How to define "work" on a movement of a celestial body? A moving celestial body (such as a planet orbiting the Sun) creates infinite distance. How to relate this to Work formula ($W=F\cdot d$)?
I know I must have wrong understanding about this formula when realize that it will be wrong when we said that the Work on that celestial body is infinite since the distance created by that celestial body is infinite.
 A: The Earth almost revolves around the sun in a circular orbit.  If it were perfectly circular, then the Sun's gravitational field would do no work on the Earth, since the force is radial and would be at an an angle of 90 degrees to the Earth's motion.
With the small eccentricity, there will be some work done (the angle between F and d is no longer exactly 90 degrees).  The Earth will have its fastest orbital speed at its closest approach to the sun (perihelion) and its slowest speed at its farthest orbital point from the sun (aphelion).
A: 
Lets assume astronaut throw a ball in space, unlike on earth which the ball will stop on certain distance, on space ball will never stop until some force stop it. On earth we use W=F•d to find Work done by thrower.

Two things:

*

*$d$ isn't the distance the ball moves, it's the distance over which the force is applied.  The astronaut can throw the ball very far, but is only applying a force to it over about a meter.  After that, the force is zero and the work done on the ball (by the astronaut) is zero.  Ignoring the effects of other forces, the speed of the ball remains constant and the energy remains constant.


*$W = Fd$ is an approximation used when the force and the velocity are in the same direction.  When they are not, the more general formula is $W = Fd \sin(\theta)$, where $\theta$ is the angle between the force and the object's velocity.  In the case of an approximately circular orbit, the gravitational force and the velocity are at right angles, $\sin(\theta)$ is zero, and the work done is zero.
A: A planet in a circular orbit around the sun would actually experience no work, since the gravitational force and its motion are always perpendicular.  The differential work done by the force $\vec{F}$ as the planet moves through the small displacement $d\vec{r}$ is $dW=\vec{F}\cdot d\vec{r}$.  The gravitational force is $\vec{F}_{g}=-(GM_{\odot}m/r^{2})\hat{r}$, pointing in the direction toward the sun, $-\hat{r}$. For a planetary orbit in the equatorial plane (or $xy$-plane), the displacement in a time $dt$ is $d\vec{r}=\vec{v}\,dt=v\,dt\,\hat{\phi}$, where $\hat{\phi}$ is the azimuthal direction.  Since $\hat{r}$ and $\hat{\phi}$ are unit vectors in perpendicular directions, $\hat{r}\cdot\hat{\phi}=0$, and consequently, $dW=0$.  In other words, no work is being done, because there is no motion in the direction of the force.  Regardless of how far the planet travels in this circular orbit, work is never done.
True planetary orbits are not circular but elliptical, but this does not change the total work done over a complete orbit.  While the planet is moving inward toward the sun, the gravitational force does positive work.  However, while the planet is moving away from the sun, there is an exactly opposite negative work done, and one the planet returns to its original position, the total work is zero.  This is one way of characterizing the a conservative force like the gravitational force.  Any closed path, that brings the planet back around to its original position, will have zero net work done (leaving the planet with exactly the same energy it started with).
A: thanks for the answer.
Maybe I should rephrase my thought about this subject. Lets assume an astronaut throws a ball in space, unlike on earth where the ball will stop on certain distance, on space the ball will never stop until any other force stop it. On earth we use W=F•d to find Work done by thrower. On space how we can find Work done by astronaut if the distance of the thrown ball reach infinity?
I read somewhere that the Work equal to kinetic energy of the ball (1/2 m v^2), but I am trying to clarify my understanding about the "d" on W=F•d. Is it really refer to the distance that I thought?
On the other words, if a car can not move let say because the tire was jammed, can we say that there is no Work done by the car even the engine was running?
