Gravitational PE Potential & Field Gravitational PE, Potential, and Field Gravity pulls, but it also “stores” the ability to do work. The pull is described by the gravitational field, a vector that gives force per unit mass at each point in space. The storage idea is captured by gravitational potential (energy per unit mass), a scalar that tells you how much energy is associated with a location if you place 1 kg there with zero reference at infinity. When you move in a gravitational field, energy shifts between kinetic and potential in a predictable way because gravity is conservative: the work done depends only on the starting and ending points, not on the path. That is why we can solve many problems without tracking forces along a complicated path; we can jump straight to energy differences using potential. The sign convention is vital: by NCERT standard we take potential to be zero at infinity. Because gravity is always attractive, you must do positive work (from outside) to take a mass from a finite distance to infinity, so the potential at any finite distance comes out negative. This negative number is not “less than zero energy” in an absolute sense; it only means the point lies in a gravitational well relative to infinity. Equipotential surfaces (surfaces where potential is constant) help visualize this: moving along one costs no work against gravity. Field lines are always perpendicular to these surfaces and point toward decreasing potential. With these ideas, we can compute the energy needed to raise satellites, the speed required to escape a planet, and even the subtle differences between “no field” and “no potential change.” remember Analogy: Think of potential like height in a landscape. Deep valleys correspond to large negative potential. The gravitational field is like the slope: it points downhill and its magnitude is the steepness. Gravitational field (intensity) Force per unit mass at a point: the acceleration a small test mass would experience. Vector quantity, usually denoted by g . Potential energy per unit mass at a point with zero taken at infinity. Scalar quantity, denoted by V . Gravitational potential Gravitational potential energy (PE) Energy of a mass m due to gravity at a point: U = mV . With zero at infinity, U=-GMm/r for a point mass M . Equipotential surface A surface where V is constant everywhere. No work is done by or against gravity when moving along it. A force for which work done between two points is path independent. Gravity is conservative. Conservative force Zero at infinity convention Standard choice V( )=0 . Under this convention, gravitational potentials at finite distances are negative. Binding energy Energy required to take a bound mass from its current position to infinity (to unbind it). Numerically = -U at that position. Field vs potential in daily thinking: the field g tells you the immediate acceleration a freely falling object would feel at a point; it directly gives forces via F =m g . Potential V tells you the energy landscape: how much kinetic energy you would gain or lose if you move between locations. Because V is scalar, potentials from multiple sources add up algebraically (superposition) without worrying about directions. Fields from multiple sources add vectorially, requiring directions and components. This split is powerful for problem-solving: use V to handle energies and multi-source setups quickly, and use g to understand directions and instantaneous accelerations. Field of a point mass Radially inward field of a spherically symmetric mass M . Potential of a point mass Scalar potential with zero at infinity. Potential energy of mass m in the field of M . Potential energy near a point mass Satellite Energy Map Energy Type Formula Sign Relation to Kinetic Energy Relation to Total Energy Kinetic is King (Positive), Potential is Pit (Negative), Total is Tied (Negative), and Binding breaks free (Positive). Kinetic Energy ( K ) G M m 2r Positive ( + ) -E Potential Energy ( U ) - G M m r Negative ( - ) -2K 2E Total Energy ( E ) - G M m 2r Negative ( - ) -K Binding Energy ( B.E. ) G M m 2r Positive ( + ) +K -E (Magnitude only) satellite energy map This potential energy U defines the scalar potential V , which is used to find the radial gravitational field g r = -dU/dr . The field points in the direction of greatest decrease of potential. Field–potential relation V(r) = - GM r Spherically symmetric mass M so exterior field is central Zero potential at infinity: V( )=0 Potential of a point mass V(r) = - GM r Why the negative sign? Because gravity attracts, you must supply positive external work to carry a test mass to infinity. Hence, at any finite r , the potential must be less than the zero at infinity, i.e., negative. The deeper the well (smaller r or larger M ), the more negative V . Importantly, differences in potential matter physically: U = m V is what sets the kinetic energy change for slow (non-relativistic) motion when only gravity acts. If you move along a path that stays on the same V , no work is exchanged with gravity. tip Applicability: V=-GM/r and g=GM/r 2 are valid outside a spherically symmetric mass (or for a true point mass). Inside a uniform solid sphere, g r and V is quadratic in r . Inside a thin shell, g=0 but V is constant and negative. At the surface of a spherical body: g=GM/R 2 . Useful as the local field magnitude for near-surface estimates and for connecting M and R to g . g = GM R 2 Surface gravity g = GM R 2 Spherical symmetry Exterior field is that of a point mass at center Test mass m is negligible Minimum speed to reach infinity with zero speed left: v e= 2GM/R . Follows from energy conservation using U=-GMm/r and K= 1 2 mv 2 . v e = 2GM R No air resistance Spherical symmetry Zero potential at infinity Escape velocity v e = 2GM R Energy language unifies many results. Because U=-GMm/r , raising a satellite from r 1 to r 2 (slowly, without changing its orbital kinetic energy) needs external work W ext = U = -GMm ,(1/r 2 - 1/r 1) = GMm ,(1/r 1 - 1/r 2) . The farther you go, the smaller 1/r , so potential becomes less negative and U is positive: you must supply energy. Escape velocity is the special case where the final potential energy is zero and the final kinetic is also zero, so all initial kinetic balances the bound energy. Use V=-GM/r and U=mV with zero at infinity. Find the gravitational potential at Earth’s surface and the potential energy of a 2.0 kg mass there. Use G=6.67 10 -11 , N ,m 2/kg 2 , M E=6.0 10 24 , kg , R E=6.4 10 6 , m . G = 6.67 10 -11 , N ,m 2/kg 2 M E = 6.0 10 24 , kg R E = 6.4 10 6 , m m = 2.0 , kg J/kg, J easy Potential V(R E) and potential energy U(R E) Interpretation: The negative sign means the mass is bound to Earth. To remove the 2 kg mass to infinity (very slowly) you’d need + 1.25 10 8 J of external work. The magnitude of V is huge compared to typical laboratory energy scales, which is why spaceflight is energetically expensive. medium External work W ext (quasi-static raise) m = 1000 , kg r 1 = 3R E = 1.92 10 7 , m r 2 = 6R E = 3.84 10 7 , m Use U = GMm ,(1/r 1 - 1/r 2) . How much work must an external agent do to slowly move a 1000 , kg satellite from r 1=3R E to r 2=6R E around Earth? Use G=6.67 10 -11 , M E=6.0 10 24 , kg , R E=6.4 10 6 , m . Always measure r from the planet’s center, not from the surface. A common mistake is to plug altitude h directly into U=-GMm/h ; the correct distance is r=R+h . neet-alert Equipotential surfaces and field lines: For a point mass, equipotential surfaces are concentric spheres ( V depends only on r ), and field lines are radial straight lines pointing inward. Because g equipotential surfaces, any displacement tangential to a sphere costs no work against gravity. In non-spherical mass distributions (like mountains or lumpy asteroids), equipotentials warp accordingly, but the orthogonality between g and equipotentials still holds. This gives an easy test in problems: if the motion is along a path where V does not change, the work by gravity is zero. Key properties of equipotentials No work is done moving along an equipotential surface: U = m V = 0 . Field lines are perpendicular to equipotentials and point toward decreasing V . Closer equipotentials imply a stronger field magnitude (steeper energy slope). Potential is scalar, so equipotentials from multiple sources overlay via simple addition of V . V (gravitational potential) J/kg Gravitational potential vs r for a uniform solid sphere. custom -GM/R Surface potential Center potential (uniform sphere) -3GM/2R control dependent Outside a uniform sphere (r ≥ R): V = -GM/r (a negative hyperbola approaching 0 from below). Inside (r ≤ R): V(r) = -(GM/2R 3)(3R 2 - r 2 ), a downward-opening quadratic continuous at r =$R. r (distance from center) g (field magnitude) m/ s 2 Field magnitude vs r for a uniform solid sphere. custom control dependent Center Surface GM/ R 2 Outside: g = GM/ r 2 falling with 1/ r 2 . Inside a uniform solid sphere: g increases linearly from 0 at center to GM/ R 2 at the surface. r (distance from center) Inside and around spheres: For a thin spherical shell of radius R , g=0 everywhere inside but the potential is constant V=-GM/R (still negative). For a uniform solid sphere of radius R , the interior field is g(r)=GMr/R 3 (linear in r ), and potential is V(r) = - GM 2R 3 (3R 2 - r 2) . Note V is continuous at r=R and is lowest (most negative) at the center: V(0)=-3GM/2R ; the surface value is V(R)=-GM/R . Interior potential: uniform sphere Continuous with V=-GM/R at r=R . This formula describes the gravitational potential at a distance 'r' from the center of a sphere, provided the sphere has a uniform mass den Field increases linearly from the center to the surface. Interior field: uniform sphere This standard form applies when calculating the field outside the mass, treating the entire body as a point source. Two masses M and 4M are fixed on a line separated by distance d . Find (i) the point between them where the net gravitational field is zero, and (ii) the gravitational potential at that point. Also, what is the work required to bring a 1 , kg test mass from infinity to that point? Use g=GM/r 2 and superposition for fields (with directions), and V=-GM/r with algebraic addition for potentials. Masses: M and 4M Separation: d Zero at infinity convention Location of zero field between masses, potential there, and the work to bring 1 kg from infinity. m, J/kg, J hard remember Zero field does not mean zero potential. At the point where forces cancel, potential from each mass still adds to a finite negative value. Gravitational field (intensity) Vector Vector sum Force/acceleration direction and magnitude Gravitational potential Scalar Algebraic sum Energy change per unit mass Potential energy U = mV Scalar Algebraic sum Work needed or given by gravity Quantity Symbol Type Adds as Gives directly V is the Valley (negative well), g is the Gradient (slope) pointing downhill. Valley–Gradient link: g =- V Superposition is easiest for potentials: for masses M 1, M 2, , at a point P with distances r 1, r 2, , the total potential is V(P)= -G i M i/r i . There are no components or angles—just add numbers with the correct signs. For fields, you must add vectors: directions matter. Many NEET questions reward recognizing when to use potential (quick scalar sums) versus when to use fields (getting directions or accelerations). Scalar addition makes multi-source problems simpler. Superposition of potentials Use this scalar potential to quickly find the total potential energy in multi-source setups without worrying about directions. Boundaries and approximations: Near Earth’s surface for small heights h R , the change in potential per unit mass is approximately V gh with g constant. This comes from a Taylor expansion of V=-GM/(R+h) about h=0 : V(R+h) -GM/R + GMh/R 2=V(R)+gh . Use this only for small h ; for satellite scales ( h comparable to R ), always use the exact 1/r formulae. Potential is zero at the Earth’s surface by definition. By NCERT convention, the zero is at infinity. Hence V(R E)=-GM E/R E is negative. You can choose other zeros, but then all answers must be consistently shifted. If the net gravitational field at a point is zero, its potential must also be zero. Field is a vector sum that can cancel; potential is a scalar sum that generally will not cancel. Zero field points usually have finite negative potential. Define the zero of potential (use V( )=0 unless the problem states otherwise). Write U=-GMm/r and identify r from the center. For multi-step moves, compute U = U 2 - U 1 = -GMm ,(1/r 2 - 1/r 1) . For escape, set final K=0 and U=0 at infinity to get v e= 2GM/R . Use superposition of V for multiple sources, then multiply by m to get U . Energy-method checklist for gravitation Strategy pointer: When a question asks for energy required, work done, or speed at a given radius in a central gravitational field, the energy route is usually fastest. When it asks for direction of acceleration or the point where net force is zero, the field (vector) route is better. If multiple bodies are involved, evaluate V first to simplify the algebra. Units and dimensions: Gravitational potential V has units of J/kg (energy per unit mass). The field g has units of m/s 2 . The universal constant G has units N ,m 2/kg 2 or equivalently m 3/(kg ,s 2) . A useful check: in V=-GM/r , the right-hand side is ( m 3/(kg ,s 2) ) ( kg )/( m )= m 2/s 2 = J/kg . Dimensional sanity checks are vital for catching algebra slips under exam pressure. Near-surface linearization: For small vertical displacements h R , one can treat g as constant and write U mgh . This matches the exact formula since V = GM ,(1/R - 1/(R+h)) GM ,(h/R 2) = gh . But as soon as h is not negligible compared to R , abandon mgh and return to U=-GMm/r to avoid sizeable errors. Potential energy diagrams: Plotting U(r)=-GMm/r against r shows a deep well, bottoming at small r and asymptotically approaching 0 from below as r . Bound orbits have total energy E<0 , meaning the sum of kinetic and potential still lies below the zero line; an impulse that lifts E to 0 delivers escape. These diagrams help you reason about what impulses (or burns) can do: raising apoapsis, circularizing, or escaping entirely. Work sign conventions: If gravity does positive work on a falling mass, its potential energy decreases (becomes more negative), while kinetic increases. For an external agent lifting slowly, the external work equals + U ; you are storing energy into the gravitational configuration. Keeping track of signs with the zero-at-infinity convention prevents double negatives: write the general formula first, then substitute radii. Comparison note (with care): The mathematics of gravitational potential resembles electrostatic potential: both follow inverse-distance laws and obey superposition. But crucial differences matter: gravity is always attractive and depends on mass (only positive); electrostatic forces can attract or repel and depend on charges of both signs. Do not copy electrostatic sign habits onto gravity problems. Common Earth numbers: With G=6.67 10 -11 , SI , M E 5.97 10 24 , kg , and R E 6.37 10 6 , m , we have g 9.8 , m/s 2 , surface potential V(R E) -6.25 10 7 , J/kg , and escape speed v e 11.2 , km/s . Memorizing these helps ballpark answers. Do not mix up r and h : U=-GMm/r uses r from the center; the near-surface approximation U mgh uses height h measured from a local reference. Use one model consistently. neet-alert Uniform solid sphere V( )=0 Values of V(0) and V(R) and their difference. J/kg medium Find the potential at the center of a uniform spherical planet of mass M and radius R , and compare it to the surface potential. Use V(0)=-3GM/2R and V(R)=-GM/R from the interior potential formula. Mathematical link g =- V : In one-dimensional radial motion, g r=-dV/dr , meaning the slope of the V –vs– r curve gives the field magnitude (with a negative sign for inward direction). Where V(r) changes rapidly with r , |g| is large; where V(r) is flat, |g| is small. This neatly matches the landscape picture: steep hills (big slope) imply strong pull. Path independence proof sketch: Since g =- V , the line integral of gravity around any closed loop is F d r = m g d r = -m V d r = -m V loop = 0 . Hence gravity does zero net work in closed loops, making energy methods consistent and powerful. When formulas fail: The simple V=-GM/r assumes spherical symmetry or a point mass. For irregular bodies or near massive extended structures (like mountain ranges), exact evaluation requires integration over mass elements: V( r ) = -G dm | r - r '| . NEET rarely demands such integrals but may give composite systems (two or three point masses), where superposition remains the best tool. Sign sanity: If you move outward ( r increases), V becomes less negative, so V>0 and U>0 . If gravity alone accelerates you inward, your U drops (more negative) and K rises by the same amount. tip Worked micro-move: For an infinitesimal outward move dr in a central field, dV = (GM/r 2)dr and dU = m ,dV . The work done by gravity is -dU = -m ,dV . Integrating from r 1 to r 2 recovers U = -GMm(1/r 2 - 1/r 1) . Keeping this differential picture in mind can prevent sign errors. Total mechanical energy in circular orbit: A satellite of mass m at radius r has K=+GMm/(2r) and U=-GMm/r , so E=K+U=-GMm/(2r)<0 . This negative total energy confirms the bound state. Doubling r halves the magnitude of E , which explains why raising an orbit requires substantial but not astronomical energy compared to escape. Height vs radius trap: Suppose you are given altitude h above Earth’s surface. The correct distance from the center is r=R E+h . Plugging h directly into U=-GMm/h produces meaningless numbers (and wrong units). Always transform heights to radii first when using inverse-distance formulas. Worked intuition check: Why is V for a shell constant inside? Each small patch on the shell has a counterpart on the opposite side; though one is closer and exerts a stronger pull, the other is farther but covers more area. Their contributions to V cancel the distance effect exactly in the sum, making V constant; for g the vector directions cancel, yielding zero field. Multiple-source shortcut: For two or three point masses, first write V(P)=-G M i/r i . If you are asked for work to bring a mass m from infinity to P , just compute U(P)=mV(P) . If instead the question asks for acceleration or direction, compute g (P) by vector addition of -GM i r i/r i 2 . Precision and significant figures: NEET answers typically expect two significant figures. Use standard constants ( G=6.67 10 -11 , R E=6.37 10 6 , m , M E=5.97 10 24 , kg ) unless otherwise instructed, and round neatly at the end, not mid-calculation. Terms recap Gravitational field (intensity) Field Intensity Force per unit mass: g =- GM r 2 r outside a spherical mass. Potential Energy per unit mass: V=- GM r (zero at infinity). Gravitational potential GPE PE U=mV=- GMm r . Gravitational potential energy Isopotential Surface of constant V ; moving along it requires no work by gravity. Equipotential surface Escape velocity v e= 2GM R from radius R ignoring drag. Escape speed v e