Lecture05-ASTC25
... rL rL ( 3 )1/ 3 a Hill applied his equations to the Sun-Earth-Moon problem, showing that the Moon’s Jacobi constant C=3.0012 is larger than CL=3.0009 (value of effective potential at the L-point), which means that its Zero Velocity Surface lies inside its Hill sphere and no escape from the Eart ...
... rL rL ( 3 )1/ 3 a Hill applied his equations to the Sun-Earth-Moon problem, showing that the Moon’s Jacobi constant C=3.0012 is larger than CL=3.0009 (value of effective potential at the L-point), which means that its Zero Velocity Surface lies inside its Hill sphere and no escape from the Eart ...
Newton`s Laws
... 7. Two horizontal forces, 225 N and 165 N, are exerted in the same direction on a crate. A. Find the total horizontal force on the crate. ...
... 7. Two horizontal forces, 225 N and 165 N, are exerted in the same direction on a crate. A. Find the total horizontal force on the crate. ...
Gravitational Potential Energy
... Definition: GPE is the potential energy stored in the gravitational fields of interacting bodies. GPE depends on height from a “zero level” (normally the ground). (If something is above the ground, it has GPE) Units: Joules (J) ...
... Definition: GPE is the potential energy stored in the gravitational fields of interacting bodies. GPE depends on height from a “zero level” (normally the ground). (If something is above the ground, it has GPE) Units: Joules (J) ...
energy - RHIG - Wayne State University
... • Where the last approx is realized for m1 << m2. • Kepler’s statement is correct only if the mass m1 of a planet can be neglected with respect to the mass m2 of the sun. • Correction needed for Jupiter given that it is 1/1000 of the mass of the Sun. ...
... • Where the last approx is realized for m1 << m2. • Kepler’s statement is correct only if the mass m1 of a planet can be neglected with respect to the mass m2 of the sun. • Correction needed for Jupiter given that it is 1/1000 of the mass of the Sun. ...
Page 1 Problem An electron is released from rest in a uniform
... and negative charges are of the same magnitude and placed symmetrically about the point where we are to find the field, the F components of both electric fields and are of the same magnitude but opposite direction. However, the G components are of the two electric fields are of the same ma ...
... and negative charges are of the same magnitude and placed symmetrically about the point where we are to find the field, the F components of both electric fields and are of the same magnitude but opposite direction. However, the G components are of the two electric fields are of the same ma ...
Chapter 22 Problem 66 † Given V (x)=3x - 2x 2
... a) Find the locations where the potential is zero. Factor the potential function as much as possible. V (x) = x(3 − 2x − x2 ) V (x) = x(3 + x)(1 − x) The only places where the potential is equal to zero is when one of its factors is equal to zero. Therefore, Either x = 0, 3 + x = 0, or 1 − x = 0 Fro ...
... a) Find the locations where the potential is zero. Factor the potential function as much as possible. V (x) = x(3 − 2x − x2 ) V (x) = x(3 + x)(1 − x) The only places where the potential is equal to zero is when one of its factors is equal to zero. Therefore, Either x = 0, 3 + x = 0, or 1 − x = 0 Fro ...
Orbital Motion
... Orbital Motion Circular Orbits Kepler’s Laws Energy in Elliptical Orbits Homework ...
... Orbital Motion Circular Orbits Kepler’s Laws Energy in Elliptical Orbits Homework ...
Solution
... W13D1-6 V-Groove Frictional Torque and Fixed Axis Rotation: Solution A cylinder of mass m and radius R is rotated in a V-groove with constant angular velocity ω 0 . The coefficient of friction between the cylinder and the surface is µ . What external torque must be applied to the cylinder to keep it ...
... W13D1-6 V-Groove Frictional Torque and Fixed Axis Rotation: Solution A cylinder of mass m and radius R is rotated in a V-groove with constant angular velocity ω 0 . The coefficient of friction between the cylinder and the surface is µ . What external torque must be applied to the cylinder to keep it ...
Homework 7
... momentum of the electron about the center of the circe is L = 4.00 · 10−25 Js. Determine (a) the radius of the circular path and (b) the speed of the electron. Angular momentum is defined as L = r × p = mr × v ...
... momentum of the electron about the center of the circe is L = 4.00 · 10−25 Js. Determine (a) the radius of the circular path and (b) the speed of the electron. Angular momentum is defined as L = r × p = mr × v ...