Universal Law & g Variations Universal Law & g Variations Every object with mass pulls on every other mass. This pull is extremely strong for planets and stars and very weak for small objects, but it is always present. On Earth, this pull creates a steady downward acceleration we call g . Although we often use g 9.8 , m/s 2 , it is not exactly the same everywhere. It changes slightly with altitude (height above Earth), with depth below the surface, and with latitude (because Earth rotates). Newton’s universal law of gravitation ties all of these together through a single, simple inverse-square rule: double the distance from a mass and its gravitational pull becomes one-fourth. From that law, we can predict how g behaves near Earth and around other worlds, calculate escape speeds, and understand why astronauts feel nearly weightless in orbit even when gravity is still acting on them. This concept is the backbone of satellite motion, planetary orbits, tides, and many NEET questions that test both your physics sense and your command of approximations. remember Big picture: Gravity is geometry-driven. Far from a spherical mass, field lines spread out, so intensity falls as 1/r 2 . Near Earth’s surface, g is almost uniform, but it must decrease with height, go to zero at the center, and be smallest at the equator due to rotation. The proportionality constant in Newton’s law of gravitation; G = 6.6743 10 -11 , m 3 ,kg -1 ,s -2 . Universal gravitational constant (G) A physical quantity that decreases in proportion to 1/r 2 with distance r from a source. Inverse-square law Acceleration due to gravity (g) The acceleration a freely falling object experiences due to a planet’s (or star’s) gravity; near Earth’s surface g 9.8 , m/s 2 . Latitude (φ) Angle measured north or south of the equator; affects effective g because of Earth’s rotation. Apparent outward acceleration in a rotating frame; on Earth 2 R 2 ! reduces effective g . Centrifugal acceleration Average radius R 6.371 10 6 , m used in near-Earth gravity calculations. Earth’s mean radius (R) Newton’s universal law of gravitation states that any two point masses M and m separated by distance r attract each other with a force whose magnitude is F = G ,Mm/r 2 . For spherically symmetric bodies, you can treat all their mass as concentrated at the center (shell theorem), so the same formula applies using r as the distance between centers. The direction of the force is along the line joining the masses, attractive in nature. Newton’s universal law (scalar form) Magnitude of gravitational force between two point masses. Direction is toward the attracting mass (negative radial direction). Vector form This law calculates the attractive force between any two masses, which determines the acceleration experienced by a test mass near a planet. On or near Earth’s surface, the gravitational pull of Earth (mass M ) provides an acceleration g to a test mass m . Equating the gravitational force to m ,g leads to g = GM/R 2 at the surface, where R is Earth’s radius. This result is independent of the test mass and explains why all bodies fall with the same acceleration in vacuum. Surface gravity: g = GM/R 2 Earth is spherically symmetric (shell theorem holds). Point outside a spherical mass distribution feels the same field as if all mass were at the center. Ignore Earth’s rotation and local density variations. g = GM R 2 Acceleration due to gravity at the surface of a spherical body of mass M and radius R; independent of the falling mass; sets the scale of free-fall near the surface. Dimensions and units: From g = GM/R 2 , we get [G] = L 3 M -1 T -2 . Numerically, G = 6.6743 10 -11 , m 3 ,kg -1 ,s -2 . Using M = 5.972 10 24 , kg and R = 6.371 10 6 , m gives g 9.81 , m/s 2 (ignoring rotation and local geology). Dimensional formula for the gravitational constant. Dimensions of G Variation of g with Height At an altitude h above Earth’s surface, the distance from Earth’s center is r = R + h . By the inverse-square law, the magnitude of g becomes g(h) = GM/(R+h) 2 . For low altitudes where h R , a binomial approximation expands (1 + h/R) -2 1 - 2h/R to give a very useful NEET-friendly estimate: g(h) g ,(1 - 2h/R) . This shows that g decreases roughly linearly with small height and more slowly at larger heights. Use the approximation only for heights much smaller than Earth’s radius. Exact and approximate g at height h This equation determines the gravitational acceleration at any altitude h , providing a precise value or a useful linear approximation for small heights. Binomial approximation for g(h) Earth is spherical and non-rotating for this calculation. Altitude h is small compared to R so that series expansion is valid. g(h) g (1 - 2h R ) ; for ;h R Boundary check: As h 0 , g(h) g . As h , g(h) 0 . Never apply the small- h approximation when h is a significant fraction of R (e.g., satellites hundreds of km up). tip Variation of g with Depth (inside Earth) Inside a spherically symmetric Earth of uniform density, only the mass enclosed within radius r contributes to gravity at that point (shell theorem). The interior field becomes proportional to r , so g(r) = (GM/R 3) ,r . If you measure depth d from the surface, r = R - d , giving g(d) = g ,(1 - d/R) . Thus, g decreases linearly with depth and becomes zero at the center. This is an idealized but very useful model for quick estimates and many exam questions. g vs r inside a uniform Earth Linear drop of g with depth; g=0 at Earth’s center. This equation models the linear decrease of gravity as depth increases, predicting zero acceleration at the planet's core. g(d) = g ,(1 - d/R) inside a uniform Earth Earth has uniform density. Spherical symmetry; shell theorem holds. Ignore rotation. g(d) = g (1 - d R ) Boundary check: At d=0 , g(d)=g ; at d=R , g(d)=0 . Real Earth is not uniform, so observed g(d) departs slightly from this ideal near the core, but the linear law is the standard NEET model. tip Variation of g with Latitude (Earth’s rotation) Because Earth rotates with angular speed 7.292 10 -5 , rad/s , an observer at latitude is in a rotating frame. The required centripetal acceleration is provided partly by gravity; the apparent (effective) weight is reduced by the outward centrifugal term 2 R 2 ! . Ignoring Earth’s slight oblateness, an excellent working formula is g eff ( ) g - 2 R 2 ! . Hence g is maximum at the poles ( =90 , no reduction) and minimum at the equator ( =0 , maximum reduction). Effective g vs latitude Rotation reduces effective g, most at the equator. Determine the effective acceleration due to gravity at a given latitude by subtracting the rotational reduction term. Common trap: Mixing the height/depth formulas with the latitude correction incorrectly. Apply g(h) or g(d) first (gravity-only), then subtract the centrifugal term 2 R 2 ! if the point is on the rotating Earth’s surface. neet-alert Putting it all together: Near Earth’s surface at latitude and small altitude h , a quick estimate is g eff g (1 - 2h R ) - 2 R 2 ! . For problems at depth d , use g eff g (1 - d R ) if the point remains within solid Earth and rotation’s effect along the local vertical is typically small for conceptual NEET questions unless asked explicitly. Escape Velocity Link The same inverse-square law gives the minimum speed to leave Earth without further propulsion: the escape velocity. From energy conservation, 1 2 m v e 2 - GMm/R = 0 at infinity, so v e = 2GM/R . Using g = GM/R 2 , this becomes v e = 2gR . This relation is frequently used to connect surface gravity to escape speed for different planets. Spherical mass M; launch from the surface of radius R. No air resistance; no additional propulsion after launch. Final speed tends to zero at infinity. Escape velocity v e = 2GM/R = 2gR v e = 2GM R = 2gR Minimum speed to escape a spherical body ignoring drag; also equals √(2 g R) at the surface. Escape velocity vs orbital velocity trap: For a circular low orbit just above the surface, v orb = gR , which is smaller than v e by a factor of 2 . Do not confuse the two. neet-alert HDL → Height, Depth, Latitude HDL rule for quick g-variations: Height small: g(h) g(1 - 2h/R) (2 in numerator); Depth: g(d) g(1 - d/R) (1 in numerator); Latitude: subtract 2 R 2 ! . Shell theorem essentials: (1) A uniform spherical shell exerts zero net gravitational force on a particle anywhere inside it. (2) Outside a uniform spherical shell, its gravitational effect is the same as if all its mass were concentrated at its center. These two facts justify treating Earth as a point mass for exterior points and explain the linear decrease of g inside a uniform Earth. Earth mass M = 5.972 10 24 , kg Earth mean radius R = 6.371 10 6 , m Standard g 9.8 , m/s 2 (varies slightly with location) Angular speed = 7.292 10 -5 , rad/s Gravitational constant G = 6.6743 10 -11 , m 3 ,kg -1 ,s -2 Earth and constants to remember g at Earth’s surface m/ s 2 easy G = 6.6743 10 -11 , m 3 ,kg -1 ,s -2 M = 5.972 10 24 , kg R = 6.371 10 6 , m Compute Earth’s surface gravity from first principles using G, M , R . Use g = GM/R 2 . Exact: g(h) = g ,(R/(R+h)) 2 . Approx: g(h) g(1 - 2h/R) (check validity: h/R 0.063 ). Find g at h = 400 , km above Earth’s surface (ISS altitude). Compare exact and approximate results. m/ s 2 medium g(h) exactly and using g(h) g(1 - 2h/R) h = 4.00 10 5 , m R = 6.371 10 6 , m g = 9.81 , m/s 2 ; ( surface ) m (as a fraction of R) easy Depth d where g(d) = g/2 g(d) = g (1 - d R ) Set g( d ) = g/2 and solve for d in terms of R . At what depth d inside a uniform Earth is g reduced to half its surface value? Ignore rotation. Value of and consistency check hard rad/s At equator: = 0 , ; 2 ! = 1 Observed reduction 0.0034 ,g (0.34 %) g 9.81 , m/s 2 , ; R = 6.371 10 6 , m A body at the equator experiences 0.34% reduction in apparent weight due to Earth’s rotation. Using g eff = g - 2 R 2 ! , estimate 2 R/g and check with actual data. From reduction: 2 R = 0.0034 ,g. Radial distance r from Earth’s center Inside Earth (0 to R): g increases linearly with r. Outside (R to 2R): g falls as 1/ r 2 . control g(r) dependent Center: g = 0 Surface: g = g max (ignoring rotation) At r = 2R: g = g/4 2R g/4 g magnitude m/ s 2 Piecewise behavior of g inside and outside a uniform spherical Earth. custom Variation of g Situation Formula Trend Description Approx Formula ( h ≪ R ) Graph Shape Going High or Deep makes Gravity weak, but stay at the Pole if weight is what you seek. At height h (above surface) g' = g ( R R+h ) 2 Gravity decreases as altitude increases due to inverse square law from the center. g' g ( 1 - 2h R ) Hyperbolic ( g 1/r 2 ) At depth d (below surface) g' = g ( 1 - d R ) Gravity decreases linearly as depth increases, becoming zero at the center. N/A (Linear formula used) Straight line (Negative slope) At latitude φ g = g - 2 R 2 Gravity increases from the Equator to the Poles due to centrifugal force reduction. g e = g - 2 R (at = 0 ) Cosine squared curve Pole vs Equator (Earth shape) g = GM R 2 Gravity is maximum at Poles and minimum at Equator because R p < R e . g 0.018 m/s 2 Ellipsoid variation At Earth's centre g = 0 Effective mass enclosed at the center is zero, leading to weightlessness. Origin point (0,0) Inside Earth (uniform density) g = 4 3 G r Inside Earth ( r < R ), gravity increases linearly with distance from the center. N/A Linear ( g r ) At infinite distance g = 0 At infinite distance from the mass, the gravitational field strength vanishes. Asymptotic to x-axis variation of g Heavier objects fall faster than lighter ones because gravity pulls more strongly. Although gravity pulls more strongly on heavier objects, acceleration g is the same because the mass cancels in F = m g vs F = G Mm/r 2 . There is no gravity in space. Gravity exists everywhere but weakens with distance. Astronauts orbit Earth in free fall; they feel weightless even with g 8.7 , m/s 2 near the ISS altitude. Approximation trap: Do not use g(h) g(1 - 2h/R) when h/R is not ≪ 1. For h 0.1R or larger, use the exact inverse-square formula. neet-alert Inside a uniform Earth, g increases linearly from the center and matches the surface value at r=R . Outside, it decays as 1/r 2 . These two simple laws cover most NEET problems on g-variation. remember Connections across planets: For a planet with mass M p and radius R p , g p = GM p/R p 2 and v e,p = 2g p R p . Comparing worlds becomes quick: larger M/R 2 means stronger surface gravity, while larger R at fixed g raises escape speed. Practical measurement notes: Local g is measured with pendulums and gravimeters. Variations arise from altitude, Earth’s rotation, latitude, subsurface density (mountains, ore bodies), and tides due to Moon and Sun. NEET questions usually isolate one effect at a time to keep algebra clean. Sign conventions and vectors: Gravitational field g points toward the mass center: g ( r ) = - ,G M r /r 3 . Magnitudes use g = GM/r 2 . Always keep track of whether a problem needs direction (vector) or only magnitude (scalar). Vector gravitational field Field lines point radially inward toward the mass. This magnitude calculation allows determining g at any distance r, noting that g is zero at the center and changes with altitude. Edge cases to test your intuition: (1) Very far from Earth, g 0 , but never exactly reaches zero; it just becomes negligible. (2) At Earth’s center, symmetry forces cancel exactly, so g=0 . (3) On a non-rotating planet, there is no latitude effect whatsoever; only shape/density matter. Is the point outside Earth? Use g = GM/r 2 with r = R + h . Is the point inside Earth (uniform model)? Use g(r) = GM r / R 3 or g(d) = g(1 - d/R) . Is rotation relevant? Subtract 2 R 2 ! from gravity-only g at the surface. Is h R ? Then use the binomial approximation; else use the exact inverse-square form. Are you asked for escape speed? Use v e = 2gR (surface) or 2GM/r (general). Checklist for choosing the right formula Dimensional and unit sanity checks Confirm [G] = L 3 M -1 T -2 from F = GMm/r 2 . In g = GM/R 2 , dimensions of RHS must be L T -2 . For v e = 2gR , units combine to (L T -2 ) ,L = L T -1 . neet-alert Do not plug depth d into g(h) or altitude h into g(d) . These are different physical regimes. At depth, use the linear interior model; at altitude, use the inverse-square exterior model. Worked proportionality example: If a planet has the same density as Earth but twice the radius, its mass scales as M R 3 , so g = GM/R 2 R . Therefore, g would be twice Earth’s. However, escape speed v e = 2gR would scale as R times larger than Earth’s factor of 2 , giving v e increased by 2 times 1.414 for the doubled R case with same density. A planet has density equal to Earth’s but radius 1.5R . Find its surface gravity and escape velocity relative to Earth. M p /M = (R p /R ) 3 = 1.5 3 = 3.375. Then g p /g = (M p /M )/(R p /R ) 2. Same density: M R 3 Planet radius R p = 1.5 R On Earth: g , v e Ratios g p / g and v e,p / v e relative factors medium Accuracy matters: Using g = 9.8 , m/s 2 vs 9.81 , m/s 2 rarely changes multiple-choice answers at 2 significant figures. However, when comparing exact vs approximate g(h) , keep at least three significant figures in intermediate steps to avoid rounding errors. Unit handling: When computing g(h) with h in km, convert to meters and add to R before squaring. A very common slip is squaring R but not converting h properly. tip Weight vs mass: Mass is intrinsic, measured in kg. Weight is the gravitational force W = m g eff and varies with location (height, depth, latitude). Many NEET items test whether you change the correct quantity in a comparison (weight changes, mass does not). Gravity near massive bodies: On Jupiter (approximate M J = 1.898 10 27 , kg , R J = 6.9911 10 7 , m ), g 25.9 , m/s 2 . On the Moon ( M moon 7.342 10 22 , kg , R moon 1.737 10 6 , m ), g 1.62 , m/s 2 . The inverse-square dependence on radius dominates across planets and moons. Conceptual tie-in: Free-fall in orbit. A satellite in a circular orbit is constantly falling toward Earth under g but moving tangentially fast enough to keep missing the surface. Gravity is not absent; it is the very cause of the orbit. This idea helps reconcile weightlessness with persistent gravitational acceleration. Contrast with escape velocity: v e = 2 ,v orb . Circular orbital speed near surface Use this speed to determine the tangential velocity needed to maintain a stable circular orbit due to gravitational pull. Worked symbolic manipulation: From g = GM/R 2 , holding g fixed while increasing R implies M R 2 . However, if density is fixed ( M R 3 ), then g R . Such dimensional and scaling checks quickly give orders of magnitude and qualitative trends in NEET problems. Limitations of the models: The uniform-density interior and perfect-sphere exterior are idealizations. Real Earth has layered density, mountains, ocean trenches, and is slightly oblate, introducing small deviations (on the order of parts in 10 3 ). NEET typically ignores these unless the question explicitly mentions them. Quick compare: Height effect is stronger (factor 2) than depth effect (factor 1) for the same fractional change in R in their linearized forms. This is easy to recall with the HDL mnemonic. remember Sanity test across regimes: If you were to dig a straight tunnel through Earth’s center (uniform model), a dropped object would oscillate through the center with simple harmonic motion, because g(r) r acts like a linear restoring force. While beyond the scope here, this reinforces the interior linearity of g . Computational tip: When a problem provides G, M , and R , compute GM once as the standard gravitational parameter = GM to reduce arithmetic. For Earth, 3.986 10 14 , m 3/s 2 , a value that appears in many orbital and gravity calculations. Common NEET question patterns on this topic Percent change in g for small h (use linear approximation). Depth for which g reduces to a given fraction (use g(d) = g(1 - d/R) ). Difference in weight between equator and poles (use latitude formula). Comparisons across planets/moons (use g = GM/R 2 and sometimes v e link). Historical note: Cavendish’s torsion balance experiment first measured G with high precision, enabling the first good estimates of Earth’s density and mass. In modern physics, G remains the least precisely known of the fundamental constants compared to c and h . Key terms recap Universal gravitational constant Constant G = 6.6743 10 -11 , m 3 ,kg -1 ,s -2 . Local free-fall acceleration g = GM/R 2 at a spherical body’s surface. Acceleration due to gravity Escape velocity v e Minimum speed to reach infinity with zero final speed: v e = 2GM/R = 2gR . Geographic angle from equator; affects effective g via rotation. geographic latitude Latitude Shell theorem Outside a spherical shell, field is as if all mass were at center; inside, net force is zero.