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Suppose that a satellite circles around a planet that exerts $2000N$ of gravitational force on the satellite.

I understand the fact that since the circular motion and the centripetal force are always perpendicular to each other, the work done by gravity is 0.

However, the satellite is being moved from one point to another, so Intuitively I am thinking "there must be some work done to move the satellite from point A to point B".

But besides the gravitational force, there is no force acting on the satellite.

Can some one explain me the gap with my logic and intuition?

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  • $\begingroup$ Think of a free object. Since there are no forces acting on it, it will move in a straight line with constant velocity. Work only comes into play when you want to change the magnitud of your velocity. $\endgroup$ – Javier Dec 21 '13 at 4:55
  • $\begingroup$ I see. So "work" is literally defined as force times distance and if there is no force acting on it, whether it is moving fast or not does not matter. $\endgroup$ – hyg17 Dec 21 '13 at 5:05
  • $\begingroup$ Precisely. Work. Inertia. You may find those useful. Remember, you don't need a force to keep an object moving. $\endgroup$ – Geoffrey Dec 21 '13 at 5:19
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I'll expand my comment here.

First, think of an object with no forces acting on it. According to $\vec{F} = m\vec{a}$ or to Newton's First Law, such an object will move in a straight line with constant velocity. This is a very important point: you do not need a force to mantain movement. Simply because an object moves from A to B doesn't mean you have to exert a force on it.

Astronauts on the ISS live in what is essentially a force-free environment (it isn't really, but it's as if it was), and if you've ever seen one of Chris Hadfield's videos, you can see that if you give anything the slightest push, it will keep on moving until it's stopped by something else.

This is all fine and dandy, but in your example there is a force acting on the object: the centripetal force which is required to mantain circular motion (remember, if the force disappeared, the object wouldn't stop; it would keep moving in a straight line). Which brings us to a subtler point: Work is defined as $\int \vec{F} \cdot d\vec{r}$, or, if you're not familiar with calculus, as $\vec{F} \cdot \vec{d}$, where $\vec{d}$ is the displacement vector.

What this tells us is that only the component of the force in the direction of motion is what matters. In the circular movement example, the force is always at right angles to the motion, and so there is no work, because two perpendicular vectors always have a dot product equal to zero.

How does this fit with the fact that the object is not moving in uniform motion but rather is being accelerated? Now we need to remember that work is the change in kinetic energy, defined as $\frac12 m v^2$. As you can see, the kinetic energy depends only on the magnitude of the velocity and not on its direction. In uniform circular motion, only the direction of the velocity is changing, because the force is at right angles to the movement. Since the speed (i.e. the magnitude of the velocity) is constant, no work is being done and the energy remains constant.

This is intuitively clear; if you put the satellite in orbit, it will keep on orbiting for all of eternity. If it was gaining energy each time it completes a full circle, very soon it would have enormous amounts of energy, and we know that's not the case, because the planets have been going around the Sun for a very long time, and we notice that their energy (in whatever form) doesn't seem to increase appreciably.

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The object is moving around in a circle, so its direction is changing all the time.

Now the centripetal force is always pointing towards the centre of the circle.

Definition of Work Done = Force X distance moved IN THE DIRECTION OF THE FORCE.

Since the distance moved by the object, and the direction of the force are always perpendicular to each other, therefore no work is done.

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    $\begingroup$ IMHO you just rephrased what OP already knew. Consider adding more details to your answer, and perhaps to explain the intuition on why no work is done when the force is perpendicular to the motion (which is what OP is asking after all). $\endgroup$ – AccidentalFourierTransform Feb 17 '16 at 13:01
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Hey I somewhat got your point, but the fact is there is no parallel motion, I meant that the centripetal force is towards the planet, but the satellite moves a distance (scalar) perpendicular to it. It is somewhat like the tangent (distance) and the radius of a Circle(force). Recall Newton's second law the acceleration ( vector quantity ) is in the direction of resultant force. But the distance is perpendicular to the force. So no work is done.

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The gravitation creates potential energy. The satellite is moving on an path that has the same energy level, which we can call an "isoenergetic path" so there is no work needed to move it to along that path. Note that the satellite is "forced" to move along that path. Eventually, Earth-satellite system has some potential energy stored in it, particularly in the kinetic energy of the satellite.

The interesting question would be if we magically increased the mass of earth (or satellite), and the gravitational potential field is changed, which would cause satellite to move to another isoenergetic path. While this change is happening, some work will be done depending on the initial and final energy levels of the satellite.

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First, you have to understand that the satellite is constantly falling toward the planet due to the planet's gravitational force. But while it is falling (approx. 1m every 8km if the planet is Earth), the planet gets "curved away" from the satellite in the same amount so that the distance between (falling) satellite and the planet remains unchanged (i.e. Earth also curves ~1m every 8km). Since there is no change of the distance during satellite's falling to the planet, the work done is zero.

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    $\begingroup$ This could be improved by relating "distance to the planet" and "potential energy", and a mention of kinetic energy. $\endgroup$ – Kyle Oman Jan 11 '17 at 22:53
  • $\begingroup$ OP states that, "Suppose that a satellite circles around a planet that exerts 2000N of gravitational force on the satellite." I.e. what happened before (launching of the satellite and kinetic energy) is of no concern (to OP). $\endgroup$ – Bran Jan 12 '17 at 23:15
  • $\begingroup$ OP states that, "Suppose that a satellite circles around a planet that exerts 2000N of gravitational force on the satellite." I.e. what happened before (launching of the satellite and kinetic energy) is of no concern (to OP). OP is interested in work being zero. He identified correctly that there is only one tangible force in play here: gravitational force, which is constant. Since the change of distance (delta r) is zero, thus we have W = Fg x deltaR = Fg x 0 = 0. OP didn't realize that the satellite is falling toward the planet (although it looks like it is orbiting), just like a projectile $\endgroup$ – Bran Jan 12 '17 at 23:25
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The work equals to the amount of energy added to the system (or subtracted if the work is negative) and since you don't need to add energy to the satellite in order for it to stay in orbit the total work is zero.

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Centripetal force is always directed towards the center of the circular path and always directed perpendicular to the direction of displacement of particle every where on the circular path. Therefore angle between centripetal force F and displacement S is 90deg. Work W= F.S $$W=FS \cos (\theta)=FS \cos90 = FS\times0 =0$$

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  • $\begingroup$ could you elaborate please ? $\endgroup$ – user46925 Feb 18 '16 at 5:28

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