phys124s11-hw04

phys124s11-hw04 - HW 4 Solutions Escape Velocity The escape...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 HW 4 Solutions Escape Velocity The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass M . The escape velocity is a function of the distance of the object from the center of the planet R , but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question "How fast does my rocket have to go to escape from the surface of the planet?"? a) The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at its escape velocity, what is the total mechanical energy E total of the object at a very large (i.e., infinite) distance from the planet? Follow the usual convention and take the gravitational potential energy to be zero at very large distances. 2 1 0 2 total Mm Em v G R = −= Consider the motion of an object between a point close to the planet and a point very very far from the planet. Indicate whether the following statements are true or false. b) Angular momentum about the center of the planet is conserved – true. c) Total mechanical energy is conserved – true. d) Kinetic energy is conserved – false (there is transformation of kin. energy into pot. energy and vice versa). e) The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the object moves from small R to large R (like a rocket being launched) or from large R to small R (like a comet approaching the earth) - true. What if the object is not moving directly away from or toward the planet but instead is moving at an angle θ from the normal? In this case, it will have a tangential velocity tan sin vv θ = and angular momentum tan Lm vR = . Since angular momentum is conserved, tan L v mR = for any R , so v tan will go to 0 as R goes to infinity. This means that angular momentum can be conserved without adding any kinetic energy at R= . The important aspect for determining the escape velocity will therefore be the conservation of total mechanical energy. f) Find the escape velocity v e for an object of mass m that is initially at a distance R from the center of a planet of mass M . Assume that planet RR , the radius of the planet, and ignore air resistance. Express the escape velocity in terms of R , M , m , and G , the universal gravitational constant. 2 1 0 2 total escape Mm v G R ⎛⎞ =− = ⎜⎟ ⎝⎠ 2 e MG v R = Does it surprise you that the escape velocity does not depend on the mass of the object? Even more surprising is that it does not depend on the direction (as long as the trajectory misses the planet). Any angular momentum given at radius R can be conserved with a tangential velocity that vanishes as R goes to infinity, so the angle at which the object is launched does not have a significant effect on the energy at large R .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 Orbital Speed of a Satellite a) Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/15/2011 for the course APHY 124 taught by Professor Eins during the Spring '11 term at SUNY Albany.

Page1 / 7

phys124s11-hw04 - HW 4 Solutions Escape Velocity The escape...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online