Nothing new here, just wanted to capture these proofs into a single location for easy reference. I've tried to arrange them into a form that is easy to understand and follow.
Proof that a Newtonian force between two masses will produce an elliptical orbit - this is a very textbook approach using polar coordinates, I'm just capturing it here for reference. There are many other approaches, and this has likely been done millions of times now.
Remember that Newton's Law is:
F=GMmr2∴F=ma
We define a unit vector in the radial direction ˆr along the angle θ:
ˆr=ˆxcosθ+ˆysinθ Therefore ˆr changes as per angle θ perpendicular to ˆr, giving us the tangential unit vector ˆθ
dˆrdθ=ˆx(−sinθ)+ˆycosθ=ˆθ And the derivative of ˆθ is another 90° rotation, giving us −ˆr:
dˆθdθ=ˆx(−cosθ)+ˆy(−sinθ)=−ˆr Find Acceleration →a in polar coordinates, first we find velocity →v as the change in radial vector over time:
→v=d→rdt=ddt(rˆr) Applying the product rule and the chain rule we get our expression for how fast you are moving radially (˙r in ˆr), and how fast you are moving in the tangential direction (r˙θ in ˆθ):
→v=drdtˆr+rdθdtdˆrdθ=˙rˆr+r˙θˆθ Acceleration is the derivative of velocity:
→a=d→vdt=ddt(˙rˆr+r˙θˆθ) →a=¨rˆr+˙r˙θˆθ+˙r˙θˆθ+r¨θˆθ+r˙θ(−ˆr)˙θ Grouping terms in ˆrandˆθ we get:
→a=ˆr(¨r−rˆθ2)+ˆθ(2˙r˙θ+r¨θ) From Newton
→F=m→a We only need the radial term from (10) since the Force is only in the radial direction, substitute along with Law of Gravity:
−GMmr2ˆr=mˆr(¨r−r˙θ2) The radial unit vector ˆr and mass term m cancel out, giving us this equation that we must show is equal:
−GMr2=¨r−r˙θ2 The angular momentum is a constant because there are no torques acting on our system, the only Force is gravity acting radially inwards so we can define ˙θ:
L=mr2˙θ∴˙θ=Lmr2 And substitute this into (13):
−GMr2?=¨r−L2m2r3 Next we need to show that the second time derivative of this equation of an ellipse works in our force equation:
r=a(1−ϵ2)1+ϵcosθ First derivative [Wolfram|Alpha]:
drdt=drdθdθdt=−a(1−ϵ2)(−ϵsinθ)(1+ϵcosθ)2˙θ Substitute for ˙θ and again for r
drdt=a(1−ϵ2)(ϵsinθ)(1+ϵcosθ)2L(1+ϵcosθ)2ma(1−ϵ2)a(1−ϵ2) drdt=Lϵsinθma(1−ϵ2) Second derivative:
d2rdt2=ddθ(drdt)(dθdt)=Lϵcosθma(1−ϵ2)Lmr2 Which we can substitute back into our force equation (15), giving us:
−GMr2r2m2?=(L2ϵcosθm2a(1−ϵ2)r2−L2m2r2r)r2m2 Cancel out the mass and radius terms and substitute for r again:
−GMm2?=L2(ϵcosθ)a(1−ϵ2)−L2(1+ϵcosθ)a(1−ϵ2) Which will now simplify further and we can solve for L:
L2=GMm2a(1−ϵ2) Both sides are constants being equal with no dependence on any term, but defined by the ellipse of semi-major axis a and eccentricity ϵ.
This shows that a Newtonian force (proportional to the product of the masses and inversely proportional to the square of their distances) acting mutually between two masses will produce an elliptical orbit.
Next we want to show that a planet in orbit would sweep out equal areas over equal time.
We can find the angular momentum:
→L=→r×→p We only need the perpendicular component, so we can drop the vectors: L=rp⊥=r(mv⊥) And v⊥=rω:
L=rm(rω) Therefore, as we also found in the First Law (14):
L=mr2ω and we find that:
r2ω=Lm
Now we need to show that for the integral:
A=∫12r2dθ that r2 is a constant. The derivative with respect to θ is:
dAdθ=12r2 and with respect to time (t):
dAdt=dAdθdθdr=12r2dθdt but dθdt is just ω: dAdt=12r2ω And we can replace r2ω from (28):
dAdt=12Lm Which is a constant.
Finally, we want to show that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
Since we know that centripetal force is mv2R we can consider what Force is needed:
GMmR2=mv2R For a nearly circular orbit we can substitute v2=4π2R2T2
GMmR2=m4π2RRRT2 And we find:
T2R3=4π2GM Therefore this ratio depends only on M and our ratio would be constant for masses in orbit around it.
Kepler's First Law: Ellipses
Proof that a Newtonian force between two masses will produce an elliptical orbit - this is a very textbook approach using polar coordinates, I'm just capturing it here for reference. There are many other approaches, and this has likely been done millions of times now.
Remember that Newton's Law is:
F=GMmr2∴F=ma

We define a unit vector in the radial direction ˆr along the angle θ:
ˆr=ˆxcosθ+ˆysinθ Therefore ˆr changes as per angle θ perpendicular to ˆr, giving us the tangential unit vector ˆθ
dˆrdθ=ˆx(−sinθ)+ˆycosθ=ˆθ And the derivative of ˆθ is another 90° rotation, giving us −ˆr:
dˆθdθ=ˆx(−cosθ)+ˆy(−sinθ)=−ˆr Find Acceleration →a in polar coordinates, first we find velocity →v as the change in radial vector over time:
→v=d→rdt=ddt(rˆr) Applying the product rule and the chain rule we get our expression for how fast you are moving radially (˙r in ˆr), and how fast you are moving in the tangential direction (r˙θ in ˆθ):
→v=drdtˆr+rdθdtdˆrdθ=˙rˆr+r˙θˆθ Acceleration is the derivative of velocity:
→a=d→vdt=ddt(˙rˆr+r˙θˆθ) →a=¨rˆr+˙r˙θˆθ+˙r˙θˆθ+r¨θˆθ+r˙θ(−ˆr)˙θ Grouping terms in ˆrandˆθ we get:
→a=ˆr(¨r−rˆθ2)+ˆθ(2˙r˙θ+r¨θ) From Newton
→F=m→a We only need the radial term from (10) since the Force is only in the radial direction, substitute along with Law of Gravity:
−GMmr2ˆr=mˆr(¨r−r˙θ2) The radial unit vector ˆr and mass term m cancel out, giving us this equation that we must show is equal:
−GMr2=¨r−r˙θ2 The angular momentum is a constant because there are no torques acting on our system, the only Force is gravity acting radially inwards so we can define ˙θ:
L=mr2˙θ∴˙θ=Lmr2 And substitute this into (13):
−GMr2?=¨r−L2m2r3 Next we need to show that the second time derivative of this equation of an ellipse works in our force equation:
r=a(1−ϵ2)1+ϵcosθ First derivative [Wolfram|Alpha]:
drdt=drdθdθdt=−a(1−ϵ2)(−ϵsinθ)(1+ϵcosθ)2˙θ Substitute for ˙θ and again for r
drdt=a(1−ϵ2)(ϵsinθ)(1+ϵcosθ)2L(1+ϵcosθ)2ma(1−ϵ2)a(1−ϵ2) drdt=Lϵsinθma(1−ϵ2) Second derivative:
d2rdt2=ddθ(drdt)(dθdt)=Lϵcosθma(1−ϵ2)Lmr2 Which we can substitute back into our force equation (15), giving us:
−GMr2r2m2?=(L2ϵcosθm2a(1−ϵ2)r2−L2m2r2r)r2m2 Cancel out the mass and radius terms and substitute for r again:
−GMm2?=L2(ϵcosθ)a(1−ϵ2)−L2(1+ϵcosθ)a(1−ϵ2) Which will now simplify further and we can solve for L:
L2=GMm2a(1−ϵ2) Both sides are constants being equal with no dependence on any term, but defined by the ellipse of semi-major axis a and eccentricity ϵ.
This shows that a Newtonian force (proportional to the product of the masses and inversely proportional to the square of their distances) acting mutually between two masses will produce an elliptical orbit.
Kepler's Second Law: Equal Areas
Next we want to show that a planet in orbit would sweep out equal areas over equal time.
We can find the angular momentum:
→L=→r×→p We only need the perpendicular component, so we can drop the vectors: L=rp⊥=r(mv⊥) And v⊥=rω:
L=rm(rω) Therefore, as we also found in the First Law (14):
L=mr2ω and we find that:
r2ω=Lm
Now we need to show that for the integral:
A=∫12r2dθ that r2 is a constant. The derivative with respect to θ is:
dAdθ=12r2 and with respect to time (t):
dAdt=dAdθdθdr=12r2dθdt but dθdt is just ω: dAdt=12r2ω And we can replace r2ω from (28):
dAdt=12Lm Which is a constant.
Kepler's Third Law: Harmonies
Finally, we want to show that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
Since we know that centripetal force is mv2R we can consider what Force is needed:
GMmR2=mv2R For a nearly circular orbit we can substitute v2=4π2R2T2
GMmR2=m4π2RRRT2 And we find:
T2R3=4π2GM Therefore this ratio depends only on M and our ratio would be constant for masses in orbit around it.
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