advantage and the gravitational potential energy of a counterweight to hurl rocks and other projectiles at or over walls. In modern times we use the gravitational potential of water to create hydroelectricity.

]]>[Total: 3 Average: 5/5] You must sign in to vote This question has popped up many times. So here is an attempt to address it. To answer this question at the elementary level, a number of assumption will be made, which will become obvious later on. Still, at this point, we will simply deal with objects with spherical symmetry and no complicated mass distribution. In other words, the mass of each object can be considered to be at their respective center of mass. For two objects with masses, the gravitational force on mass 1 due to the gravitational force of mass 2 can be written as F_{12} = \frac{Gm_1 m_2}{h^2} where m_1, m_2 are the masses of object 1 and 2 respectively, G is the universal gravitational constant, and h is the distance between the two masses. Of course, this equation is symmetric, and the force on 2 due to 1 also has the identical form. Let m_1 be any mass and m_2 = M, the mass of the earth. Then the force on object 1 due to the earth is F_{1M} = \frac{Gm_1 M}{h^2} But since this is the force on m_1, we can equate the dynamics of this object with F_{1M} = \frac{Gm_1 M}{h^2} = m_1 a Simplifying, we get \frac{GM}{h^2} = a which we define to be “g” at the surface of the earth, and is a constant if we always put different masses at the same location. This means that no matter what m_1 is, the acceleration is always a constant g. It shows that the acceleration due to gravity is a constant. Now, in deriving this, I’ve made an implicit assumption of one thing, that “M” really doesn’t get affected much by m_1. Recall that I said earlier that the gravitational force equation is symmetric. One can also talk about the force on the earth due to the mass m_1. So that mass certainly exert a force on M. However, this is where we have to consider a little bit about the center of mass of the M + m_1 system. r_{cm} = \frac{m_1 r_1 + MR}{m_1 + M} where r_1 is the location of the center of mass of m_1, and R is the location of the center of mass of M, i.e. the earth. If we put our origin at the center of mass of the earth, then R = 0. So let’s do that. The center of mass location is then simplified to r_{cm} = \frac{m_1 r_1}{m_1 + M} Now, for ordinary, terrestrial objects, the kind that Galileo (and you and I) would test in our intro physics classes, the mass of the object m_1 is much, much smaller than the mass of the earth, i.e. m_1 < M. In that case, we can simplify the center of mass equation to r_{cm} = \frac{m_1 r_1}{M} It means that the location of the center of mass of ordinary object + Earth system, for an object placed near or on the surface of the earth (i.e. r_1 ~ R_E), depends predominately on the ratio of m_1/M. An ordinary object weights in the range of kg. The earth weighs on the order of 10^{24} kg, and the radius of the earth is of the order of 10^6m. If you do that computation, you’ll get a very, very, small number for r_{cm}. It means that the center of mass of this system is practically the same as the center of mass of the Earth, which is situated at r=0. What I’m trying to show here is in this scenario, the earth does not make any significant motion due to the force exerted by m_1. This means that for this case, it is perfectly fine to consider the earth as the fixed object, and only consider that it is the smaller mass that falls towards the earth. So when you consider that, then all the simplification that is done to allow us to deduce that the acceleration due to gravity of ALL objects falling on the surface of the earth to be a constant, independent of the mass of the object.

]]>It is one of the most famous anecdotes in the history of science. The young Isaac Newton is sitting in his garden when an apple falls on his head and, in a stroke of brilliant insight, he suddenly comes up with his theory of gravity. The story is almost certainly embellished, both by Newton and the generations of storytellers who came after him. But from today anyone with access to the internet can see for themselves the first-hand account of how a falling apple inspired the understanding of gravitational force. The Royal Society in London is making available in digital form the key original manuscript that describes how Newton devised his theory of gravity after witnessing an apple falling from a tree in his mother's garden in Lincolnshire, although there is no evidence to suggest that it hit him on the head. It was 1666, and the plague had closed many public buildings and meetings. Newton had to abandon Cambridge for Woolsthorpe Manor, near Grantham in Lincolnshire, the modest house where he was born, to contemplate the stellar problems he had been pursuing at the university. He was particularly obsessed by the orbit of the Moon around the Earth, and eventually reasoned that the influence of gravity must extend over vast distances. After seeing how apples always fall straight to the ground, he spent several years working on the mathematics showing that the force of gravity decreased as the inverse square of the distance. But what evidence is there that Newton was really inspired by a falling apple? He left no written account suggesting this, although there were other documents suggesting that he had spoken to others about it when he was an old man. Historians point to the one particular account written by one of Newton's younger contemporaries, an antiquarian and proto-archaeologist called William Stukeley, who also wrote the first biography of Britain's greatest scientist, entitled Memoirs of Sir Isaac Newton's Life. Stukeley was also born in Lincolnshire, and used this connection to befriend the notoriously cantankerous Newton. Stukeley spent some time in conversation with the older man, and the pair met regularly as fellows of the Royal Society, and talked together. On one particular occasion in 1726, Stukeley and Newton spent the evening dining in London. "After dinner, the weather being warm, we went into the garden & drank thea under the shade of some apple tree; only he & myself, " Stukeley wrote in the meticulously handwritten manuscript released by the Royal Society. "Amid other discourse, he told me, he was just in the same situation, as when formerly the notion of gravitation came into his mind. Why sh[oul]d that apple always descend perpendicularly to the ground, thought he to himself; occasion'd by the fall of an apple, as he sat in contemplative mood. "Why sh[oul]d it not go sideways, or upwards? But constantly to the Earth's centre? Assuredly the reason is, that the Earth draws it. There must be a drawing power in matter. And the sum of the drawing power in the matter of the Earth must be in the Earth's centre, not in any side of the Earth. "Therefore does this apple fall perpendicularly or towards the centre? If matter thus draws matter; it must be proportion of its quantity. Therefore the apple draws the Earth, as well as the Earth draws the apple." This is the most detailed account of the apple anecdote, but it is not the only one from Newton's day. He had also used it to entertain John Conduitt, the husband of Newton's niece and his assistant at the Royal Mint, which Newton had run in his later years. Conduitt wrote: "In the year 1666 he retired again from Cambridge to his mother in Lincolnshire. Whilst he was pensively meandering in a garden it came into his thought that the power of gravity (which brought an apple from a tree to the ground) was not limited to a certain distance from Earth, but that this power must extend much further than was usually thought. "Why not as high as the Moon said he to himself & if so, that must influence her motion & perhaps retain her orbit, whereupon he fell a calculating what would be the effect of that supposition."

]]>The Higgs Field is an energy field that is hypothesized to exist everywhere in the universe. The field is accompanied by a fundamental particle called the Higgs Boson, which the field uses to continuously interact with other particles. As particles pass through the field they are "given" mass, much as an object passing through treacle (or molasses) will become slower. Mass itself is not generated by the Higgs field- the creation of matter or energy would conflict with the laws of conservation. However, mass is "imparted" to particles from the Higgs field, which contains the relative mass in the form of energy. Once the field has endowed a formerly massless particle the particle slows down because it has become heavier. If the Higgs field did not exist, particles would not have the mass required to attract one another, and would float around freely at light speed. The process of giving a particle mass is known as the Higgs Effect . The Higgs Effect was first theorized in 1964 by writers of the PRL symmetry breaking papers. In 2013 the Higgs Boson, and implicitly the Higgs effect, were tentatively proven at the Large Hadron Collider. The effect was seen as finding a missing piece of the Standard Model. According to gauge theory- a branch of the Standard Model dealing with force-carrying particles, all force-carrying particles should be massless. However-the force-particles that mediate the weak force have mass. This is due to the Higgs Effect. Scientifically, the Higgs Effect breaks SU(2) symmetry; (SU stands for special unitary, a type of matrix, and 2 refers to the size of the matrices involved). A symmetry of a system is an operation done to a system, such as rotation or displacement, that leaves the system fundamentally unchanged. A symmetry also provides a rule for how something should always act unless acted on by an outside force. An example is a Rubik's Cube. If we take a Rubik's cube and scrambled it by making whatever moves we want, it is still possible to solve it. Since each move we make still leaves the Rubik's cube solvable, we can say that these moves are 'symmetries' of the Rubik's cube. Together, they form what we call the symmetry group of the Rubik's cube. Making any of these moves doesn't change the puzzle, always leaving it solvable. But, we can break this symmetry by doing something like taking the cube apart, and putting it back together in a completely wrong way. No matter what moves we try now, it is not possible to solve the cube. Breaking the cube apart and putting it back together in the wrong way is the 'outside force': Without this outside force, nothing we do to the cube makes it unsolvable. The symmetry of the Rubik's cube is that it stays solvable whatever moves we make, as long as we don't take apart the cube.

]]>This section is optional and covers the physics and mathematical details of the two-body problem and three-body problem. The two-body problem Consider two point masses, m₁ and m₂, moving within a plane (that is, the 2-dimensional case). Let the position of m₁ be denoted by (x₁, y₁) and m₂ by (x₂, y₂). Because the masses are moving with respect to time, their positional coordinates (x, y) are actually functions of time t, that is (x(t), y(t)). Figure 1 shows such a situation for a particular value of t (a moment frozen in time): Figure 1 In figure 1, we see that m₁ and m₂ are a distance r apart and are mutually attracted to one another through the force of gravity F₁ and F₂ (whose magnitudes are the same). This force of gravity (F) is given by Newton's law of universal gravitation: The force vector F₁ acting on the first point mass m₁ has the following x- and y-components (Fₓ and Fy, respectively), as can be seen here: Figure 2 Using trigonometry, we see that: And, using the larger triangle, shows that: If we focus on Fₓ and substitute for F and cos θ, we have: Where F(t) and (t) are both vector functions of time. If, for now, we consider just the x-direction of m₁, we see that: Where, in the last expression, we drop the implied function of time variable t. We now have two expressions for Fₓ, the first from Newton’s second law and the second from Newton’s law of universal gravitation: By using the Pythagorean theorem (see figure 2), the distance r between m₁ and m₂ is given by: Therefore (for m₁ in the x-direction), we have: Now given the initial positions (x₁, y₁) and (x₂, y₂) at t = 0, we can calculate the acceleration of m₁ due to m₂ in the x-direction using equation 14. We can use the same reasoning as above to show that the y-component of m₁’s acceleration (due to m₂) is given by: From m₂’s perspective, we see that the acceleration equations are identical except for a change in sign (indicating opposite force directions) as provided by the swap in positional terms in the numerator: Thus, we have the following set of equations, where the numeric subscripts indicate the mass m₁ or m₂: Where: (when α = 0, an infinite acceleration is imparted when the two masses exactly collide, which is physically impossible). Now we have acceleration as a function of each mass’s current position but how do we approximate its velocity and position at some small amount of time h (that is, Δt) later? This is where numerical integration comes in. In particular, we’ll use leapfrog integration (the position Verlet algorithm) in that it’s relatively simple but reasonably accurate and stable. Consider N celestial masses. Let i indicate one of the masses (i = 1, …, N) and h a small time interval. For the position Verlet algorithm, the ith mass’s next position and velocity values are calculated as follows: Where, for a given directional component, xⁱ is the position, vⁱ the velocity, and aⁱ the acceleration of the ith mass. In order to improve accuracy, on first use, equation 22 is replaced with the following: This information leads to the following algorithm: Choose a small value for h (Δt). Choose initial (component-wise) position and velocity values for all masses. Using equations 18 through 21, calculate initial acceleration values. Using the values from the prior step, use equation 25 to calculate the initial values. Using the previously calculated values, use equations 18 through 21 to calculate the values. Using the values from the prior step, calculate the values using equation 23 Using the values from the prior step, calculate the values using equation 24. Using the values from the prior step, use equation 22 to calculate the values. Go to step 5 until the user chooses to stop. The three-body problem When there are three masses, any one mass has two forces acting on it from the other two masses. For example, m₁ has the following forces (F₂ and F₃) acting on it: Figure 3 To start, we note that the net force F₁ acting on m₁ is the sum of F₂ and F₃. That is, F₁ = m₁₁ = F₂ + F₃ Now applying the same trigonometry used in figure 2 to figure 3, the magnitude of the net force F₁ on m₁ can be broken into its x- and y-components as follows: Using the red and green triangles in figure 3, we see that: From Newton’s law of universal gravitation, F₂ and F₃ can be expressed as: Substituting equation 28 through 33 into equations 26 and 27 results in: Simplifying 34 and 35 yields: Replacing r₂ and r₃ with their Pythagorean formulae yields the following x- and y-component acceleration equation for m₁:

]]>