What is accelerating the moon from a straight line course? If gravity was turned off, wouldn't objects on the surface depart in a straight line path tangent to the surface, since the earth is rotating?
Sure but that is a force. Force and energy are different things. The change in energy due to a force is F x d, distance gone under the influence of that force. In this case, the distance between the moon and Earth does not change so no work is done and the energy stays constant.
Even I know that is a stupid question The speed (not accelerating but constant) of the moon is counter balanced by the Earth's gravity pull The moon is in a constant state of free fall Constantly falling over the edge of Earth's horizon Sure turn off Earth gravity and it will head off in a straight line and not fall to wards Earth As I understand the current situation the distance between Earth and Moon is lengthening by about 4 cm per year Don't know how long it will be before it leaves us Gravity off, float away at a angle not straight up sure Earth stops spinning with gravity in place you would scrap along what ever surface you were on until you hit your personal non movable object Please Register or Log in to view the hidden image!
w=work, f=force, m=mass, d=distance, a=acceleration w=fd=mad v=velocity, t=time v=at d=at2/2 =v2/2a w=mv2/2 =m v2/2 g=gravitational acceleration, h=height To move a mass vertically a distance h, w=mgh In the graphic, the moon has two components, translational motion at v, and acceleration at g for the distance h.
Fd work is distance gone in the direction of the applied force. Since the force always acts perpendicular to the velocity, there is no Fd work done.
We'll need to slow the moon down, just a tad. Maybe attach a parachute to its trailing end. The moon doesn't rotate, so that would work just fine, even without an atmosphere.
Great idea Go one better, put up a solar panel (dual power and solar wind) sail which activates when Moon moving towards the sun Power provides energy to the batteries for the service unit on the Moon getting ready for the Mars launch Solar wind pushes against sail slowing Moon down so it's still around when the Mars explorers come back a few years later Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!
The moon has no propulsion units to alter its straight line course (Newton), the g-field does that. The moon mass is displaced a distance h. That is work. I consulted a physics reference written by Issac Asimov. If I'm wrong, so is he!
Asimov would not make such a foolish error so it follows that you have misunderstood. If you can post a reference to it that I can read, I can go through it and help you understand it properly. I repeat: if the force is at right angles to the motion, there is no motion in the direction of the force and thus no work is done.
In post 64, the distance h = 1.3 mm/sec of moon motion in a straight line (no gravity). It's a tiny displacement but continuous to produce the orbit. Here's the page... Please Register or Log in to view the hidden image!
More generally, the work done by a force is $W=\bf{F}\cdot\bf{d}=|\bf{F}||\bf{d}|\cos \theta$, where $\theta$ is the angle between the force and displacement vectors. In the case of the (approximately) circular orbit of the moon, the gravitational force on the moon is always perpendicular to the Moon's displacement, so no work is done by gravity. Even if we take into account the slight ellipticity of the orbit, the total mechanic energy of the Earth-Moon system is still conserved, and for each complete orbit gravity does no work.
As I thought, this is a discussion of a scenario of motion in a straight line under the influence of a force. Suggest you go on a chapter or two and look at motion in a circle. If you do that you will find that what James and I are telling you is correct.