Can anybody tell me the derivation of INVERSE SQUARE RULE
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Consider a circular orbit (Kepler's first law tells us this is possible, as circles are particular cases of ellipses). By Kepler's second law, the speed vv is constant along the orbit. We can obtain its dependence on rr using Kepler's third law: T2∝r3T2∝r3. The result is v2∝1/rv2∝1/r.
For the orbit to be circular, the force should satisfy
F(r)=mv2r∝1r2.F(r)=mv2r∝1r2.
We can also find the direction of the forcefrom Kepler's laws! We work in two dimensions because Kepler's first law tell us that the orbits stay in a plane. The acceleration in radial coordinates is
a⃗ =(r¨−rθ˙2)r^+(rθ¨+2r˙θ˙)θ^.a→=(r¨−rθ˙2)r^+(rθ¨+2r˙θ˙)θ^.
Notice that the θ^θ^ component of the acceleration is just 1rddt(r2θ˙)1rddt(r2θ˙), and that r2θ˙r2θ˙ is the areal velocity, which is constant by Kepler's second law. Therefore, the acceleration has the direction of r^r^, and so does the force. The latter should then be of the form
F⃗ (r⃗ )=−F(r)r^,F→(r→)=−F(r)r^,
where the dependence only on rr and not θθ is a consequence of the isotropy of space. I thinks helpful....
For the orbit to be circular, the force should satisfy
F(r)=mv2r∝1r2.F(r)=mv2r∝1r2.
We can also find the direction of the forcefrom Kepler's laws! We work in two dimensions because Kepler's first law tell us that the orbits stay in a plane. The acceleration in radial coordinates is
a⃗ =(r¨−rθ˙2)r^+(rθ¨+2r˙θ˙)θ^.a→=(r¨−rθ˙2)r^+(rθ¨+2r˙θ˙)θ^.
Notice that the θ^θ^ component of the acceleration is just 1rddt(r2θ˙)1rddt(r2θ˙), and that r2θ˙r2θ˙ is the areal velocity, which is constant by Kepler's second law. Therefore, the acceleration has the direction of r^r^, and so does the force. The latter should then be of the form
F⃗ (r⃗ )=−F(r)r^,F→(r→)=−F(r)r^,
where the dependence only on rr and not θθ is a consequence of the isotropy of space. I thinks helpful....
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