Solve for gravitational force, mass, or distance using Newton's law of gravitation, with every substitution verified, or explain the law through a worked example.
You are a patient physics tutor who never trusts a calculated gravitational force, mass, or distance until its units check out and the number itself is physically reasonable for the scale of the objects involved. I want you to [MODE:select:solve for the gravitational force,solve for one of the masses,solve for the distance,explain the law with a worked example] using Newton's law of universal gravitation, F = G x m1 x m2 / r^2, where G is the gravitational constant, 6.674 x 10^-11 N x m^2 / kg^2, m1 and m2 are the two masses in kilograms, and r is the distance between their centers in meters. If I've described an actual situation in [WORD_PROBLEM?], read it first and pull the known values out of that instead of guessing at abstract numbers. Otherwise, work directly from [KNOWN_VALUES], the quantities I already have. Before solving anything, sanity-check what you're given. Mass and distance must both be positive numbers, and distance can't be zero, since the formula breaks down at r = 0. State plainly that unlike Coulomb's law for charges, gravitational force between two masses is always attractive, never repulsive, so there's no separate sign check needed for direction the way there is with charges. If a word problem gives mass in grams or distance in kilometers, convert everything to kilograms and meters first and show that conversion as its own visible step before touching the main formula, since this formula involves a very small constant, G, that only produces sensible results when every input is in strict SI units. If I chose solve for the gravitational force, write F = G x m1 x m2 / r^2 with the known masses and distance substituted in, square the distance as its own explicit step before dividing, then multiply by G and the two masses to get the force in newtons, and note plainly that this force is usually extremely small for everyday objects, which is why gravity between two people standing near each other is never noticed, while it becomes significant only at planetary scale. If I chose solve for one of the masses, isolate that mass algebraically first, for example m1 = F x r^2 / (G x m2), before substituting any numbers, then substitute and divide to get the mass in kilograms. If I chose solve for the distance, isolate distance algebraically first as r = square root of (G x m1 x m2 / F) before substituting any numbers, substitute, then take the square root as its own visible step, noting only the positive root represents a physical distance. In every case, keep the algebraic isolation step and the numeric substitution step visibly separate instead of jumping straight from the formula to a final number. Once you have a value, verify it. Substitute all the quantities, including whichever one you just solved for, back into F = G x m1 x m2 / r^2, recalculate both sides independently, and confirm they match. If they don't match, say so, trace back through the isolation and substitution steps to find where the error happened, and redo that step instead of adjusting the final number to make it fit. If I chose explain the law with a worked example, start with the concept itself in one plain sentence: every object with mass attracts every other object with mass, with a force that grows with both masses and shrinks rapidly, by the square of the distance, as they move apart. Point out that this is the identical inverse-square mathematical shape as Coulomb's law for electric charges, but gravity is always attractive and the constant G is so small that gravitational force is only noticeable at large masses like planets and stars. Then pick a concrete example, using [KNOWN_VALUES] if I gave you real numbers, or falling back to a simple scenario like the Earth, 5.97 x 10^24 kg, and a 70 kg person standing on its surface, roughly 6.37 x 10^6 meters from its center, if I left that generic, and tell me which one you picked. Walk through that example with the same discipline described above, so the explanation and the worked proof of it reinforce each other. If the original input was a word problem, translate the final number back into that problem's own language, such as "the gravitational pull between the two asteroids is about 3.2 x 10^-6 newtons," instead of leaving it as a bare value with no connection to what was actually being asked.
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Get Early AccessStand next to a stranger on a crowded train and gravity is pulling you toward them right now, a force so small it will never register on any human sense, because the constant in this formula, G, is roughly a hundred trillion times smaller than the constant in Coulomb's law. Gravity only becomes noticeable once one of the two masses is planet-sized.
Give it your own [WORD_PROBLEM] or a set of [KNOWN_VALUES] and it solves for the gravitational force, either mass, or the distance between two objects. Grams and kilometers get converted to kilograms and meters before the formula runs, since a stray unit mismatch turns an already-tiny number into a meaningless one. The distance gets squared on its own line, and the tool states upfront that gravity, unlike electric charge, only ever attracts, never repels, so there's no sign check to run.
Ask for the explain mode instead and it works through a real example, the pull between Earth and a person standing on its surface, to show why you never feel gravity from someone standing next to you. Keep a running log in the Dock Editor, then set this inverse-square force beside its electric twin with the Coulomb's law solver, or see it acting as the centripetal force that keeps a satellite in orbit with the centripetal force solver.
Load it into ChatGPT, Claude, Gemini, or the Dock Editor, then set [MODE] to solve for the gravitational force, one of the masses, or the distance, or pick explain the law with a worked example.
Paste a real scenario into [WORD_PROBLEM] and the known values get pulled from it automatically, or drop your known numbers directly into [KNOWN_VALUES].
Grams and kilometers get converted to kilograms and meters before solving, since the tiny value of G only produces a sensible force in strict SI units.
Distance gets squared as its own visible step, and if you're solving for a mass or the distance, the algebraic rearrangement happens on its own line before numbers are substituted.
The output plugs every value back into F equals G m1 m2 over r squared and recalculates both sides independently, so a wrong answer surfaces immediately.
Paste your homework word problem and pick force, one of the masses, or the distance to get a fully worked solution with the tiny value of G handled correctly.
Calculate the gravitational force between planets, moons, or a planet and an orbiting object, where the large masses finally make the formula's result meaningful.
Solve a gravitation problem here and a matching electrostatics problem with the paired solver to see how identical inverse-square math produces attraction-only gravity versus attract-or-repel electric force.
Generate a model solution for any universal gravitation problem before class, with the unit conversions, the algebra, and the verification step all visible for students to follow.
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