Gravitation Concept Page - 9

Example

Conservation of Angular Momentum of masses in orbits

Two ordinary satellites are revolving round the Earth in same elliptical orbit, comment on the angular momentum of the system of masses. As in the absence of external torque, L$L$ (Angular Momentum) is conserved.Thus, L$L$ should be conserved for the two satellites revolving in same elliptical orbit round the Earth and angular momentum for both must be same.

Example

Condition on minimum and maximum distance of approach

Example: A planet A$A$ moves along an elliptical orbit around the Sun. At the
moment when it was at the distance r0$r_0$ from the Sun its velocity was equal to v0$v_0$ and the angle between the radius vector r0$r_0$  and the velocity vector v0→$\overrightarrow{v_0}$ was equal to a$a$. Find the  maximum and minimum distances that will separate this planet from the Sun during its orbital motion.

Solution:
From the conservation of angular momentum about the Sun,
mv0r0sinα=mv1r1=mv2r2$m{v}_0{r}_0\sin \alpha =m{v}_1{r}_1=m{v}_2{r}_2$ or, v1r1=v2r2=v0r0sinα$v_1{r}_1=v_2{r}_2=v_0{r}_0\sin \alpha$ (1)

From conservation of mechanical energy,

12mv20−γmsmr0=12mv21−γmsmr1${\displaystyle \frac{1}{2}m{v}_0^2-\frac{\gamma m_{s}m}{r_0}=\frac{1}{2}m{v}_1^2-\frac{\gamma m_{s}m}{r_1}}$

or,

v202−γmsr0=v20r20sin2α2r21−γmsr1${\displaystyle \frac{v_0^2}{2}-\frac{\gamma m_s}{r_0}=\frac{v_0^2{r}_0^2{\sin }^2\alpha }{2r_1^2}-\frac{\gamma m_s}{r_1}}$ (Using 1)

or, (v20−2γmsr0)r21+2γmsr1−v20r20sin2α=0${\displaystyle (v_0^2-\frac{2\gamma m_s}{r_0})r_1^2+2\gamma m_s{r}_1-v_0^2{r}_0^2{\sin }^2\alpha =0}$

So,

r1=−2γms±4γ2m2s+4(v20r20sin2α)(v20−2γmsr0)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2(v20−2γmsr0)${\displaystyle r_1=\frac{{\displaystyle -2\gamma m_s\pm \sqrt{4{\gamma }^2{m}_s^2+4(v_0^2{r}_0^2{\sin }^2\alpha )(v_0^2-\frac{2\gamma m_s}{r_0})}}}{{\displaystyle 2(v_0^2-\frac{2\gamma m_s}{r_0})}}}$

=1±1−v20r20sin2αγms(2r0−v20γms)−−−−−−−−−−−−−−−−−−−−−−−⎷(2r0−v20γms)=r0[1±1−(2−η)ηsin2α−−−−−−−−−−−−−−−√](2−η)${\displaystyle =\frac{{\displaystyle 1\pm \sqrt{1-\frac{v_0^2{r}_0^2{\sin }^2\alpha }{\gamma m_s}(\frac{2}{r_0}-\frac{v_0^2}{\gamma m_s})}}}{{\displaystyle (\frac{2}{r_0}-\frac{v_0^2}{\gamma m_s})}}=\frac{r_0[1\pm \sqrt{1-(2-\eta )\eta {\sin }^2\alpha }]}{(2-\eta )}}$

where η=v20r0γms${\displaystyle \eta =\frac{v_0^2{r}_0}{\gamma m_s}}$, (ms$m_s$ is the mass of the Sun).

Example

Condition for elliptical, parabolic and circular orbits

Example

Problem on Angular momentum conservation in elliptical orbits

Example: A planet of mass m$m$ revolves in elliptical orbit around the sun so
that its maximum and minimum distances from the sun are equal to
ra$r_a$ and rp$r_p$ respectively. Find the angular momentum of this planet relative to the sun.

Solution:
For a planet moving around the sun in an orbit, angular momentum (L)$(L)$ is constant.
L=mvara=mvbrb⇒va=vbrbra$L=m{v}_a{r}_a=m{v}_b{r}_b\Rightarrow v_a=v_b{\displaystyle \frac{r_b}{r_a}}$
By conserving energy between the two points, farthest and nearest.
−GMmrb+12mv2b=−GMmra+12mv2a${\displaystyle \frac{-GMm}{r_b}}+{\displaystyle \frac{1}{2}}m{v}_b^2={\displaystyle \frac{-GMm}{r_a}}+{\displaystyle \frac{1}{2}}m{v}_a^2$
⇒GM(1ra−1rb)=12(v2a−v2b)$\Rightarrow GM({\displaystyle \frac{1}{r_a}}-{\displaystyle \frac{1}{r_b}})={\displaystyle \frac{1}{2}}(v_a^2-v_b^2)$
Substituting va$v_a$ from momentum equation.
⇒GM(1ra−1rb)=12[v2b(r2br2a)−v2b]$\Rightarrow GM({\displaystyle \frac{1}{r_a}}-{\displaystyle \frac{1}{r_b}})={\displaystyle \frac{1}{2}}[v_b^2\left({\displaystyle \frac{r_b^2}{r_a^2}}\right)-v_b^2]$

⇒GM[rb−rararb]=v2b2[(rb−ra)(rb+ra)r2a]$\Rightarrow GM\left[{\displaystyle \frac{r_b-r_a}{r_a{r}_b}}\right]={\displaystyle \frac{v_b^2}{2}}\left[{\displaystyle \frac{(r_b-r_a)(r_b+r_a)}{r_a^2}}\right]$

⇒v2b=2GMra(ra+rb)rb$\Rightarrow v_b^2={\displaystyle \frac{2GM{r}_a}{(r_a+r_b)r_b}}$

∴L=mvbrb=m2GMra(ra+rb)rb−−−−−−−−−√rb=m2GMrarbra+rb−−−−−−−−√$\therefore L=m{v}_b{r}_b=m\sqrt{{\displaystyle \frac{2GM{r}_a}{(r_a+r_b)r_b}}}{r}_b=m\sqrt{{\displaystyle \frac{2GM{r}_a{r}_b}{r_a+r_b}}}$

Example

Use angular momentum conservation in problems where nature of orbit of earth changes

Example:
A space vehicle approaching a planet has a speed v$v$, when it is very far from the planet. At that moment tangent of its trajectory would miss the centre of the planet by distance R$R$. If the planet has mass M$M$ and radius r$r$, what is the smallest value of R$R$ in order that the resulting orbit of the space vehicle will just miss the surface of the planet?

Solution:
Consider conservation of angular momentum,mvR=mv0r$mvR=m{v}_{0}r$⟹v0=Rrv$⟹v_0={\displaystyle \frac{R}{r}}v$From conservation of energy,12mv2−0=12mv20−GMmr${\displaystyle \frac{1}{2}}m{v}^2-0={\displaystyle \frac{1}{2}}m{v}_0^2-{\displaystyle \frac{GMm}{r}}$
12mv2−0=12mv2(Rr)2−GMmr${\displaystyle \frac{1}{2}}m{v}^2-0={\displaystyle \frac{1}{2}}m{v}^2({\displaystyle \frac{R}{r}}{)}^2-{\displaystyle \frac{GMm}{r}}$Rearranging to get the value of R gives,R2r2=2GMrv2+1${\displaystyle \frac{R^2}{r^2}}={\displaystyle \frac{2GM}{r{v}^2}}+1$
⟹R=rv[v2+2GMr]1/2$⟹R={\displaystyle \frac{r}{v}}[v^2+{\displaystyle \frac{2GM}{r}}{]}^{1/2}$

Example

Energy of satellites

Example: The energy required to remove an Earth's satellite of mass m$m$  from its orbit of radius r$r$ to infinity?

Solution: For a satellite under motion,
P.E.=−GMmr$P.E.=-{\displaystyle \frac{GMm}{r}}$
K.E.=GMm2r$K.E.={\displaystyle \frac{GMm}{2r}}$
T.E.=−GMm2r$T.E.=-{\displaystyle \frac{GMm}{2r}}$
The energy required to remove an Earth's satellite of mass m$m$ from its
orbit of radius r$r$ to infinity is called as Binding Energy,
AS  B.E.=−T.E$B.E.=-T.E$.
∴B.E.=+GMm2r

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