A satellite is in an elliptical orbit around the earth with the center of the earth at one focus. The height of the satellite above the earth varies between 140 mi and 440 mi. Assume the earth is a sphere with radius 3960 mi. Find an equation for the path of the satellite with the origin at the center of the earth.
136592
This is an application of the equation of an ellipse.  The standard equation of an ellipse is:\n(x/a)^2 + (y/b)^2 = 1 where a and b are the distances from the center in the x and y direction.\n\nIf the earth is a sphere of radius 3960 mi, and the satellite is above the earth varying from 140 mi to 440 mi, then the satellite is going to be at most 3960+440 miles away, and at least 3960+140 miles away... this gives the 'a' and 'b' values of 4400 and 4100 respectively (actually, these can be interchanged since the problem didn't give any other details).  This leaves the final equation as:\n\n(x/4400)^2 + (y/4100)^2 = 1\n\nFrom here you can use algebra to solve for x and y if you have to, but be careful and don't lose any solutions if you do!\n--Ron
22884
The first consideration is how high do you want it to orbit --- higher means lower satellite velocity. Also, do you want a circular orbit or an elliptical orbit? How about the orbit's inclination from the Equator?  Once decided, the satellite's mass is the major parameter of concern.
217475
Equatorial diameter 6,804.9 km (4228.4 Miles)\n(0.533 Earths)\nPolar diameter 6,754.8 km (4197.2 Miles)\n(0.531 Earths)\nOblateness 0.007 36\nSurface area 1.448×108 km2 (55.907 Million Square Miles)\n(0.284 Earths)\n\n\n"Mars-5-inch-thing" Sure... but in which category of questions are we?
609497
An object orbiting another in space can be considered to be in freefall around the center object.  You have probably heard someone explaining circular motion like that of an astronaut or space station or the moon as being in freefall towards the earth.  It's a balancing act between the gravity between the two objects (causing the free fall) and the inertia of the planets, moons, satelites, etc that keep them moving in a direction perpinduclar to the inward falling direction.  \n\nEventualy the orbiting object will have to fall toward the center object or away from it, debending on whether the inertia or the gravity is stronger.  \n\nOur moon is moving away from us.\n\nAll the stars in the galaxy are moving around the center of the galaxy and all the galaxies are moving around the center of the universe but we will never fall in because not only is the universe still expanding, the rate at which all the galaxies are moving away from the center of the universe is getting faster.
7697
The distance between the earth and the sun is not constant.  The earth's\norbit around the sun is nearly circular, but not perfectly so.  The earth\nis actually closest to the Sun in the northern hemisphere winter, but the average distance between the earth and the sun is 93,000,000 miles.
71432
If an observer is on the surface of earth, he cannot see objects below the horizon, which is the tangent plane to the earth at his position. \n\nBut if the observer is at an elevated position the new horizon is called visible horizon or offing.\n\n\nThe angle between directions of ordinary horizon and visible horizon is called Dip of horizon.\n\n\nThe dip is given by the formula D = square root of (2h/a) radian, where ‘h’ is the height from the surface of earth and ‘a’ is the radius of earth.\n\n\nThe radius of the circular area is given by the formula x= a * sin D since D is small we can take as "aD".
463256
Simply? Gravity is the force responsible for accelerating you towards the earth's center. Thus, when considering the forces exerted during launch, there is a constant m(9.8m/s^2) force directed straight down to the center of the earth. Dividing all forces out by mass, then no matter what acceleration into the heavens the rocket is given, it will face a near-constant 9.8m/s^2 (~ closer to 8.6 at altitude (not that much different from on earth)). What this means pertains to energy.\n\nThe rocket must get form one level to another, so the energy well decreases, in the sense of its kinetic energy must at least be (GMm/r1 - GMm/r2), where r1 is on the ground and r2 is at altitude. If you didn't factor in gravity, no force would be acting downwards on the rocket. Technically, this means any acceleration could place it at any altitude; i.e., flicking it would put it in orbit.
262841
First, the speed of the rotation of the earth relative to where you are on the earth.  At the equator, the circumference of the Earth is roughly 24000 miles.  This is convenient, because the Earth rotates about its axis once every 24 hours.  Therefore an hour = 1000 miles of equatorial rotation.  Understand the fastest moving part of the Earth is the equator, since an object on the equator travels 24000 around the earth's Axis every day.  If you are closer to the poles, you are on a "thinner" slice of the earth, where the circumferene may only be 20000miles.  In that case you are moving at 833 miles per hour.  The fact that the Earth moves faster at the Equator than at the poles is responsible for the Coriolus Effect which is really important to people who study meteorology, or who wonder why their toilets get all swirly when flushed.  Why we don't "feel" the rotation is, simply put, because we are rotating exactly the same way as the earth.  In physics, this is called a "frame of reference".  For most intents and purposes, the earth is at "rest" relative to us.  As a human, you make up part of the mass of the earth, and you spin with it in a constant rotation.  It's only when we look to things that are not directly connected to the earth that we really see the effects of it's rotation (ie Tides of water, moments of clouds and weather, etc. . .)  Hope that helps.
469825
Here is the math for finding our solar system's barycenter. (It's under the heading "Barycenter").\nhttp://en.wikipedia.org/wiki/Center_of_mass
416131
To get the Centripetal force(the force acting inwards to the center of the radius in this case the sun)\n\nwe use the formular angular force=mass*Acceleration\n=mass*(angular velocity^2/radius)\nsince you have already provided the nessesary requirements lets get started>\n\n=5.9742*10^24*(29.8/93,000,000 Miles)\nsince the sun is93,000,000 Miles away from the earth we use the distance from the earth to the sun as the angular radius.\n from mile to kilometers we convert to get\n93 000 000 mile = 149 668 992 000 meters \n\nnow centripetal force=5.9742*10^24*(29.8/149 668 992 000 )\n\n=1.991060379427156160709627816562e-10kg\nthat is the force that acts on the earth inwards
48428
What is the speed of the Earth's rotation?\n\n--------------------------------------------------------------------------------\n\n\nAt the equator, the circumference of the Earth is 40,070 kilometers, and the day is 24 hours long so the speed is 1670 kilometers/hour ( 1070 miles/hr). This decreases by the cosine of your latitude so that at a latitude of 45 degrees, cos(45) = .707 and the speed is .707 x 1670 = 1180 kilometers/hr. You can use this formula to find the speed of rotation at any latitude.
43713
>>>> what is the altitude of a right triangle?\n\nI have several here, and I am at 12 metres above sea level.\nSeriously, you haven’t provided enough information. You need at minimum the length of 2 sides or a side and an angle as well as the orientation to be able to calculate height (altitude).\n\n>>>>>>if the equation is d=2r what is the d and the r what exectly \n>>>>>>>am I finding with this equation \n\nd = diameter. r= radius. You are finding either how far it is from one side of a circle to another (d), or how far it is form the edge of a circle to the centre (r).
470107
The earth has a center of mass and you also have a center of mass (although a much smaller mass).  Two free bodies (you and the earth) attract each other as though their masses were concentrated at their centers of mass.  As you climb a mountain or fly in a plane you weigh slightly less than at sea level because your center of mass is further away from the center of the earth.  No matter where you are on the globe (earth) you are attracted towards its center.
366442
(a) The angular speed of the earth is the circumference of the orbit divided by the time taken to complete one orbit.\n\nAngular speed = (2*pi rad) / 3.16x10^7s = 1.987x10^-7 rad/s. \n\n(b) The tangential speed is the same as the angular speed, but in different units (probably m/s).\n\n1 rev/yr * (1/3.16x10^7) yrs * (2*pi*r)m/rev = 29810 m/s\n\n(c) The magnitude of the centripetal acceleration is the radius times the angular speed squared.\n\na = r*w^2 = 1.50x10^11 * (1.987x10^-7)^2 = 5.924x10^-3 m/s^2\n\nThe direction is toward the sun.
91121
depends.....at the equator, from the surface to sea level, it is 6,378.135 km. from the centre to the north or south pole, at sea level, it is 6,356.750 km. remember, both of those are to sea level, and they differ because the world is not a perfect sphere.
404563
Consider a mass m on the surface of \nEarth. The equation governing this mass\ncan be written as\nmg=G*m*M/R^2\nm=mass of the object=kg\ng=acceleration due to gravity=9.81m/sec^2\nM=mass of Earth kg ?\nR=radius of Earth=6.4*10^6m[given]\nG=universal constant of Gravitation\n=[6.6742+-.0010]*10^-11Nm^2kg^-2\nCanceling m we can write\nM=g*R^2/G\nYou have all the values\nSubstitute and get Mass of Earth.
40530
work is done to change the height of an object.  If the object is always the same distance above the center of mass, then there is no change in height in which case no work is done.  In an elliptical orbit the distance between the objects changes so work is done.