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Specific angular momentum
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<h2 id="definition">Definition</h2> <p>The specific relative angular momentum is defined as the <a href="/facts/Cross_product/uRBJzkcW">cross product</a> of the relative <a href="/facts/Orbital_position_vector/9PlwJYYP">position vector</a> r {\displaystyle \mathbf {r} } and the relative <a href="/facts/Orbital_velocity_vector/9PlwJYYP">velocity vector</a> v {\displaystyle \mathbf {v} } . h = r × v = L m {\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} ={\frac {\mathbf {L} }{m}}} </p><p>where L {\displaystyle \mathbf {L} } is the angular momentum vector, defined as r × m v {\displaystyle \mathbf {r} \times m\mathbf {v} } . </p><p>The h {\displaystyle \mathbf {h} } vector is always perpendicular to the instantaneous <a href="/facts/Osculating_orbit/hA7omegY">osculating</a> <a href="/facts/Orbital_plane_(astronomy)/xPdMmVKP">orbital plane</a>, which coincides with the instantaneous <a href="/facts/Perturbation_(astronomy)/og5hPpjQ">perturbed orbit</a>. It is not necessarily perpendicular to the average orbital plane over time. </p> <h2 id="proof-of-constancy-in-the-two-body-case">Proof of constancy in the two body case</h2> <p>Under certain conditions, it can be proven that the specific angular momentum is constant. The conditions for this proof include: </p> <ul><li>The mass of one object is much greater than the mass of the other one. ( m 1 ≫ m 2 {\displaystyle m_{1}\gg m_{2}} )</li> <li>The coordinate system is <a href="/facts/Inertial_frame_of_reference/1U6TrdG3">inertial</a>.</li> <li>Each object can be treated as a spherically symmetrical <a href="/facts/Point_particle/4MPToN8W">point mass</a>.</li> <li>No other forces act on the system other than the gravitational force that connects the two bodies.</li></ul> <h3>Proof</h3> <p>The proof starts with the <a href="/facts/Two-body_problem/YCG4geSH">two body equation of motion</a>, derived from <a href="/facts/Newton%27s_law_of_universal_gravitation/F6qH2XC3">Newton's law of universal gravitation</a>: </p><p> r ¨ + G m 1 r 2 r r = 0 {\displaystyle {\ddot {\mathbf {r} }}+{\frac {Gm_{1}}{r^{2}}}{\frac {\mathbf {r} }{r}}=0} </p><p>where: </p> <ul><li> r {\displaystyle \mathbf {r} } is the position vector from m 1 {\displaystyle m_{1}} to m 2 {\displaystyle m_{2}} with scalar magnitude r {\displaystyle r} .</li> <li> r ¨ {\displaystyle {\ddot {\mathbf {r} }}} is the second time derivative of r {\displaystyle \mathbf {r} } . (the <a href="/facts/Acceleration/UVDmjYvR">acceleration</a>)</li> <li> G {\displaystyle G} is the <a href="/facts/Gravitational_constant/GdQ8tcBD">Gravitational constant</a>.</li></ul> <p>The cross product of the position vector with the equation of motion is: </p><p> r × r ¨ + r × G m 1 r 2 r r = 0 {\displaystyle \mathbf {r} \times {\ddot {\mathbf {r} }}+\mathbf {r} \times {\frac {Gm_{1}}{r^{2}}}{\frac {\mathbf {r} }{r}}=0} </p><p>Because r × r = 0 {\displaystyle \mathbf {r} \times \mathbf {r} =0} the second term vanishes: </p><p> r × r ¨ = 0 {\displaystyle \mathbf {r} \times {\ddot {\mathbf {r} }}=0} </p><p>It can also be derived that: d d t ( r × r ˙ ) = r ˙ × r ˙ + r × r ¨ = r × r ¨ {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)={\dot {\mathbf {r} }}\times {\dot {\mathbf {r} }}+\mathbf {r} \times {\ddot {\mathbf {r} }}=\mathbf {r} \times {\ddot {\mathbf {r} }}} </p><p>Combining these two equations gives: d d t ( r × r ˙ ) = 0 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)=0} </p><p>Since the time derivative is equal to zero, the quantity r × r ˙ {\displaystyle \mathbf {r} \times {\dot {\mathbf {r} }}} is constant. Using the velocity vector v {\displaystyle \mathbf {v} } in place of the rate of change of position, and h {\displaystyle \mathbf {h} } for the specific angular momentum: h = r × v {\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} } is constant. </p><p>This is different from the normal construction of momentum, r × p {\displaystyle \mathbf {r} \times \mathbf {p} } , because it does not include the mass of the object in question. </p> <h2 id="keplers-laws-of-planetary-motion">Kepler's laws of planetary motion</h2> <p class="note">Main article: <a href="/facts/Kepler%27s_laws_of_planetary_motion/jYQoIBjM">Kepler's laws of planetary motion</a></p> <p>Kepler's laws of planetary motion can be proved almost directly with the above relationships. </p> <h3>First law</h3> <p>The proof starts again with the equation of the two-body problem. This time the cross product is multiplied with the specific relative angular momentum r ¨ × h = − μ r 2 r r × h {\displaystyle {\ddot {\mathbf {r} }}\times \mathbf {h} =-{\frac {\mu }{r^{2}}}{\frac {\mathbf {r} }{r}}\times \mathbf {h} } </p><p>The left hand side is equal to the derivative d d t ( r ˙ × h ) {\textstyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\dot {\mathbf {r} }}\times \mathbf {h} \right)} because the angular momentum is constant. </p><p>After some steps (which includes using the <a href="/facts/Triple_product/46u8v6SV">vector triple product</a> and defining the scalar r ˙ {\displaystyle {\dot {r}}} to be the radial velocity, as opposed to the norm of the vector r ˙ {\displaystyle {\dot {\mathbf {r} }}} ) the right hand side becomes: − μ r 3 ( r × h ) = − μ r 3 ( ( r ⋅ v ) r − r 2 v ) = − ( μ r 2 r ˙ r − μ r v ) = μ d d t ( r r ) {\displaystyle -{\frac {\mu }{r^{3}}}\left(\mathbf {r} \times \mathbf {h} \right)=-{\frac {\mu }{r^{3}}}\left(\left(\mathbf {r} \cdot \mathbf {v} \right)\mathbf {r} -r^{2}\mathbf {v} \right)=-\left({\frac {\mu }{r^{2}}}{\dot {r}}\mathbf {r} -{\frac {\mu }{r}}\mathbf {v} \right)=\mu {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\mathbf {r} }{r}}\right)} </p><p>Setting these two expression equal and integrating over time leads to (with the constant of integration C {\displaystyle \mathbf {C} } ) r ˙ × h = μ r r + C {\displaystyle {\dot {\mathbf {r} }}\times \mathbf {h} =\mu {\frac {\mathbf {r} }{r}}+\mathbf {C} } </p><p>Now this equation is multiplied (<a href="/facts/Dot_product/tNz8MLjT">dot product</a>) with r {\displaystyle \mathbf {r} } and rearranged r ⋅ ( r ˙ × h ) = r ⋅ ( μ r r + C ) ⇒ ( r × r ˙ ) ⋅ h = μ r + r C cos θ ⇒ h 2 = μ r + r C cos θ {\displaystyle {\begin{aligned}\mathbf {r} \cdot \left({\dot {\mathbf {r} }}\times \mathbf {h} \right)&=\mathbf {r} \cdot \left(\mu {\frac {\mathbf {r} }{r}}+\mathbf {C} \right)\\\Rightarrow \left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)\cdot \mathbf {h} &=\mu r+rC\cos \theta \\\Rightarrow h^{2}&=\mu r+rC\cos \theta \end{aligned}}} </p><p>Finally one gets the <a href="/facts/Orbit_equation/SvJdAIxM">orbit equation</a><a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> r = h 2 μ 1 + C μ cos θ {\displaystyle r={\frac {\frac {h^{2}}{\mu }}{1+{\frac {C}{\mu }}\cos \theta }}} </p><p>which is the <a href="/facts/Conic_Sections_in_Polar_Coordinates/teqgLptX">equation of a conic section in polar coordinates</a> with <a href="/facts/Semi-latus_rectum/teqgLptX">semi-latus rectum</a> p = h 2 μ {\textstyle p={\frac {h^{2}}{\mu }}} and <a href="/facts/Eccentricity_(mathematics)/SzgpW5R7">eccentricity</a> e = C μ {\textstyle e={\frac {C}{\mu }}} . </p> <h3>Second law</h3> <p>The second law follows instantly from the second of the three equations to calculate the absolute value of the specific relative angular momentum.<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a> </p><p>If one connects this form of the equation d t = r 2 h d θ {\textstyle \mathrm {d} t={\frac {r^{2}}{h}}\,\mathrm {d} \theta } with the relationship d A = r 2 2 d θ {\textstyle \mathrm {d} A={\frac {r^{2}}{2}}\,\mathrm {d} \theta } for the area of a sector with an infinitesimal small angle d θ {\displaystyle \mathrm {d} \theta } (triangle with one very small side), the equation d t = 2 h d A {\displaystyle \mathrm {d} t={\frac {2}{h}}\,\mathrm {d} A} </p> <h3>Third law</h3> <p>Kepler's third is a direct consequence of the second law. Integrating over one revolution gives the <a href="/facts/Orbital_period/94f56uh2">orbital period</a><a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a> T = 2 π a b h {\displaystyle T={\frac {2\pi ab}{h}}} </p><p>for the area π a b {\displaystyle \pi ab} of an ellipse. Replacing the semi-minor axis with b = a p {\displaystyle b={\sqrt {ap}}} and the specific relative angular momentum with h = μ p {\displaystyle h={\sqrt {\mu p}}} one gets T = 2 π a 3 μ {\displaystyle T=2\pi {\sqrt {\frac {a^{3}}{\mu }}}} </p><p>There is thus a relationship between the semi-major axis and the orbital period of a satellite that can be reduced to a constant of the central body. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Specific_orbital_energy/TvB2yk8a">Specific orbital energy</a>, another conserved quantity in the two-body problem.</li> <li><a href="/facts/Classical_central-force_problem/jU7Ys65c">Classical central-force problem § Specific angular momentum</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Vallado, David A. (2001). Fundamentals of astrodynamics and applications (2nd ed.). Dordrecht: Kluwer Academic Publishers. pp. 20–30. ISBN 0-7923-6903-3. <a href="0-7923-6903-3" target="_blank">0-7923-6903-3</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Vallado, David A. (2001). Fundamentals of astrodynamics and applications (2nd ed.). Dordrecht: Kluwer Academic Publishers. pp. 20–30. ISBN 0-7923-6903-3. <a href="0-7923-6903-3" target="_blank">0-7923-6903-3</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>Vallado, David A. (2001). Fundamentals of astrodynamics and applications (2nd ed.). Dordrecht: Kluwer Academic Publishers. pp. 20–30. ISBN 0-7923-6903-3. <a href="0-7923-6903-3" target="_blank">0-7923-6903-3</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>Vallado, David A. (2001). Fundamentals of astrodynamics and applications (2nd ed.). Dordrecht: Kluwer Academic Publishers. pp. 20–30. ISBN 0-7923-6903-3. <a href="0-7923-6903-3" target="_blank">0-7923-6903-3</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> </ol>