assuming the earth to be a sphere of uniform density. PDF Problem Set 1: Structure of the Earth. Updated On: 12-03-2022 This browser does not support the video element. jpg A particle of mass mmoves under the in uence of a central attractive force F= k r2 e r=a. The weight of the fluid is equal to its mass times. (c ) Acceleration due to gravity is independent of mass of the earth/mass of the body. JEE Main & Advanced Physics Gravitation. The opposite effect occurs for falling leaves, as their radial distance from the earth's axis. Step 3: The surface charge density of the sphere is uniform and given by 2 QQ A4a σ π == (5. Let's assign this liquid some additional properties: We know that the average density of the earth is about. After many millions (or billions) of years, the surface hardened into a thin crust. HC Verma Solutions Chapter 10. Assuming, the earth to be sphere of uniform density, what is the value of acceleration due to gravity at a point 100 km below the earth surface? Given R = 6380 × 10 3 m ) 1138 72 VMMC Medical VMMC Medical 2014 Report Error. Download Construction Project Worksheet, Assumption Worksheet and Cost Element Worksheet. 5 \times 10^{11} \textrm{ m}$has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun). 2) should be true with respect to any chosen origin. 1 kg (Total 1 mark) 8 Page 3 of 20. 1st PUC Physics Question Bank Chapter 8. Then = 400, where R is the Earth radius of 6378 kilometers. Take the equatorial radius as 6400 km. Note: The above formula of acceleration due to gravitational force is only valid for a depth below the earth surface not for a height or above the surface. 00 cm, centered at the origin, has a uniform volume charge density 4. ∴ Linear velocity of point on the rim =. A uniformly solid sphere of mass M and radius R is fixed a distance h above a. The traveler would pop up on the opposite side of the Earth after a little more than 42 minutes. Assuming that the Earth is a sphere of uniform mass density, find the percentage decrease in the weight of a body when taken to the end of a tunnel 32 k m below the surface of Earth. Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration. Assuming the earth to be a uniform sphere of radius 6400km and density 5. But since the earth is flatter (not a perfect sphere) there are some deviations in the direction of gravity because of that. A sphere of radius R has a uniform distribution of electric charge in its volume. Help me answer: Distance of the centre of mass of a solid uniform cone from its vertex is z0. Compute the density of the earth, using the power of 10 notations and the correct number of significant figures. (Hint: You may neglect the curvature of Earth. x^2 + y^2 + z^2 - 8x + 2y +6z + 1 = 0. By combining these two cases, and introducing the additional assumption that the elastic moduli are functions of the distance from the center, we should have. It rotates on its own axis once per day. 0 x 10 24 kg, find the gravitational field strength g at a point: (a) on the surface, Calculate the acceleration due to gravity on a planet which has the same density but twice the radius of Earth. "Nucleus Research: Unit4 shows its X-Factor with ERPx". Assuming the earth to be sphere of uniform density, the acceleration due to gravity- A At a point outside the earth is inversely proportional to the square of its distance from the centre B At a point outside the earth is inversely proportional to its distance from the centre C At a point inside is zero D. Acceleration that the Earth imparts to objects on or near its surface Earth's gravity measured by NASA GRACE mission, showing deviations from the theoretical gravity of an idealized, smooth Earth, the so-called Earth ellipsoid. We can still try to mimic the uniform plane wave source on a sphere by assuming external electrical currents flowing making a natural assumption that current density is a solenoidal field, one can. The cutting board loosely tented with aluminum housing also to import custom sound effects. 0° slope and a uniform surface. A vertical U-tube of uniform cross-section contains water up to a height of 20 cm. Instead of separating the core as a sphere and the mantle as a hollow shell, the same answer can be calculated if you treat the earth as the sum of two spheres. The Sun is considered to produces a constant amount of energy. 6 in the textbook), make the unrealistic assumption that the density of the earth is uniform. (If this seems too ﬂippant, consider the problem of an object from outter space striking the earth - it would be reasonable a priori to assume a uniform distribution for the strike point). Potential outside uniform spheroid. Assuming that the Earth is a uniform sphere (which is not correct, but is close enough to get an order-of-magnitude estimate) with M = 5. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the earth if it weighed 250 N on the surface? Answer: Weight of body = 250 N Now, at a depth d, let acceleration due to gravity be g d, g d = g(1-1/2) = g/2 New weight = mg d = mg/2 = $$\frac{250 \mathrm{N}}{2}$$ = 125 N. Assuming earth to be a uniform sphere finds an expression for density of earth in terms of g and G? gravitation; class-11; Share It On 1 Answer. At the surface of the Sun the intensity of the solar radiation is about 6. PDF Earth Coverage by Satellites in Circular Orbit. 16 Lateral Heterogeneity in the Mantle. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface - Physics - Gravitation. Due to the change in weather conditions, the surface temperature of the earth suddenly changes to 268 K. Question: Assuming Earth is a sphere with uniform density, what is the (initial speed) of a m = 1. ux at the surface of the earth) is about 0. The time taken to fall along a straight line between any two points is no longer independent of. 93-kg object at the surface of the white dwarf. (a) assuming the sphere's charge is uniformly distributed, what is the charge density inside of it?. Assuming that the Earth is a uniform sphere of radius 6. With this approximation, the gravitational force on an. What is the value of acceleration due to gravity at a point 100 km below the earths surface? (Given, R = 6380 × 1 0 3 m ). Assuming the earth to be a sphere of uniform density, the acceleration due to gravity inside the earth at a distance of r from the centre is proportional to 1753 60 AIEEE AIEEE 2012 Gravitation Report Error. Thus the weight of the body is:- or. Let$\vec u\in\mathbb{R}^d$be a random unit vector, with uniform distribution on the surface of the unit sphere. Two capacitors A and B with capacities 3 mF and 2 mF are charged to a potential difference of 100 V and 180 V respectively. The moment of inertia of a sphere with uniform density about an axis through its center is 2/5 MR = 0. What happens to the level of water in the glass as a result of the ice melting? c. For solid objects of uniform density, I think it's simply a question of the mass. The Earth gives off IR radiation in all directions. 97219 x 10^24 kg and radius = 6,371. A skier starts from rest and slides down from a high hill and then, without losing energy, up a smaller hill. The gravitational potential energy of a sphere of uniform density and radius ris U sphere 3 5 GM2 sphere r 5 (a)507 m/s (b)643 m/s (c)894 m/s (d)1021 m/s (e)771 m/s. The electric potential within the conductor will be: V = 1 4πϵ0 q R V = 1 4 π ϵ 0 q R. We need to find the gravitational field at the center of the flat surface of the hemisphere of uniform density ρ and radius R. The sphere has a uniform volume charge density (. We solve the escape velocity equation for R: vescape = (2GM/R)0. I hadn't noticed the ice reference on a quick re-read, but it does sound sort of odd, given that it's currently about 80 and sunny in Schenectady. The plates of the capacitors are connected as shown in the figure. If density is doubled then mass will also double, radius being constant. Region 1: Consider the first case where ra≤. case, the physical charge is inside the sphere, while the image charge lies outside. After converting kilograms and cubic kilometers to grams and cubic centimeters, Earth density is about 5. a whole-space background with conductivity σ 0. If volume is constant then density is proportional to mass. If your grid coarseness is about the same as the size of a triangle, the smallest of the three cells will contain a. The Rotational Kinetic Energy of the Earth. Body of mass m is located at depth, Where, = Radius of the Earth. At a depth d below the surface of the earth. Concentric with this sphere is an uncharged, conducting hollow sphere whose inner and outer radii. Gravity prevents the atmosphere from escaping. Recent simulations suggest that a dark matter area, also containing some visible stars, may extend up to a diameter of almost 2 million light-years. Furthermore, inside a uniform sphere the gravity increases linearly with the distance from the center; the increase due to the additional mass is 1. 5: Newton's Law of Universal. 80\,{\text{m/s}}^{2}) [/latex], we see that the astronauts in the International Space Station still have 88% of their weight. Thus, the net electric flux through the area element is. (b) Suppose that this energy could be harnessed for our use. The atmosphere is a fluid, and the pressure in this fluid is a function of the altitude. Thus, ˙T4(R =d )2 = 0:1, where R = 7:0 105 km is the radius of the sun and d = 1:5 108 km is the distance between the earth and the sun. Originating in theoretical physics, the metaphor refers to physicists' tendency to reduce a problem to the simplest form imaginable in order to make calculations more feasible, even if the simplification hinders the model's application to reality. A solid, insulating sphere of radius ahas a uniform charge density ˆand a total charge Q. A satellite is revolving is a circular path close to a planet of density P. Then the surface is deﬁned by gz(ϕ)=C+ 1 2 Ω2r2 refcos 2 ϕ,. The charge distribution divides space into two regions, 3. This applies only to spherical shells of uniform density, which happily is a condition well met by stars and many other objects. A) Inside a uniform sphere of density p there is a spherical cavity whose centre is at a distance $\ell$from the centre of the sphere. The gravitational force is conservative. Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). Then, for the uniform distribution, the theoretical probability for any number i in the range r out will be p out (i) = 1/r out. The corresponding gravitational field G =×6. For a uniform sphere of mass M and radius R and an axis through its center, 2 5 I MR. Solution: Chapter 19 Electric Charges, Forces, and Fields Q. This equation can be solved iteratively: Try R=aa for various values of a until n/a is determined to four significant. Two very interesting special cases occur. 00 km and density (mass per unit volume) 2. Initially the satellite is traveling to the right with speed v and rotating clockwise with angular speed wi. assuming that the earth is a solid sphere with uniform mass density, the force of attraction exerted on a body inside the earth varies as the distance of the body from the earth's centre, i. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh 800 km below the surface of the earth, if it weighed . Radius of the earth is 6380 km, and g = 9. Find the electric eld at distance rfrom the axis where rPart 2: Solar Energy Reaching The Earth's Surface. b) the induced surface-charge density; The induced surface-charge density is given by ˙= 0 @ @n S where in this case the unit normal is pointing into. Weight of a body of mass m at the Earth's surface, W = mg = 250 N. PDF A spherical planet of uniform density 1 has radius R. 0 km, and assuming you could walk around the equator on land, you would exert a torque on the earth slowing down its rotation if you walked in the right direction. Assuming also that the Earth is spherical, (it is actual an oblate spheroid, that is to say, wider at the equator than at the poles). (a) Calculate its rotational kinetic energy. Suppose a shaft Assuming the bullet goes beyond the earth's surface, calculate how far it will go before it stops. Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration in mine 100 km below the. The whole point of the question was about a uniform distribution of stars over the surface of a sphere. Determining the Mass and Density of the Earth:. (Hint: Replace the cavity with spheres of equal. 37 x 10 6 /7905 = 5063 seconds = 1. Figure 1 [3] c)The displacement of an object is determined by the following function: s= 2t3 9t2 + 12t+ 4. Assuming earth to be a sphere of a uniform density, what is val. Table of Contents ()Chapter 1: Introduction ()Chapter 2: Flow Past a Sphere I: Dimensional Analysis, Reynolds Numbers, and Froude Numbers ()Chapter 3: Flow Past a Sphere II: Stoke's Law, the Bernoulli Equation, Turbulence, Boundary Layers, Flow Seperation (PDF - 1. When the average density had been determined as approximately 5:510 3kgm it became clear, from the lower density of surface. Equating these two calculations of speed we get: T = 2π x 6. Show that the period of the oscillation is about 84 min. Estimate the rotational kinetic energy of Earth by treating it as a solid sphere with uniform density. Assuming the earth to be a sphere of uniform mass density, how much would it weigh half way down to the centre of. Then the total mass of air in the. Homework Statement The Earth can be approximated as a sphere of uniform density, rotating on its axis once a day. Calculate the volume of a sphere (such as the earth, a marble) using many different dimensions for both input and output. The earth is a sphere of radius RE. Since the mass density of the earth is assumed to be constant, we have. how much would it weigh half way down to the centre of the earth?. If radius of earth is 6400km, what will be the weight of 1 quintal body if taken to the height of 1600 km above the sea level? 28. Assume the earth is radially symmetric and composed of 2 layers - a uniform outer mantle and a uniform core. where p (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter. In other words, each area of the sphere would tend to have the same amount of stars as each other area. No such object is found to be spinning with a period shorter than found by this analysis. 156 10 s L I mr S ZZ §· u u u ¨¸. 5 × 1011 m from the Sun and takes a year to make one complete orbit. Rough Estimates Of The Solar Energy Available At The Earth's Surface. However, the condition for an equilibrium state is that the potential be constant over the surface. The volume of the Earth is considered to be 1. • b) Calculate the surface free-fall acceleration. A uniformly solid sphere of mass M and radius R is . Body of mass m is located at depth, d = 1/2R_e. 23, Young: The electric field at a distance of 0. Assuming that the Earth is a sphere of uniform mass class 11 physics CBSE. Now, we can find the speed of rotation by looking at distance travelled (circumference of Earth) and time for one revolution - the period T. The surface charge density of the plane, assuming it to be an infinite sheet, is :- (1) 8. Body of mass m is located at depth, d=1/2 R e. (i) and (ii), g=G/ (R^ (2))xx (4pi)/ (3)R^ (3)p= (4piGRp)/ (3) or p= (3g)/ (4piGR). 80 kg starts from rest and rolls without slipping down a 30. How many Earth radii must this same object be from the center of Earth if it is to weigh the same by a plumb line. This relationship is easy to verify for a sphere of uniform density, but Equation 2. 9 × 10−3 N on a point charge of +1. Project your point to each of the xy-plane, the xz-plane, and the yz-plane and snap it to a grid. Graduate certificate in regard the illustration portfolio. How does g vary inside this model of the earth as a function of the distance r from c. Therefore, a projection that preserves equal areas is necessary. 1 Buried sphere Gravity measurements are made on a surface profile across a buried sphere. The earth can be approximated as a sphere of uniform density, rotating on its axis once a day. (a) Calculate its - askIITians. 3 The Poisson and Laplace Equations 2. Simplify the expression by assuming typical atmospheric conditions and common units. Imagine the Earth to be a homogeneous, self-gravitating sphere of radius R and a uniform density ρ a. 15 One-dimensional Earth's Structure 1. Assuming that the Earth is a sphere of uniform density (which it is not, but is close enough. More than 2,000 years ago, the ancient Greeks figured out that Earth was round rather than flat. Permit me to replace the assumption of a water-filled pipe with another assumption that will produce the desired result: Let's assume that the earth is a uniform sphere of some hypothetical liquid having an overall radius of 6,400 km. Assuming earth to be a sphere of uniform density. (b) Determine the gravitational field at a distance r from the center of the Earth where r < R E, assuming the Earth's mass density is uniform. If the mass of the sphere is 0:5kg, calculate the tension, T, in the thread. Answer the following questions about a sphere of radius R which has a non-uniform volume mass density, which depends on the distance r from the sphere's center as ρ = ρ0 (r/R)2, where ρ0 is a constant density. This is an average of all of the material on the planet. per cubic foot (3 grams per cubic centimeter), Earth's center has a density of about 811 lbs. Acceleration due to gravity increases with the increase of depth [assuming earth to be a sphere of uniform density] Acceleration due to gravity decreases with the increase of latitude: Acceleration due to gravity is independent of the mass of the Earth ^ ^. 1) where A is the surface area of the sphere. 0 cm in radius carries a uniform volume charge density.$\begingroup\$ To clarify your question, this is an estimate for the mass of the earths atmosphere based on the assumption the earth's atmosphere is dry. The Earth can be considered to be a sphere made of a large number of concentric uniform spherical shells. We know this can't be accurate because we know that our planet isn't a uniform sphere. Thus the equation of hydrostatic equilibrium will be dP dr = − GMr r2 ρ (22) Since the density is not constant, we must treat Mr with care. ) If the surface of the sphere has a uniform mass-density σ, the mass contained on the infinitesimal surface is given by dM. The first part of this expression is our old friend, the original equation for gravitational potential energy. If the Earth is a uniform sphere with a density ρ, the gravitational acceleration will be. If we assume that the Earth has a constant density, we can work out the mass from the equation mass= density x volume. The assumption of a uniform density contrast thus might yield large errors in the estimated Moho depths. • c) Calculate the gravitational potential energy associated with a 4. Solution: $$\begin{array}{l}g’ = g (1-\frac{d}{R})\end{array}$$ Therefore,. Sensitive meters that measure the local free-fall acceleration. Here G is the gravitational constant, M is the mass of the uniform sphere, m is the mass of the object in the sphere's field, and r is the distance between the mass and the center of the sphere:. Best model obtained by forward modelling of. Solution: Here, Depth = 100km = 100× 103m Radius of the earth R = 6380× 103m Acceleration due to gravity below the earth's surface is given by g′ = g(1− Rd ) = 9. answered Aug 28, 2018 by SunilJakhar (89. (a) Assuming the earth to be a sphere of uniform density, calculate the value of acceleration due to gravity at a point (i) 1600 km above the earth, (ii) 1600 km below the earth, (b) Also find the rate of variation of acceleration due to gravity above and below the earth's surface. 50 m plumb line at ••25 A solid uniform sphere has a mass of 1. 0 N rolls up an incline at an angle of 30. Scientists have found evidence that Mars may once have had an. For the web resources, the density is treated as uniform throughout the sphere. ∫ V ∇ ⋅ e d V = ∫ V ρ ε 0 d V = Q. find the value of g on the surface where G=6. A model for the density δ of the earth's atmosphere near its surface is. (A) the solid shell of inorganic materials on the surface of the Earth (B) the thin shell of organic matter on the surface of earth comprising of all the living things (C) the sphere which occupies the maximum volume of all the spheres (D) all of the above. A body weighed 250 N on the surface assuming the earth to be a sphere of uniform mass density. 58 An infinitely long insulating cylinder of radius R has. (b) Suppose that this energy could be harnessed for our use . A 3D Poisson distribution consists in drawing random positions in space independently for the 3 axis and independently. 250 kg is placed next to the spring, sitting on a A skier starts from rest at the top of a frictionless incline of heigh. sphere of which the density is a given function of the distance from the center, the elastic moduli being assumed to be uniform in value throughout the body. As a (very) rough approximation, one may assume that in a given line of sight, there is locally a uniform, independent and identically distributed density of stars. We will use this integral later in the section on interpretation to determine the excess Assuming a spherical earth with a mean value of observed g, the experimental measurements of. Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m, when it is taken from the earth's surface to a height 3R above its surface, is. Of particular interest for planetary structure is the how the moment of inertia changes as one goes from a solid sphere with a uniform density to a sphere with higher density material at is core. If the sphere is uniform, then the density ρ is constant throughout. 5 g cm^(-3) , find the value of g on its surface. The bulk density of the earth is about $$\rho_{bulk}=5550$$ kg/m$$^3$$ and the average moment of inertia is $$0. 16 gravitation 2 See answers Advertisement abhi178 We know, accⁿ due to gravity at depth h from earth's surface , g = go ( 1 - h/r) here, h = r/2 g = go (1 - r/2r) = go/2. In this way, the sphere is kept at a constant location with respect to the external frame. Find an expression for Earth's Density in terms of g and G, assuming Earth is a uniform sphere. Each region is tatVolume charge density of earth. Assuming that the earth is a sphere of uniform mass density, what is the percentage decrease in the weight of a body when taken to the end of a tunnel 32 km below the surface of the earth? Radius of earth = 6400 km. The Milky Way is a barred spiral galaxy with an estimated visible diameter of 100,000–200,000 light-years. S center 04 IS d f hag I voar — 7th 3 t; q where co ovio-tt01z1 cl per (v. A rocket is fired vertically with a speed of 5 km s-1 from the earth's surface. where G is the gravitational constant, M is the mass of the sphere, and R is its radius. The method explores a model, in which the planet earth is assumed to be composed of crust and an inner sphere representing the mantle and the core together (i. Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine 100 km. We have made use of the fact that the attractive force at radial distance r inside a uniform density sphere with radius R involves only the mass lying below. part a what is the moment of inertia of the earth? use the uniform-sphere approximation described in the introduction. However, gravity varies because the density varies within the Earth and the Earth is not a perfect sphere. you assume Earth has uniform density. A positively charged solid sphere of radius 100 mm has a uniform volume charge density of 250 nC/m^2. Assuming earth to be a perfect sphere of radius R and uniform density p, mass of Earth=volume of Earth xx denisity,. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the . Equation 1: where is the mass of the earth and is the radius of the earth. 2 The earth's atmosphere can be modeled as a uniform layer of air of thickness 20 km and average density 0. 1 Outside the sphere these currents would add a dipole magnetic ﬁeld to the uniform external magnetic ﬁeld. per cubic foot (13 grams per cubic. 2 m/s 2 Explanation: Consider the spherical shell idea: only the mass of the earth in the inner shell con- tributes to g ; the mass. You can put this solution on YOUR website! Let x be the difference of latitudes (the larger minus the smaller), in degrees. Likewise, assume a uniform distribution on r base for the underlying source of random numbers, i. Consider a self-gravitating spheroid of mass , mean radius , and ellipticity , such as a star or a planet. Determine the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh as much as at present. Find the potential at the center of the sphere, if its radius is R and its dielectric constant is K. Replacing M E M E with only the mass within r, M = ρ × (volume of a sphere) M = ρ × (volume of a sphere), and R E R E with r, Equation 13. There is a spherical cavity cut out of its center of radius 2. (a) Find an equation of the sphere that passes through the point (6, -2, 3) and has center (-1, 2, 1. An insulating solid sphere of radius a has a uniform volume charge density and carries a total positive charge Q. The moment of inertia of a solid sphere of radius r and mass m about its center is 2 5 mr 2. the origin of coordinate system coincides with the center of the sphere. (a) What is the new charge density on the outsize of the sphere?(b)Calculate the strength of the electric eld just outside the. Assuming that the Earth is a solid sphere of uniform density, with mass M_Earth = 5. mg = G M m R - d 2 M = 43π R - d3 ρ⇒ g = GM R - d2 g = 43 Gπ R - d ρ On the surface of the earth g = G mR V g = 43πGRρ ∴ g = g2∴ The body weighed 250 N on the surface of the earth would weigh 12 × 250 = 125 N, halfway down towards the centre of the earth. The moment of inertia for this can be approximated by assuming it is a uniform sphere, but it isn't. Show that the electric field in the cavity is uniform and is given by Ey = 0, Ex = (b/3(0. CBSE NCERT Notes Class 11 Physics Gravitation. Question Physics Class 11 Assuming Earth to be a uniform sphere, find an expression for mean density of Earth in terms of\ ( g$$ and \ (G\). In the case of the uniform spherical shell, equation 8. Calculate the volume of a cube or brick (such as a dice or an iPhone) using plenty of different input measurements and get. Show that the value of k for which du/d1=0 satisfies the equation 51(1 - exp(-a/ñ)) = a. Calculate C/(MR2) for the equal thickness model in c above. Assuming a uniform, solid disk, its moment of inertia about a perpendicular axis through its center is , so gives. Calculate the volume charge density of the atmosphere assuming it to be uniform between 250 and 400 m. Assuming earth to be sphere of uniform density what is the value of acceleration due to gravity at a point 100km below the earth surface (Given R=6380×103m). The entire spherical surface (roughly speaking) is at a P ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =. Satellite observations show that the earth's moment of inertia is 0. If you tried to travel towards the edge, you'd be. Assuming the earth to be a sphere of uniform mass density, calculate the value of acceleration due to gravity in a mine 100 km deep. How far from the earth does the rocket go before returning. Show that the total energy stored in its electric field is U = kQ 2=2R. I thought we were to assume a spherical cow of. Since the Earth is a sphere, the formula 4/3πr 3 is used to find the volume. in mine 100 km below the earth surface ms-2. the surface assuming the earth to be a sphere of uniform mass density. Assuming that the Earth has a uniform mass density, that there is no friction or air resistance in the tunnel, and that the subway car is moving solely under the influence of gravity with no other forces, the time required for a one-way Manhattan-to-Sydney trip in the subway car is an amazing 43 minutes. where p is the density of the planet, assumed to be homoge- neous. And that tells us that the force of gravity. easily quantified by regarding θ as a uniform random variable on [0, 2π] and deriving the corresponding density of the random latitude ψ. Weight of a body of mass m at the Earth’s surface, W = mg = 250 N. In spherical coordinates, a small surface area element on the sphere is given by (Figure 4. Q: Assuming the earth to be sphere of uniform mass density, how much would a body weight half way down the center of the earth, If it weighed 100 N on the surface? (a) 75 N (b) 50 N (c) 60 N (d) 62…. Which of the curves shown below correctly illustrates the dependence of the magnitude of the electric field of the sphere as a function of the distance r from its centre? ents real and 0 R R 1 hen Electric field magnitude Electric field magnitude (IV) 0 R 1 0 R (a) I (b) II (c) III (d) IV. 5 times the decrease due to the larger. The theoretical method is based on gravitational potential from which gravity field can be calculated at the surface of the earth. Anchordoqui February 24, 2015. 38×106 m , and the period of rotation for the earth is 24. A straight tunnel is cut through P, but not necessary though the plant's center. PDF PHYSICS 111 HOMEWORK SOLUTION #13. A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rad/s. Find the separation between the point charge and the center of the sphere. Determine the electric field (a) 20 mm, (b) 90 mm, and (c) 110 mm from the center of the sphere. U = − 3 G M 2 5 R {\displaystyle U=- {\frac {3GM^ {2}} {5R}}} where G is the gravitational constant, M is the mass of the sphere, and R is its radius. The volume of a sphere is V = 4/3 π r E 3. Assuming that the Earth is a sphere of uniform density (which it is not . 4k points) selected Aug 28, 2018 by faiz. I do not hold any copyright or anything. 0 cm from the sphere's center has magnitude 3 9 kN/C. This leads to the interpretation of the Hubble law as implying an expanding universe. As  g= (GM)/ (R^ (2)`, therefore.