Quick question on Newton, gravity and planetary movement

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Contadoren11
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Homework Statement


Hi,

I've been tasked with showing that the length of vector F_gravity is inversely proportional with radius squared (i.e. |F_vector|=c/r^2) and is central, i.e. consistently directed toward the same point. Apparently, this is the same as (<=>) the acceleration of a particle following an elliptical path being radial (and thus, directed towards the same point as F) *and* it being equal to

a_r = - C/r (1)

where C is a universal constant.

C ends up being 4*π^2*(a^3/T^2), i.e. positive.

My questions are: Why is there a minus in (1), and what is the explanation for the <=> being valid? I realize it's Newtons 2nd law, but the constants C and c could conceivably differ by more than m, no?

Apologies for the shoddy notions, not used to doing physics in english (or writing formulas outside of Word...)

Homework Equations


|F_vector|=k*1/r^2 and central
<=>
a_r = - C*1/r (1)

where C is a universal constant. C ends up being 4*π^2*(a^3/T^2), i.e. positive.

The Attempt at a Solution


Can't figure it out.
 
Last edited:
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The minus sign is because the acceleration is towards the origin, the point r is measured from, so tends to reduce r.
I don't see that it matters that the constants can be different. You are only asked to show that there exist constants that make the two equivalent.
 
Gah, I messed up - forgot to put r^2 rather than just r in the formula for a_r.
 
Hm. So the force exerted on the Earth by the Sun would be considered negative as well, or? (Also, I don't understand what you mean by "so tends to reduce r").

I reckon I get the constants thing.
 
Contadoren11 said:
Hm. So the force exerted on the Earth by the Sun would be considered negative as well, or? (Also, I don't understand what you mean by "so tends to reduce r").
r is the distance from the Sun to the Earth, so is positive. ##\dot r## is the rate of increase of r, so is speed measured away from the Sun. The acceleration is the rate of increase of velocity away from the Sun, but it will be towards the Sun in practice, so its value will be negative. That is, over time, it acts to reduce the velocity (as a signed quantity) and hence to reduce r; in short, gravity attracts.
 
Of course! Thanks a lot for the elaboration.