1

The paragraph below is my understanding of how the moon revolves.

When the moon revolves around the earth, the moon has a fixed velocity and a direction (inertia), but the Earth exerts a force on the moon as gravity. Any force causes a mass to accelerate, so either the velocity or the direction has to change. In this case the direction. The direction changes but the speed remains constant.( please assume the orbit to be circular for God's sake, I've seen some eccentric answers that seem too hard to wrap my mind around. I'm not an expert.) The following are my questions

  1. The change in direction of the moon is obviously towards the Earth's core, right? So why doesnt it crash towards the earth? In some answers I see that force influences motion to be a bit more in that direction than before Why doesn't the Moon fall onto the Earth? but what is the formula for it? What is the direction exactly?

  2. It is said that there is no work done by gravity on the moon. But the force exerted by the moon changes the direction and moon moves in the direction for a split second,right? So in the case there is a force and displacement in (almost) the direction of force. Mathematically it should be like W=F×d cos(90-x). I know I am wrong somewhere and can someone help me clear it?( I don't know trigonometry even though I put some cos symbols there.) :/

Qmechanic
  • 220,844
Incognito
  • 29

2 Answers2

0

My favorite description of orbiting is a mis-applied statement from The hitch-hiker's guide to the Galaxy - in order to fly one should throw oneself at the ground and miss. This is actually a good description of how to be in orbit - an object in orbit does continually fall, but keeps on missing.

To follow your mathematical approach, as an approximation we could consider a force acting in a constant direction perpendicular to the original motion, while the path of the orbiting object subtends an angle $\delta\theta$ at the centre of the circle. Suppose the force is $F$ and the radius of the circle is $r$, making the length of the subtended chord $r\sin\delta\theta$ Then the incremental work done would be $F.r\sin\delta\theta.\sin\delta\theta$. Over the entire circle this works out to be $$F.r\sin\delta\theta.\sin\delta\theta.\frac{2\pi}{\delta\theta}$$ As the approximation is improved $\delta\theta$ becomes closer and closer to zero, and this expression for the work done also becomes closer and closer to zero (it tends to zero as $\delta\theta$ tends to zero).

Peter
  • 2,055
0

You think that when there is a force, there is also a displacement in the direction of the force. That is not true. According to Newton's law a force does not cause a displacement, but an acceleration in the direction of the force. So a force does not cause displacement, not even the first derivative of a displacement, but a second derivative of a displacement. Since the force is perpendicular to the velocity, it causes only a change in direction of the velocity. Similar to a ball attached to a rope that you swing around in a circle. The force is along the rope, perpendicular to the velocity. That force does not cause a displacement towards the centre of the circle. The force causes the ball to continue in its circular orbit.

Fred. Zwarts
  • 551