I have a problem in solving this task which orbits of the satellite with a mass of 2.5 tonnes, whic

I have a problem in solving this task which orbits of the satellite with a mass of 2.5 tonnes, which encircles the earth in 143min and 20s? The radius of the earth is 6400, the gravitational constant of 6.67 * 10 -11Nm/kg seenly the square mass of the earth is 6 * 10 24 Looking for the height of the satellite ... teja33 Posts: 2 Joined: 18/09/2007 16:09
That we all once wrote, enclose a link that will help you: http://forum.kvarkadabra.net/viewtopic.php?t=1616 first task is different from yours, except that instead of searching radius of gyration seenly of patrol time. Zdravapamet Posts: 2827 Joined: seenly 08/16/2004 18:41
To say. We know mass of the satellite and the satellite circling time. Radius of the Earth to be, gravitational constant and the mass of the Earth. The satellite orbits, away from the surface of the earth, that is, the radius of the track and that we were looking for. The satellite operates gravitational force (the Earth works the same force only in the other direction) according seenly to Newton's law of gravitation: On the other hand, the rotation of mechanics know that the force that the body awarding the necessary acceleration for uniform circular motion, size: When calling a word and equation množiš s, you get: Upoštevši immediately followed by: and sought after height: zdravapamet Posts: 2827 Joined: 16/08/2004 18:41
Greetings! I need help with the following tasks: What is the circling time satellite that orbits the Earth at low altitude? Note: The centrifugal force equal to the force of gravity, seenly the satellite! Thank you for any help. Bojan 0bojan0 Posts: 2 Joined: 01/14/2008 16:03
Greetings! I need help in solving the following tasks: 10 m long railway wagon is moving steadily at a rate of 20 m / s. The bridge, which is 10 m above the line, the stone thrown horizontally at the moment when the front part of the wagon runs under the bridge. What should be the maximum, and what minimum initial speed of the stone, the stone falls on the wagon? Solution: v1 = 13 m / s, v2 = 20 m / s Thanks seenly in advance for the help in the task. Aurora Posts: 2 Joined: 12/03/2007 19:21
Aurora wrote / a: Greetings! I need help in solving the following tasks: 10 m long railway wagon is moving steadily at a rate of 20 m / s. The bridge, which is 10 m above the line, the stone thrown horizontally at the moment when the front part of the wagon runs under the bridge. What should be the maximum, and what minimum initial speed of the stone, seenly the stone falls on the wagon? Solution: v1 = 13 m / s, v2 = 20 m / s Thanks in advance for the help in the task. In the case of horizontal stone throw in a wagon by a steady movement. These are two examples: 1 Stone falls to the back of the wagon. 2 Stone fall to the front of the wagon. In the second case, the range of the stone the same way, by a wagon:. In the first case, the range of the stone for the length of vehicle () shorter than the path made by the wagon: When inserted, we get:. shrink Posts: 7918 Joined: 09/04/2004 17:45
Please help with one task: at the equator of a spherical planet body weighs less than twice the pole. The density of the planet is 3x10na3 kg / m 3 Calculate the time in which the planet made one revolution around the axis. The solution is: 2 h 41.6 min. Anya Posts: 162 Joined: 5/13/2009 15:14
On the pole weighs at the equator is: Note that the resultant of all forces acting seenly on the body (the same as shown by the scale), mass times the radial acceleration of the body: This yields: Masi body other out, we get: considering that, and to give : The term bypass time, if something is not clear, ask Naoki Posts: 66 Joined: 12.11.2008 23:02
Anya wrote / a: ask for help with one task: at the equator of a spherical planet body weighs less than twice the pole. The density of the planet is 3x10na3 kg / m 3 Calculate the time in which the planet made one revolution around the axis. The solution is: 2 h 41.6 min. Have you tried yourself? Tip: On the equator due to the centrifugal effect of gravity centrifugal force counts. This difference (based on data) corresponding to half the weight. Weight over the course write Newton's law of gravitation, mass planet you get to množiš density of the volume of a sphere. From here you get a ratio that is multiplied seenly by a flat circling time of rotation around the axis. PS I see you've already got a complete solution. seenly shrink Posts: 7918 Joined: 09/04/2004 17:45
Back to School playground Jump to: Select a forum ------------------ Questions & Answers seenly About the history of time, the universe and everything about the world of the Moon Fire, water, earth, air Live load faster, higher, stronger What holds the world together? Questions for Einstein X-Files From zero to infinity seenly power of pure reason science in the kitchen Heaven and Earth Kvarkaflanc science in the news, books and movies school playground Doing science at home and abroad Tenders event notifications Interesting Websites New Books All other information and working The cities Cartography & nbs