Why is work done by the force of gravitation on a satellite when it moves from one part of an elliptical orbit
962875
The speed of the satellite changes in an elliptical orbit. It slows as it approches the farthest part of the orbit and speeds up as it leaves the farther parts and comes closer to the planet. When it speeds up, its kinetic energy is increasing. For its kinetic energy to increase, work must have been done.
1041594
Things that are orbiting our planet are in a state of constant "free fall". That is why they appear to be weightless. The force of gravity is still very high on the astronaut.
359330
shouldn't you subtract the work done by gravity\nfirst?\n\nwork By Horizontal Force = 55Nm - work By Gravity
398654
In the special case where centrifugal force exactly balances gravity, the orbit IS circular.  But say you have a planet in a circular orbit.  Give it a little extra speed in the direction it's going.  Now it has too much centrifugal force, and it begins to gain distance from the star -- but that means it's gaining potential energy and so must be losing kinetic energy.  So it slows down.  The detailed math shows that by the time it's gone 180 degrees around it has lost so much speed that, even though it is farther from the star, now it doesn't have enough centrifugal force to counteract gravity, so it begins to speed up.  Again, if you do the math, you see that it will return to the exact position it was at when you gave it the extra push, and with the exact same velocity it had just after that push.  So it will just keep repeating that orbit, which is highest (and slowest) at one point and lowest (and fastest) 180 degrees around; in other words, an elliptical orbit.
246461
No more than the earth does work orbiting the sun. An orbit has constant energy, and if it is not circular, there is a continuous process of exchanging kinetic energy with potential energy while the sum stays the same.
1169631
(G* m1*m2)/r.^2=force on the satellite due to earth in the radial direction ==m1* accelaration of the satellite in radial direction\n\nm1=mass if the satellite\nm2 is the mass of earth\n\naccelaration of satellite= v.^2/r\n\n===>G*m2/r=v.^2\n\n==>r=G*m2/v.^2\nonce r is calculated here by substituting the values og G ,m2,& v ==5950 m/s\n\ntime for one revolution == (2* pi *r)/v
256930
dba_gregg is correct.\n\nThe gravity pulls the front while the back is still supported by normal reaction force. Thus a torque is acting so as to rotate the bus with the front down. The angular momentum produced is conserved and the bus begins to rotate about the free axis passing through its centre of mass.
70887
they can't.  That would be a perpetual motion machine.  \nConservation of energy says that you can't get something for nothing.  for a terrestrial body to do work continuously it would need to be given a continuous supply of energy.\n\nIn an elliptical orbit, two boddies can take turns doing work on each other and energy is conserved.
894961
The gravitational force (Fg) exerted by Saturn on its moon is\n\nFg = G*M*m/r^2\n\nwhere G is the universal gravitation constant, M is the mass of Saturn, m is the mass of the moon, and r is the radius of the orbit.\n\nThe gravitational force is balanced by the centripetal force (Fc) due to the motion of the moon in a 'circular' orbit around Saturn.\n\nFc = m*v^2/r \n\nwhere v is the orbital speed, which can be expressed as\n\nv = 2*pi*r/T\n\nwhere T is the period of the orbit. Inserting this into the equation for the centripetal force you have\n\nFc = 4*pi^2*m*r/T^2\n\nNow since\n\nFg = G*M*m/r^2 = 4*pi^2*m*r/T^2 = Fc\n\nThe mass of the moon, m, cancels leaving\n\nG*M/r^2 = 4*pi^2*r/T^2\n\nwhich can be rearranged to give Kepler's third law that says that the square of the orbital period is proportional to the cube of the orbit's semimajor axis (taken as the radius for this problem, which assumes a circular orbit). Solving for the mass of Saturn\n\nM = 4*pi^2*r^3/(G*T^2).
941619
In simple English...It means that the pull of gravilty is less at 2 points so the orbit expands at those points before being pulled back in by gravity, and so the eliptical shape occurs.
1175221
I do not understand why you specified standing on the "Galapadous Island [Galapagos Islands?]" as opposed to anywhere else on Earth.\nNever-the-less,\n\nSince you are on Earth, you would experience an acceleration due to gravity.  Similarly, on Earth, one would experience a centripetal acceleration caused by the force of gravity.  Gravity acts as the centripetal acceleration, these two accelerations are not really separate.\n\nAt the same time, the Earth (with you on it) is orbiting the sun, which is intern, orbiting the galaxy, which is intern orbiting the center of a galactic cluster, and so on.  All this orbiting implies a force of gravity and a centripetal acceletation (again, the same acceleration).\n\n\nEDIT:\nYour question was about acceleration.  Acceleration requires a net force to be present.  In most of the examples you listed, those force acting on a body are very weak and are easily canceled out by forces of friction.  If the force is canceled out and no net force remains, there is no acceleration.
845902
Let’s consider from viewpoint of physics. F=G*M*m/r^2 is gravitation force holding the sat on its orbit. On the other hand F=m*a – centripetal force. Thus a=G*M/r^2, where M is mass of Earth (I don’t remember), r=R+h, R=6.4*10^6m is radius of Earth, h=0.615*10^6m is height of the sat, a is gravitational acceleration at h (like that ‘g’ on the ground), G is gravitational constant  (?? 6.67*10^-11 ??  I don’t remember!), m was mass of the satellite. Another formula for centripetal acceleration for circular track a=r*w^2, w is angular velocity or w=2pi/T, T being period. G*M/r^2=r*(2pi/T)^2. Now T=2pi*r*sqrt(r/(G*M)) seconds
823207
gravity is a special case  of gravitation.It is the force by which earth pulls a one kg object towards it.\n\nie g=GM/R^2\n M=mass of earth\nR=Radius of earth\n\nnow force acting on a object  of mass 1 kg at a height h(from earths surface) will be \nf=GM/(R+h)^2\nwe can take out R out of bracket so that (R+h)^2 can be R^2{1+h/R}^2\nNow\nGM/R^2=g\nSo we can easily WRITE it as \nf= g/{1+h/R}^2\n\nThis f is that something we usually write as g'\n\nNow g' can be written as g{1+h/R}^(-2)\n\nFOR NOT TOO SOPHISTICATED CALCULATION\nIT CAN BE REDUCED USING BINOMIAL THEORM TO \ng'=g{1-2h/R}
512623
I will state this here, in yet another question: THERE IS NO SUCH THING AS CENTRFUGAL FORCE!! Look it up in a basic physics textbook like Giancoli or Serway. Or even do a google search. You have to understand what reference frame you're talking about. The fact that your body tends to want to move to the outside of the curve when your car goes around in a circle has nothing to do with anything other than the fact that your body wants to keep going along in a straight line (conservation of momentum), but the car gets in front of it at every turn (semi pun intended). What is actually happening is that the car is an accelerating reference frame.\n\nAs for the moon/earth question... imagine a man 6' tall throws a tennis ball horizonatally with respect to the earth's surface. The ball will go forward a certain distance (because of it's velocity tangential to the earth's surface) as it falls the ~6' to the ground (because gravity pulls it down). The faster he throws the ball (horizontally), the further it will go before it hits the ground. If he throws it fast enough then the amount that the ball travels forward and the amount that it drops towards the earth's surface can match the curvature of the earth, and the ball will orbit the planet (neglecting air resistance, as we physicists always do) and come around and bonk the thrower in the back of the head. The moon is just a much larger tennis ball, thrown by a much larger man.\n\nYour eqn. is correct, but it only describes the force acting upon the moon. It is not the equation of motion, which detirmines what actually happens to the moon. The equation of motion is:\n\nF = m * a  (F = force, m = mass of moon, a = acceleration of moon around earth)\n\nIn the case of the circular motion of the moon around the earth, "a" is defined by:\n\na = (v^2)/r (where v is the magnitude of the velocity of the moon and r is the center-to-center distance from the earth to the moon)\n\nPlugging one into the other you get:\n\nG * M * m/r^2 = m * (v^2)/r (where M is the earth's mass)\n--> G * M/r = v^2\n\nSo, as long as the tangential velocity, v = sqrt(G * M/r), the moon falls just enough towards the earth as it travels just enough tangentially second by second, that it continues to orbit around the earth. (Note: there is NO CENTRIFUGAL FORCE anywhere in any of these eqns.!!)
56969
the work-energy theorem says that work is equivalent to energy.\n\nso, if you lift a 1 kg object 1 meter, you do some amount of work against gravity.  if you do the math, you'll find that the work is W = 9.8 kg*m^2/s^s.\n\nthe work-energy theorem says that this is equivalent to a change in energy, so delta_E = W = 9.8 Joules.\n\nbasically it is a simple statement of the conservation of energy, where a change in the potential energy equals the work put into the system.\n\nnote that it only applies for conservative forces, like gravity or the electrostatic repulsion.  if you are pushing something against friction, then there is an additional loss of energy to heat at the frictional surface, and the work-energy theorem does not apply.
1081934
There are too many formulas to calculate that answer. You have to take in account for example, position of object to find potential energy, mass and the weight due to the gravitational force, also Force that is going to be exerted the object and the amount of that energy that will be lost. SO it is a lot to  find that out. But their is not one formula for it. Sorry, But if you do find out Let me know.
919025
Let's clear up our terms a little bit.\n\nGravity only generally occurs between things of mass.  And it is pretty much independant of what the mass is doing.  And I say pretty much because there are things that can create or destroy mass... but in a general sense it all comes down to the mass.\n\nNow, as astronomical bodies form, they tend to get a spin.  Because of the conservation of angular momentum, things that suck up more mass tend to have more spin than others.  These are, however, tendancies, NOT necessities.\n\nCase in point - the Earth.  Because of the tides, the spin of the Earth is slowing down.  We used to have twelve-hour days millions and millions of years ago, but now they're twice as long.  Has this affected the force of gravity?  Not a bit.  The same mass is present, and it's causing the same amount of gravity.\n\nWhat your friend may be confused by is what is sometimes called centrifugal force (a physicist would probably just call it inertia, though).  Although gravity pulls us down to the centre of the Earth, the motion of spin carries us away from the centre, counteracting gravity.  This means if you take a tool to measure gravity near the equator, you'll get a lower reading then you will when you measure it at the axis.  The force of gravity hasn't changed at all, but the spin is creating another force that is mitigating gravity's effects.  On Earth, this force is pretty small, but it's concievable that it could amount to a huge amount on a fast-enough spinning object... say, a space station with almost no gravity, but which was spinning fast to simulate gravity but pulling out instead of in.\n\nSo in a literal sense, you are completely right.  Gravity has ONLY to do with mass (well, mass and distance).  In a squishier sense, your friend is sort of right, in that spin can change the APPARENT force of gravity in local areas and that objects with large gravities also tend to have large spins.\n\nHope that helps!