Correct option is B) The gravitational potential due to solid sphere at a distance r from the centre of the sphere is given by V= rGM. Let's find the \( \vec{g} \) field due to a sphere of constant mass density \( \rho \), using the gravitational potential \( \Phi \). A corollary is that inside a solid sphere of constant density, . Get a quick overview of Gravitational potential due to solid sphere from Gravitational Potential - L2 and Gravitational Potential in just 3 minutes. (Gravitational Potential = potential energy per unit mass) Let's assume that the Earth is a uniform solid sphere of mass M and radius R. We define gravitational potential difference as the work done to slowly move a unit mass from one point to another. It will not be pulled in all directions; it will experience no stress . Nobody has given the derivation using integration from scratch and it's a straightforward and elegant piece of maths, so here goes. Orodruin told me in post#12 that potential inside the sphere is constant and the same as at the surface is only for spherical shell so I tried to find formula for solid sphere. Why can't we do it like for a point outside or on the surface where we just need to integrate from . This can be seen as follows: take a point within such a sphere, at a distance from the center of the sphere. which is the same as the gravitational attraction of the mass interior to the observation point. Its unit is N m kg1. The gravitational potential at the centre of the Earth is -1.5GM/R. Contents 1 Potential energy 2 Mathematical form 3 Spherical symmetry Example: gravitational potential of a sphere. The matter above it (since it is inside its shell) exerts no influence on it. Solve any question of Gravitation with:-. Sphere Gravitational Potential Energy For a self-gravitating sphere of constant density , mass M, and radius R, the potential energy is given by integrating the gravitational potential energy over all points in the sphere, (1) (2) (3) where G is the gravitational constant, which can be expressed in terms of (4) as (5) According to Equation ( 897 ), the gravitational potential inside a uniform sphere is quadratic in . I'll assume the question is referring to a ball (solid sphere) with uniform d. The gravitational force inside a hollow sphere shell of uniform areal mass density is everywhere equal to zero, and may be proved by the following argument: Let the sphere have a radius a. Gravitational Potential and attraction of solid sphere outside from the sphere have been derived in previous video. Place a point P inside the sphere at a distance r from the center where r < a; i.e., r is strictly less than a. In the case of the gravitational field of a massive sphere, the quadratic term in the potential is proportional to the gravity gradient. Place a point P inside the sphere at a distance r from the center where r < a; i.e., r is strictly less than a. It turns out that this is a general result for any finite spherically symmetric mass distribution. But, for solid sphere, gravitational potential inside it is a hectic deal. Figure %: Forces exerted on a particle inside a solid sphere. . Answer: For a point outside the sphere : V = -GM/r For a point on the surface : V = -GM/R For a point inside the sphere : V = -(GM/2R^3)*(3R^2-r^2) Considering the solid sphere be homogeneous and having mass M & radius R G be the universal gravitational constant r be the distance of the poin. Gravitational Potential and attraction of solid sphere outside from the sphere have been derived in previous video. So inside of a sphere, there is no gravitational force at all! The potential at P is due to two parts: The inner part of the sphere with radius less that 1 Gauss' law states that, for a closed surface S, S g d A = 4 G m e n c l where m e n c l is the total mass enclosed by the surface. An approximation of its density is (r)=(ABr), where A=12700kg/m 3,B=1.50 10 3kg/m 3 and r is the distance from the centre of earth. I am considering a point a position vector $\textbf{r}$, and a small mass element of the sphere within, at a position vector $\textbf{r}_m$. . For a solid sphere this means that for a particle, the only gravitational force it feels will be due to the matter closer to center of the sphere (below it). where G is the gravitational constant, and F is the gravitational force. It is a scalar quantity. The gravitational force inside a hollow sphere shell of uniform areal mass density is everywhere equal to zero, and may be proved by the following argument: Let the sphere have a radius a. Gravitational potential inside a solid sphere. These suprisingly simple results are actually consequences of some powerful math at work, which we . Hence, the gravitational potential inside a solid sphere is given by Advertisement Derivation of gravitational field outside of a solid sphere. Or J kg-1 dimensional formula M0L2T-2 Mathematically V = W/m By the definition of potential energy U = W So V = U/m In mathematics, the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory. There are three steps to proving Newton's shell theorem. Figure 6: Observation point inside a solid sphere. we allow a new class of solutions given by the Yukawa potential, . MS 2053.7-0449: Confirmation of a bimodal mass distribution from strong gravitational lensing arXiv:0704.3012v1 [astro-ph] 23 Apr 2007 T. Verdugo and J.A. The potential has units of energy per mass, e.g., J/kg in the MKS system. It may also be used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies. First, the equation for a gravitational field due to a ring of mass will be derived. For a solid sphere this means that for a particle, the only gravitational force it feels will be due to the matter closer to center of the sphere (below it). To calculate the gravitational potential at any point inside a solid sphere, why do we need to separately integrate gravitational field from infinity to radius and then from radius to the point? This implies that if a narrow shaft were drilled though the center of the sphere then a test mass, , moving in this shaft would experience a gravitational force acting toward the center which scales linearly in . Electric Potential of a Uniformly Charged Solid Sphere Electric charge on sphere: Q = rV = 4p 3 rR3 Electric eld at r > R: E = kQ r2 Electric eld at r < R: E = kQ R3 r Electric potential at r > R: V = Z r kQ r2 dr = kQ r Electric potential at r < R: V = Z R kQ r2 dr Z r R kQ R3 rdr)V = kQ R kQ 2R3 r2 R2 = kQ 2R 3 . Note: Conventionally gravitational potential energy on the surface of the earth is considered to be zero. Answer (1 of 5): > What is the derivation for gravitational potential inside a sphere? Gravitational potential Gravitational potential (V) at a point is defined as the amount of work done in moving unit mass from the point to infinity against the gravitational field. Since the gravitation field within the hollow sphere is zero, the object will experience no gravitational force at all. Gravity Force Inside a Spherical Shell For application of the law of gravity inside a uniform spherical shell of mass M, a point is chosen on the axis of a circular strip of mass. The object inside a hollow shpere will experience a pull in all directions, expanding the object slightly (assuming that the object isn't decompressable). Where G is universal gravitation constant and M is the mass of the earth and r is the distance of the body from the centre of the earth. By projecting the matter in a plane, one obtains the circularly-symmetric surface mass distribution () = v 2G 1 (19) The mass inside . Say we want to find g at radius r; then let S be the sphere of radius r. Due to the symmetry of the sphere, this can be rewritten as S g d A = 4 G m e n c l giving us g 4 r 2 = 4 G M r 3 R 3 For the given case, r < R where R is the distance outside the surface and r is the distance inside from the surface of the sphere. Its derivation involves complex steps. de Diego Instituto de Astronoma, UNAM, AP 70-264, 04510 Mexico DF tomasv, jdo@astroscu.unam.mx and Marceau Limousin Dark Cosmology Centre, Niels Bohr Institute, University of Copenhagen, Juliane Marie Vej 30, 2100 Copenhagen, Denmark . . A corollary is that inside a solid sphere of constant density, the gravitational force within the object varies linearly with distance from the center, becoming zero by symmetry at the center of mass. Indeed, from the previous analysis, it is clear that and for a spherically . According to Equation ( 898 ), the gravitational potential outside a uniform sphere of mass is the same as that generated by a point mass located at the sphere's center. If this setup is placed in a gravitational field represented by a Newtonian potential , the total potential is given by V=V trap +m . PG Concept Video | Gravitational Field | Gravitational Potential due to a Solid Sphere by Ashish Arora Students can watch all concept videos of class 11 Grav. Its derivation involves complex steps. 2.4 Non-Singular isothermal sphere Here we assume singular isothermal sphere with a three-dimensional density distribution = 2 v 2G 1 r2 (18) where v is the one-dimensional velocity dispersion. The BEC 38 is therefore spherically symmetric, and possesses a radius 0. Draw a line through P to intersect the sphere . The product GM is the standard gravitational parameter and is often known to higher precision than G or M separately. The matter above it (since it is inside its shell) exerts no influence on it. This result shows that Poisson's equation holds in a sphere of uniform density. For the given case, r < R where R is the distance outside the surface and r is the distance inside from the surface of the sphere. . Suppose that P is in a narrow, spherical cavity of radius r and thickness indicated in the following figure: Solid spherical sphere . Gravitational potential Continuing from last time, we defined the gravitational potential (not the potential energy!) >. But, for solid sphere, gravitational potential inside it is a hectic deal. Let's derive it then. Figure %: Forces exerted on a particle inside a solid sphere. Why can't we do it like for a point outside or on the surface where we just need to integrate from . So,as 'r' increases the magnitude of V decreases. clearly illustrates this fact. I am struggling to derive the gravitational field strength within a solid sphere. After replacing the dot product, the value of gravitational potential for a point inside the solid sphere becomes, On solving the integration, the expression for the potential becomes, . The potential at a point inside the solid sphere will be the sum of both potentials On solving the dot product, we get After replacing the dot product, the value of gravitational potential for a point inside the solid sphere becomes, On solving the integration, the expression for the potential becomes, to define quantitatively what we mean by the strength of a gravitational field, which is merely the force experienced by unit mass placed in the field. I shall use the symbol g for the gravitational field, so that the force F on a mass m situated in a gravitational field g is F = mg. 5.2.1 It can be expressed in newtons per kilogram, N kg-1. For a self-gravitating sphere of constant density , mass M, and radius R, the potential energy is given by integrating the gravitational potential energy over all points in the sphere, (1) (2) (3) where G is the gravitational constant, which can be expressed in terms of. Example The earth does not have a uniform density; it is most dense at its centre and least dense at its surface. By convention, it is always negative where it is defined, and as x tends to infinity, it approaches zero. Gravitational potential inside a solid sphere. The correct one is ##V=-\frac{GM(3R^2-r^2)}{2R^3}## because the question is about solid sphere, not shell. So , total Potential at Point A = (- GM/L) + (-3GM/2L) = - 5GM/2L To calculate the gravitational potential at any point inside a solid sphere, why do we need to separately integrate gravitational field from infinity to radius and then from radius to the point? The Laplacian of U is: 2U = 2 3 G 3a2 (x2 +y2 +z2) (38) = 4G (39) which is Poisson's equation. The derivation in the case of P being inside the sphere requires a bit more understanding. Solution - Gravitational Potential at point A due to mass M is = - GM/L and due to mass 3M is = - G (3M)/2L We know that Gravitational potential is a scalar quantity so for total Gravitational potential at point A , we can simply add the potential due to both masses separately. Answer: For a point outside the sphere : V = -GM/r For a point on the surface : V = -GM/R For a point inside the sphere : V = -(GM/2R^3)*(3R^2-r^2) Considering the solid sphere be homogeneous and having mass M & radius R G be the universal gravitational constant r be the distance of the poin. Let's derive it then. For PDF Notes and best Assignments visit @ http://physicswallahalakhpandey.com/Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, . which is related to the gravitational field as \vec {g} = -\vec {\nabla} \Phi g = . Hence, the gravitational potential inside a solid sphere is given by. The problem is envisioned as dividing an infinitesemally thin spherical shell of density per unit area into circular strips of infinitesemal width. Gravitational Potential due to a solid sphere. clearly illustrates this fact. Patterns of problems. For a source mass M M at the origin, the potential takes the form \begin {aligned} \Phi (r) = -\frac {GM} {r} \end {aligned} (r) = rGM so that Thus gravitational potential energy = Gravitational potential at a point x Mass of the body at that point.