The parallel axis theorem relates the moment of inertia $I$ about an axis to $I_{\mathrm{CM}}$ about a parallel axis through the centre of mass, separated by perpendicular distance $d$:In the defining mass integral, expanding the squared distance produces a cross term whose first mass moment about the centre of mass is zero.

For a thin planar rigid body, the perpendicular axis theorem applies to two perpendicular axes $x,y$ in its plane and the perpendicular axis $z$, all meeting at the same point:Indeed, $I_x=\int y^{2}dm$, $I_y=\int x^{2}dm$, and $I_z=\int (x^2+y^2)dm$. The planar condition is essential; this is not an identity for an arbitrary three-dimensional rigid body.

Take Jupiter's heliocentric velocity as ${\mathbf{V}}_J=13\hat{\mathbf{x}}\mathrm{km}{\mathrm{s}}^{-1}$. Conservation of energy in the Jupiter frame makes the incoming and outgoing asymptotic speeds both $10\mathrm{km}{\mathrm{s}}^{-1}$. For the $90^{\circ }$ encounter, an outgoing velocity in Jupiter's direction requires the incoming Jupiter-frame velocity to be perpendicular to it:Use a Galilean transformation back to the Sun frame:The heliocentric speeds are thereforeThe incoming direction makes an angle $\arctan (10/13)\simeq 37.6^{\circ }$ with Jupiter's motion; its two signs describe mirror-image encounter planes. The gravitational assist transfers orbital energy from Jupiter to the spacecraft: the spacecraft's specific kinetic energy increases by $\frac{1}{2}(23^2-16.4^2)\mathrm{k}{\mathrm{m}}^2{\mathrm{s}}^{-2}$, or, using the unrounded squared incoming speed, exactly $130\mathrm{k}{\mathrm{m}}^2{\mathrm{s}}^{-2}=1.30\times 10^8\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. There is no gain in asymptotic speed in the Jupiter frame; the gain is a consequence of changing frames while the planet is moving.

A $90^{\circ }$ gravitational scattering angle has $\cot (\varphi /2)=1$. ThusThe denominator uses the squared speed in $\mathrm{m}{\mathrm{s}}^{-1}$, not its numerical value in $\mathrm{km}{\mathrm{s}}^{-1}$. From b, the corresponding periapsis distance is $r_{\min }=(\sqrt{2}-1)b\simeq 5.53\times 10^8\mathrm{m}$; this differs substantially from the impact parameter because of gravitational focusing.

Let $h=J/m=b{v}_0$ be the specific angular momentum, and measure the orbital angle $\vartheta$ from periapsis. Set $w=1/r$. From $\dot{\vartheta }=h/r^2$, differentiating gives $\dot{r}=-hdw/d\vartheta$. The radial equation of motion $\ddot{r}-r{\dot{\vartheta }}^2=-{\mu }_J/r^2$ becomes the Binet equationIts solution symmetric about periapsis isFor example, the expression for the orbital eccentricity follows by substituting $w$ and $\dot{r}=-h{w}^{\prime }$ into the energy equation: $E/m={\mu }_J^2(e^2-1)/(2h^2)=v_0^2/2$. Since $e>1$, this is a hyperbolic Kepler orbit.

The outgoing asymptote occurs at ${\vartheta }_{\infty }$ with $\cos {\vartheta }_{\infty }=-1/e$, where $\pi /2<{\vartheta }_{\infty }<\pi$. The incoming velocity points inward along the incoming radial asymptote, whereas the outgoing velocity points outward. Therefore the change in velocity direction is $\varphi =2{\vartheta }_{\infty }-\pi$, not $2{\vartheta }_{\infty }$. It follows thatFor $b>0$, this gives $0<\varphi <\pi$: a large impact parameter produces a small deflection, and the ideal point-mass limit $b\to 0^{+}$ approaches a reversal.

Put ${\mu }_J=G{M}_J$. At the periapsis distance, the radial velocity vanishes. Insert $E=m{v}_0^2/2$ and $J=mb{v}_0$ into a:The positive root isRationalizing gives the numerically stable alternative $r_{\min }=b^2{v}_0^2/[\sqrt{{\mu }_J^2+b^2{v}_0^4}+{\mu }_J]$. Attraction makes $r_{\min }<b$. This is a point-planet scattering calculation; a noncolliding physical encounter must also have $r_{\min }$ outside Jupiter's surface. At $b=0$, the limiting formula gives a radial collision, not a flyby.

Use a reference frame moving with Jupiter, neglect the spacecraft's effect on the planet and neglect the Sun during the short encounter. The Newtonian central force makes the orbit planar and conserves the angular momentum $J=m{r}^2\dot{\vartheta }$. In plane polar coordinates, the kinetic energy is $m({\dot{r}}^2+r^2{\dot{\vartheta }}^2)/2$. Choosing zero gravitational potential energy at infinity givesThe terms are respectively radial kinetic energy, transverse kinetic energy, and attractive gravitational potential energy. The second term is often combined with the third into the radial effective potential. Far from Jupiter,Here $b$ is the perpendicular impact parameter of the incoming asymptote, not the eventual periapsis distance.

Take positive $x$ horizontally toward the downhill end of the plane, with $0\le \theta <\pi /2$. The normal force is $N=mg\cos \theta$, since there is no acceleration perpendicular to the stationary plane. Along the plane, the only tangential force is $mg\sin \theta$, so Newton's second law gives $a_{\parallel }=g\sin \theta$.

Its horizontal projection is thereforeEquivalently, the normal force has horizontal component $N\sin \theta$ while gravity has none, giving $m{a}_x=N\sin \theta$. For a plane whose height increases in the positive $x$ direction, the answer instead carries a minus sign. A horizontal plane gives zero acceleration; the horizontal projection also tends to zero for a vertical plane.

For independent generalized coordinates $q_i$, their generalized velocities ${\dot{q}}_i$ and a Lagrangian $L(q,\dot{q},t)$, the Euler-Lagrange equations areHere the partial derivatives treat $q_i,{\dot{q}}_i,t$ as independent arguments, whereas $d/dt$ follows the motion. For a conservative mechanical system, choose coordinates satisfying the mechanical constraints, express the kinetic energy $T$ and potential energy $V$ in those coordinates, set $L=T-V$, and evaluate these equations to obtain the equations of motion. If nonconservative generalized forces $Q_i$ are present, they appear on the right-hand side.

A cyclic coordinate is absent explicitly from $L$: $\partial L/\partial q_i=0$, even though $L$ may depend on ${\dot{q}}_i$. Its canonical momentum is consequently conserved,This does not require ${\dot{q}}_i=0$ or even that ${\dot{q}}_i$ be constant; the relation between canonical momentum and generalized velocity depends on the Lagrangian.

Let $u>0$ be the projectile's final speed, taking its deflection above the $x$-axis. Momentum conservation givesSubstituting these into kinetic energy conservation, $V^2=u^2+2|{\mathbf{v}}_2{|}^2$, gives $3u^2-2Vu\cos 30^{\circ }-V^2=0$. Its positive root is $u=V(\sqrt{3}+\sqrt{15})/6$. HenceThese vectors satisfy ${\mathbf{v}}_1+2{\mathbf{v}}_2=(V,0)$ and $|{\mathbf{v}}_1{|}^2+2|{\mathbf{v}}_2{|}^2=V^2$. For the reflected deflection, reverse both $y$ components.

For masses $m$ and $2m$, the constant centre-of-mass velocity is $(V/3){\mathbf{e}}_x$. Initially the laboratory velocities are $V{\mathbf{e}}_x$ and zero, while the centre-of-mass frame velocities are $(2V/3){\mathbf{e}}_x$ and $-(V/3){\mathbf{e}}_x$. After the elastic collision they areAdding $(V/3,0)$ produces the lower-left panel's laboratory velocity vectors. The given $30^{\circ }$ laboratory angle impliesThe centre-of-mass frame scattering angle is therefore different from the laboratory angle. Reflection across the $x$-axis gives the equally possible negative-angle configuration.

For equal masses $m$, the reduced mass is $m/2$. The kinetic energy of their centre of mass cannot be removed by an isolated collision, since total momentum is fixed. Therefore the maximum energy available for conversion to internal forms is the initial relative kinetic energy,It is attained by a perfectly inelastic collision in which the final relative velocity is zero. This concerns the convertible initial kinetic energy, not an arbitrary pre-existing internal energy reservoir.

For two particles, let $M=m_1+m_2$ and ${\mathbf{V}}_{\mathrm{cm}}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/M$. The centre-of-mass frame velocities are ${\mathbf{v}}_i^{\ast }={\mathbf{v}}_i-{\mathbf{V}}_{\mathrm{cm}}$, so $m_1{\mathbf{v}}_1^{\ast }+m_2{\mathbf{v}}_2^{\ast }=0$. Thus before and after an isolated collision the two momentum vectors are equal and opposite. For an elastic collision their common magnitude is also unchanged: only their direction rotates.

The kinetic energy decomposition about the center of mass separates unchanged bulk translation from relative motion:The reduced mass $\mu$ reduces the calculation to one relative velocity. After solving the geometry in the centre-of-mass frame, add ${\mathbf{V}}_{\mathrm{cm}}$ to each velocity to return to the laboratory. The four panels below illustrate this for the subsequent example.

The specified rocket fuel fraction gives $m(T)=2m_0/3$, so $\alpha T=m_0/3$. Imposing $v(T)=0$ yieldsThen $h(T)=0$ fixesNumerically,The associated constant mass loss rate is $\alpha =m_0/(3T)\simeq 88.78\mathrm{kg}{\mathrm{s}}^{-1}$. The initial upward thrust acceleration already exceeds $g$ and increases as mass falls, so the downward speed decreases monotonically to zero. This calculation assumes constant lunar gravitational acceleration, constant relative rocket exhaust speed, and negligible drag.

Height decreases at rate $\dot{h}=-v$. Set $m=m_0-\alpha t$; an antiderivative needed for the rocket equation isConsequentlyDifferentiating this expression recovers $\dot{h}=-v$ and substituting $t=0$ gives $h(0)=h_0$, which check both the sign and the integration constant.

Integrating the rocket equation with $v(0)=v_0$ givesThis formula holds during the constant-flow burn, while $m_0-\alpha t>0$ and fuel remains. The negative logarithmic term represents the slowing due to upward rocket thrust.

Choose downward velocity $v$ positive. The remaining rocket mass is $m(t)=m_0-\alpha t$, and the upward rocket thrust has magnitude $\alpha u$, where $u>0$ is the relative rocket exhaust speed. The rocket equation including gravity isThis follows by applying momentum conservation to the rocket and its just-expelled rocket propellant, with gravity supplying the external impulse. Applying $d(mv)/dt=mg$ to the rocket alone would omit the exhaust momentum flux.

The inward radial kick also leaves angular momentum unchanged and increases orbital energy by $m{\delta }^2/2$. Its eccentricity and semi-major axis equal those in case $b$, but the orientation of its eccentricity vector is reflected about the initial radius. The satellite initially moves inward on its ellipse, toward pericentre. Neither radial kick decreases the orbital energy.

For a small opposing tangential kick,The new tangential speed is less than the circular orbital speed, so the kick point is the apocentre of an ellipse. The orbit falls inward from the original circle. The signs assume $0<\delta \ll v_c$; a large velocity reversal is outside the stated small-impulse regime.