Keppler's 1st Law: Examining a Proof

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The discussion centers on understanding the proof of Kepler's First Law, specifically the application of Newton's laws of motion and universal gravitation. A participant questions the presence of an extra vector 'r' in the equation, which is clarified as representing the direction of the gravitational force. The negative sign in the equation is explained as indicating the attractive nature of gravitational force. Additional requests for clearer explanations or links to more straightforward proofs are made, highlighting the complexity of the topic. Overall, the conversation emphasizes the need for a deeper understanding of the physics involved in Kepler's First Law.
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Homework Statement


i am examining a proof of kepplers first law for my conics project and I am a bit rusty on the physics, it begins withe the following line


Homework Equations


To begin with, we will start off by applying Newton's law of motion and Newton's law of universal gravitation together to find that

m a = (-G m M/r^2) r

i think there is an extra r in this equation? if no where does it come from??

here is the link if it helps
http://members.kr.inter.net/joo/physics/curriculum/kepler/proof.html

The Attempt at a Solution

 
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The r is in bold which indicates its a vector. So the force of gravity has a magnitude of (-G m M/r^2) along the unit vector r.
 
thanks, altho the more I am looking at this the more i don't understand it,
can anyone explain where the negative comes from?
 
or if anyone knows of a link to a clearer proof, that would be appreciated also
 
The negative is there because the force of gravity is attractive.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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