Urgent! Please take this challenge! time required for 1 revolution for a satellite of circular orbit?
106990
Period is proportional to R^3/2.\nThis is an answer I posted to bombay2pac2000 under Astronomy & Space 8 hours previously, and it applies here as well.\nKepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)\nwhere R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body.  In orbit, centrifugal acceleration CA balances gravitational acceleration GA.  CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T). So CA = (2pi/T)^2*R. From Newton we know GA = GM/R^2. Put 'em together, \nCA = GA\nR*(2pi)^2/T^2 = GM/R^2 \nT^2 = R^3*(2pi)^2/GM\nT = 2*pi*(R^3/2)/SQRT(GM)
407005
I'd say 17000-19000 miles per hour.\n\nIts at a inclined geo-stationary orbit probably.
810893
Oscar. You have written too much. This 125m tower stuff does not matter compared to r=6.38*10^6. Why is this talk about 1.00 km and angle theta? See what “zipper” above says. Besides I answer your last question << Why is the trajectory of the cannonball not parabolic?>>. Why should it be if both you and Newton demand it to be circular? Assume that V1 of cannonball is the 1st cosmic speed corresponding to a circular track. Let V2 be > than V1, then you receive an oval track. Take V3 > than V2, then the track may become parabolic, so that cannonball would never return back to the cannon to linger out among planets and stars. If you are intrigued calculate a set of velocities and corresponding orbits of the cannonball. If cannon is not available consider a football! I like your inventive wits – to be honest!
864302
I'll assume it is your typical physics problem, which states some highly simplified conditions to demonstrate some physics principal and not a real engineering or natural science problem. We will assume the Earth increases its radius by 30 meters without any change in total mass, and that the Earth is a sphere of uniform density. In that case, the rotation rate would be proportional to the change in the moment of inertia of the Earth, I. For a uniform sphere I=2/5MR^2. You need to calculate the existing I and the new I after the radius change. Since M does not change, and we want only the ratio of the two Is, we can ignore the 2/5M and just compare radius. The radius of the Earth is about 6,370km or 6,370,000 meters. Adding  30 meters makes it 6,370,030 meters and the ratio of 6,730,030^2/6,730,000^2 is about 1.00001. That is about a 0.001% change, and about 0.001% of 24 hours is about 0.8 seconds. \n\nAs an angular momentum problem it is neat, but as a real life Earth problem it is totally bogus. There are all sorts of complications, like the sea only covers 70% of the Earth, the density of water is less than the average density of the Earth, the ice that melted started out far above sea level in many cases, and so on.
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
841046
Let's do that from the beginning.\nLets take M for the mass of Mars, m - for the (irrelevant) mass of the satellite, R for the distance from the centre of Mars to the satellite, T - the rotational period)\nThe centripetal force is given by m*v^2/R\nThe gravitational force is G*m*M/(R^2). G- gravitational constant\nThe forces are the sme so we get\nv^2=GM/R, but v=2*PI*R/T  because the satellite des a full circle with the radius of R in the time T.\nSolving for R, R= v*T/(2*PI), into our formule and we get\nv^2=2GM*PI/(v*T) so v^3= 2GM*PI/T and we get the velocity, if we put in the correct mass for Mars. \nBtw. it should't be called a geosynchronous satellite, because the prefix geo- suggests Earth. Probably Martosynchronous...
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
432119
It takes about 365.2422 days to orbit the sun (i.e. a solar year). If we have a year be 366 days every four years, that averages out to 365.25 daysper year. This is still a bit large, so we declare 3 of the four century marks to *not* be leap years. That then gives an average of 365.2425, which is close enough for most purposes.\n\nSpecifics: Every year that is divisible by 4 is a leap year *except* those ending in 00. These are leap years only if divisible by 400.
882266
Using Kepler's laws (third law):\n\nThe ratio between the T*T and a*a*a is a constant for our Sun and it applies to all objects rotating around it. where T is the period and a is the average distance. \nSo:\nT2=T1*(a2/a1)^(3/2) ;  ^ is used for power \nT2=1year *[  (4.5*10^9)  /  (1.5*10^9)  ]^(3/2)\nT2=164.3 years     \nTo make sure...see the links
233745
The mean radius is 6371.0 km.  The radius at any latitude is equal to that times the cosine of the latitude.  The Earth turns once on it's axis in 23 hr 56 min 4.0989 sec, or 23.93447 hrs.\n\nIt takes that long to travel 2(Pi)(6371)(cos(lat)).  Divide that distance by the time above and you're in the ball park (neglecting the difference in polar and equatorial radii).\n\nNote: The figure given for the Earth's rotation is correct according to Astrophysical Data: Planets and Stars.  The 24-hour day is measured from noon to noon - in relation to the Sun.  The Earth kind of rolls around its orbit, so after it has turned thru one rotation it has also moved along its orbit.  It has to turn for another 53.9 sec past the full rotation for the Sun to be back to due south again.
697340
first: The mass of the satellite doesnt matter.  it NEVER does in orbit problems.  No really knows why beyond that the m's cancel.\n\nT^2=4 pi^2 r^3\n       -----------\n            GM\n\nr= radius of the orbit(radius of the earth)\npi= 3.14\nG= gravitational constant= 6.67E-11 ..i think...\nM= mass of earth\n\nthe answer is .0581 days or 83.7 min.
308171
this is an answer to a slightly different question (how long does the moon take to make one rotation?): 27 days, 7 hours, and 43.2 minutes.  someone must remember that the moon revolves around the earth (earth-moon center of mass) at the same time that it rotates.
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
211464
Live forever.\nSlow the Earth's rotation so each revolution takes 36 hours.
16582
to tell you with approximate data the earth completes it revolution in 356 and a quarter days. This 1/4 adds up to one full day every 4 years. Thus we need to have this compensatory extra day every four years.
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
626709
Four hours.  The earth rotates through a full 360 degree circle in 24 hours, this means it rotates at the rate of 15 degrees and hour (360 divided by 24).  So it will take 4 hours to rotate through 60 degrees.