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So, we know that the gravity is responsible for pulling the Moon towards the Earth. But because it moves in an orbit, it makes me think that there must be a force that is causing the moon to travel in the direction other than the Earth's. So it doesn't fall straight to the Earth.

Qmechanic
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Anonymous
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    possible duplicate of Why doesn't the Moon fall upon Earth? – Javier Sep 28 '15 at 11:30
  • Tie a soft weight on a string and whirl it around your head. The string is pulling the weight but the weight does not fall on your head ( unless you stop turning). It is called the centrifugal force and is equal and opposite to the centripetal which your hand gives. – anna v Sep 28 '15 at 11:40
  • A body in motion stays in motion does not require a force to stay in motion. – WillO Sep 28 '15 at 13:12
  • @WillO Yeah but what's keeping the body in motion. In this case: what's keeping the moon in motion? – Anonymous Sep 28 '15 at 15:29
  • @Anonymous: Perhaps my earlier comment was unclear because I inadvertently omitted the word "and". A body in motion stays in motion and does not require a force to stay in motion. – WillO Sep 28 '15 at 16:12
  • @WillO So, you are saying that the moon was moving forever? – Anonymous Sep 28 '15 at 16:32
  • What the earlier comments do not make clear is that a body, such as the Moon, obeys Newton's laws of gravitation and Kepler's laws of planetary motion, which provide that "a moving body shall continue to move unless acted upon by an outside force". Whatever it was that gave the Moon its momentum (i.e. its velocity) is - for current purposes - irrelevent. It has velocity, and angular motion. The laws of physics predict that it will retain that angular momentum unless acted upon by an external force. If no such (gravitational) force is applied, it will retain its current motion -- perpetually. – Ed999 May 23 '19 at 15:58

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Although the force is radial, the direction of motion is not the direction of the force, rather it is the direction of the velocity at any time $t$. In order to find out the dependence $\mathbf{v}(t)$ one must solve the equations of motion $\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}})=m\mathbf{a}$.

Doing so with the gravitational potential $V(r) = -G\frac{mM}{r}$ gives back trajectories which happen to be conic sections. The only case when such trajectories can degenerate in straight lines is when the initial velocity is zero.

gented
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  • It might be useful to expand a little on what is meant by 'conic sections', as the asker is unlikely to recognise these as circles, ellipses or hyperbolas. – Gert Sep 28 '15 at 12:50