Gravitational force due to a mass The gravitational force on a point mass m due to a point mass M at the origin is a gradient field with potential
a. Find the components of the gravitational force in the x-,y-, and z-directions, where F(x, y, z) = –▿U(x, y z).
b. Show that the gravitational force points in the radial direction (outward from point mass M) and the radial component is
c. Show that the
Want to see the full answer?
Check out a sample textbook solutionChapter 14 Solutions
Calculus: Early Transcendentals (2nd Edition)
Additional Math Textbook Solutions
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (4th Edition)
Glencoe Math Accelerated, Student Edition
University Calculus: Early Transcendentals (4th Edition)
- Heat transfer Fourier’s Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k > 0 is called the conductivity, which has metric units of J/(m-s-K). A temperature function for a region D is given. Find the net outward heat flux ∫∫S F ⋅ n dS = -k∫∫S ∇T ⋅ n dS across the boundary S of D. In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume k = 1. T(x, y, z) = 100 + x + 2y + z;D = {(x, y, z): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1}arrow_forwardHeat transfer Fourier’s Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k > 0 is called the conductivity, which has metric units of J/(m-s-K). A temperature function for a region D is given. Find the net outward heat flux ∫∫S F ⋅ n dS = -k∫∫S ∇T ⋅ n dS across the boundary S of D. In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume k = 1. T(x, y, z) = 100 + e-z;D = {(x, y, z): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1}arrow_forwardHeat transfer Fourier’s Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k > 0 is called the conductivity, which has metric units of J/(m-s-K). A temperature function for a region D is given. Find the net outward heat flux ∫∫S F ⋅ n dS = -k∫∫S ∇T ⋅ n dS across the boundary S of D. In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume k = 1. T(x, y, z) = 100 + x2 + y2 + z2;;D = {(x, y, z): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1}arrow_forward
- Heat transfer Fourier’s Law of heat transfer (or heat conduction) states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k > 0 is called the conductivity, which has metric units of J/(m-s-K). A temperature function for a region D is given. Find the net outward heat flux ∫∫S F ⋅ n dS = -k∫∫S ∇T ⋅ n dS across the boundary S of D. In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume k = 1. T(x, y, z) = 100e-x2 - y2 - z2; D is the sphere of radius a centered at the origin.arrow_forward3-find the gradient of the function v if w=(xy)/z at the point(1,-2,1), find the directional derivative of w in the direction of vector v = 2i + j − 2k, then find the maximum value of the directional derivative.arrow_forward1aarrow_forward
- Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y) = 3e* sin(y), (0, π/3), v = (-10, 24) D f(0, π/3) = (72-10√3) 2√ 676arrow_forwardFind maximhm rate of changearrow_forwardA seasoned parachutist went for a skydiving trip where he performed freefall before deploying the parachute. According to Newton's Second Law of Motion, there are two forcës acting on the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa) as shown in Figure 1. Fa = -cv ITM EUTM FUTM * UTM TM Fg= -mg x(t) UTM UT UTM /IM LTM UTM UTM TUIM UTM F UT GROUND Figure 1: Force acting on body of free-fall where x(t) is the position of the parachutist from the ground at given time, t is the time of fall calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration, v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the position is given by the equations below: EUTM PUT v(t) = mg -et/m – 1) (Eq. 1.1) x(t) = x(0) – Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the critical position to deploy the parachutes is at 762 m from the ground…arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage