PHYS 1010 Chapter Notes - Chapter 10: Ellipse, Heavy Object, Orbital Period

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maroonchinchilla495

22 Sep 2016

School

University of Virginia

Department

Physics

Course

PHYS 1010

Professor

Neu Christopher

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Document Summary

Any object that continues in motion by its own inertia. The trajectory of a projectile that accelerates in only the vertical direction while moving at a constant horizontal velocity. When air resistance is small enough to ignore. Ex) a heavy object with great speed. A projectile that falls around the earth rather than into it. Its speed must be great enough to ensure that its falling distance matches the earth"s curvature. The earth drops a vertical distance of 5m for every 8000m in tangent to the surface. Launched to altitudes of 150 km or higher to be above the atmosphere and free of air resistance. Newton said that the moon was a projectile circling earth under the attraction of gravity. Remain in orbit because they are pulled by gravity. Move perpendicular to the force of gravity (at a right angle) Moving in or against the direction of the force would cause it to speed up or slow down.

Related Questions

A conductor bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and constant, uniform, vertical magnetic field B fills the region between the rails (as shown in Fig.)

1) Find the magnitude of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance.

2) What is the direction of the net force? to the left to the right

3) If the bar has mass m, find the distance d that the bar must move along the rails from rest to attain speed v.

4) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s). Let B = 0.8 T , I = 2x10³ A, m = 25 kg, L = 50 cm. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

$B =$

ochrekoala614

MAC, Inc. (the company you work for) has decided to move forwardwith a feasibility
study for a product it wishes to manufacture and sell. This productis a particle launcher that should have the capability, for everyactivation, of launching two particles into the air at twodifferent times and at two different angles of elevation, in anautomated fashion so that they both land at the same distance fromthe launch point at the same time. For example, let P1 and P2 betwo particles, let T1 and T2 be two times, and let theta1 andtheta2 be two angles of elevation. Once activated, the launchershould be able to operate so that particle P1 is launched at timeT1 with an angle of elevation theta 1, and particle P2 islaunched
at time T2 with an angle of elevation theta2.
The launcher must be configured by a technician, prior to eachactivation, so that the
initial speed is set for the launcher, and the distance from thelauncher to where both
particles will land is set. For each activation of the launcher,the initial speed will be the
same for both particles. Once the technician has set the initialspeed and desired distance,
the launcher must calculate the angles of elevation to use and howmuch time to wait in
between particle launches. Then, when the technician activates thelauncher, both
particles will be launched at the appropriate angles of elevationand at the appropriate
times so that they both land at the configured distance from thelauncher at the same time.
The launcher, based on its calculations, will change the angle ofelevation to be ready for
the second particle launch. The launcher, however, must flash awarning for the
technician if the information the technician used to configure thelauncher does not result
in a two-particle launching solution.
Certain assumptions must be made. It is assumed that that launchercan always change
the angle of elevation to get ready for the second particle launchfaster than the time
delay calculated for the second particle launch. It is assumed thatthe units for speed are
meters per second, that the units for distance are meters, and thatgravity is 9.8 meters per
second squared. It is assumed that gravity is the only force actingon the motion of the
particles after launch, and that there is no wind to be considered,so that the particles land
at a point exactly in line with the direction the launcher ispointed.
The questions that must be answered and written up in anengineering report are as
follows:
1. Given the initial speed and desired landing distance for theparticles, find the
mathematical expressions that must be used to find 1, 2, and thetime delay
between particle launches.
2. Determine under which conditions the launcher will not be ableto perform the
operation requested, and will consequently flash a warning to thetechnician.
3. Each particle launched will follow a path of motion that isparabolic. Show that
the vertex of the parabola for each particleâs path occurs at thesame distance from
the launcher.
4. Along the path of each launched particle, there will be amaximum curvature.
Find that maximum curvature for each launched particle.

Use the result (9.73) of Problem 9.26 io do the following: A naval gun shoots a shell at colatitude theta in a direction that is a above the horizontal and due east, with muzzle speed v0. Ignoring the earth's rotation (and air resistance), find how long (l) the shell would be in the air and how far away (R) it would land. If v0 = 500 m/s and alpha = 20degree, what are t and R? A naval gunner sports an enemy ship due east at the range R of part (a) and, forgetting about the Coriolis effect, aims his gun exactly as in part (a). Find by how far north or south, and in which direction, the shell will miss the target, in terms of ohm, v0, alpha, theta, and g. (It will also miss in the east-west direction but this is perhaps less critical.) If the incident occurs at latitude 50degree north (theta = 40degree). What is this distance? What if the latitude is 50degree south? This problem is a serious issue in long-range gunnery: In a battle near the Falkland Islands in World War I, the British navy consistently missed German ships by many tens of yards because they apparently forgot that the Coriolis effect in the southern hemisphere is opposite to that in the north. In Section 9.8, we use a method of successive approximation to find the orbit of an object that is dropped from rest, correct to first order in the earth's angular velocity ohm. Show in the same way that if an object is thrown with initial velocity v0 from a point O on the earth's surface at colatitude theta, then to first order in ohm its orbit is (First solve the equation of motion(9.53) in zeroth) order. that is ignoring ohm entirely. Substitute your zeroth-order solution for and into the right side of equation (9.53) and integrate to give the next approximation. Assume that v0 is small enough that air resistance and that g is a constant throughout the flight) Use the result (9.73) of Problem 9.26 io do the following: A naval gun shoots a shell at colatitude theta in a direction that is a above the horizontal and due east, with muzzle speed v0. Ignoring the earth's rotation (and air resistance), find how long (l) the shell would be in the air and how far away (R) it would land. If v0 = 500 m/s and alpha = 20degree, what are t and R? A naval gunner sports an enemy ship due east at the range R of part (a) and, forgetting about the Coriolis effect, aims his gun exactly as in part (a). Find by how far north or south, and in which direction, the shell will miss the target, in terms of ohm, v0, alpha, theta, and g. (It will also miss in the east-west direction but this is perhaps less critical.) If the incident occurs at latitude 50degree north (theta = 40degree). What is this distance? What if the latitude is 50degree south? This problem is a serious issue in long-range gunnery: In a battle near the Falkland Islands in World War I, the British navy consistently missed German ships by many tens of yards because they apparently forgot that the Coriolis effect in the southern hemisphere is opposite to that in the north.

PHYS 1010 Chapter Notes - Chapter 10: Ellipse, Heavy Object, Orbital Period

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