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OWAO 2025, Теория — Задача 4. Tides on a Rod

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Всички материали от OWAO 2025

2025 ◇ Theoretical Round ◇ Problems

Внимание: Бележка към темата

All eight whole original tasks and 29 lettered parts, complete official worked solutions, alternatives and marking schemes are retained. Constants are native question page 5 and repeated key page 31, fully transcribed once as a shared reference. No paper/time/task/part total is printed, so no totals are inferred from marking criteria. Source mathematical/subscript/endpoint/equality discrepancies are preserved with attributed local notes. The separately printed cooler-star Criterion (x) remains inside original task 8; no extra problem is created.

Константи

Universal

Gravitational constant	$G=6.67\cdot10^{-11}\ \mathrm{m^3/(kg\cdot s^2)}$
Speed of light	$c=3.00\cdot10^8\ \mathrm{m/s}$
Planck constant	$h=6.63\cdot10^{-34}\ \mathrm{J\cdot s}$
Boltzmann constant	$k_B=1.38\cdot10^{-23}\ \mathrm{J/K}$
Gas constant	$\mathfrak R=8.314\ \mathrm{J/(mol\cdot K)}$
Proton mass	$m_p=1.673\cdot10^{-27}\ \mathrm{kg}$

Astronomical

Astronomical unit	$1\ \mathrm{au}=149.6\cdot10^6\ \mathrm{km}$
Parsec	$1\ \mathrm{pc}=206265\ \mathrm{au}$
Hubble constant	$H_0=70\ \mathrm{(km/s)/Mpc}$

Earth

Radius	$R_\oplus=6371\ \mathrm{km}$
Mass	$\mathfrak M_\oplus=5.97\cdot10^{24}\ \mathrm{kg}$
Obliquity of ecliptic	$\varepsilon=23.4^\circ$
Surface gravity	$g=9.81\ \mathrm{m/s^2}$
Orbital period	$T_\oplus=365.26\ \mathrm{days}$
Orbital eccentricity	$e_\oplus=0.0167$

Hydrogen spectrum

Lyman $\mathrm L\alpha$	$\lambda_{\mathrm L\alpha}=1215.7\ \text{Å}$
Balmer $\mathrm H\alpha$	$\lambda_{\mathrm H\alpha}=6562.8\ \text{Å}$

Jupiter

Radius	$R_J=6.99\cdot10^4\ \mathrm{km}$
Mass	$\mathfrak M_J=1.90\cdot10^{27}\ \mathrm{kg}$
Orbital radius	$a_J=5.20\ \mathrm{au}$
Orbital period	$T_J=11.86\ \mathrm{yr}$

Sun

Radius	$R_\odot=6.96\cdot10^5\ \mathrm{km}$
Mass	$\mathfrak M_\odot=1.99\cdot10^{30}\ \mathrm{kg}$
Absolute magnitude	$M_\odot=4.74^{\mathrm m}\ \text{(bol.)}$
Effective temperature	$T_\odot=5.8\cdot10^3\ \mathrm K$
Luminosity	$L_\odot=3.828\cdot10^{26}\ \mathrm W$
Color index	$(B-V)_\odot=+0.65^{\mathrm m}$

Emission constants

Stefan-Boltzmann	$\sigma=5.67\cdot10^{-8}\ \mathrm{W/(m^2\cdot K^4)}$
Wien’s displacement	$b=2898\ \mathrm{\mu m\cdot K}$

UBV… system — Mean wavelengths

U band	$\lambda_U=364\ \mathrm{nm}$
B band	$\lambda_B=442\ \mathrm{nm}$
V band	$\lambda_V=540\ \mathrm{nm}$
K band	$\lambda_K=2190\ \mathrm{nm}$

Условие

An experimental gradiometer satellite consists of a thin, rigid, massless rod of length $L=40\ \mathrm m$ with small masses $m=500\ \mathrm{kg}$ at its ends. Its center moves in a circular orbit at an altitude $h=400\ \mathrm{km}$ above the Earth’s surface. In equilibrium, the rod is oriented radially toward the Earth’s center.

Assume that the Earth’s gravitational field is the same as that of a point mass:

\mathbf g(\mathbf r)=-\frac{G\mathfrak M_\oplus}{r^3}\mathbf r.

Complete native tilted thin rod with both point masses m, rod length L and both endpoints.

a) Determine the orbital period of the satellite, neglecting the rod’s length.

b) Find a small correction to the satellite’s circular orbital speed that comes from the fact that it isn’t a point, but has a finite length $L$ :

v\approx\sqrt{\frac{G\mathfrak M_\oplus}{R_\oplus+h}}\left(1+[\text{SOME NUMBER}]\frac{L^2}{(R_\oplus+h)^2}\right).

c) Find the period of small oscillations of the satellite about the equilibrium in the orbital plane, assuming that the satellite’s center continues to move in a circular orbit.

Отговори

Решение

a) The radius of satellite’s orbit is $r_0=R_\oplus+h$ . For a circular orbit, the centripetal acceleration is provided by gravity:

\frac{v_0^2}{r_0}=\frac{G\mathfrak M_\oplus}{r_0^2}\quad\Longrightarrow\quad v_0=\sqrt{\frac{G\mathfrak M_\oplus}{r_0}},\quad\omega_0=\sqrt{\frac{G\mathfrak M_\oplus}{r_0^3}};

T_0=\frac{2\pi r_0}{v_0}=2\pi\sqrt{\frac{r_0^3}{G\mathfrak M_\oplus}}=2\pi\sqrt{\frac{(R_\oplus+h)^3}{G\mathfrak M_\oplus}}=

=2\times3.14\times\sqrt{\frac{(6.371\cdot10^6+0.4\cdot10^6)^3\ \mathrm{m^3}}{6.67\cdot10^{-11}\ \mathrm{m^3/(kg\cdot s^2)}\times5.97\cdot10^{24}\ \mathrm{kg}}}=5.5\cdot10^3\ \mathrm s=92^{\mathrm m}\approx1.5^{\mathrm h}.

b) Let $\ell=L/2$ be the half-length of the rod. The positions of its endpoints are $\mathbf r_\pm=\mathbf r_{\mathrm{COM}}\pm\boldsymbol\rho$ , where $|\boldsymbol\rho|=\ell$ . By definition, their average gives the center of mass (COM):

\mathbf r_{\mathrm{COM}}=\frac{\mathbf r_++\mathbf r_-}2,\qquad|\mathbf r_{\mathrm{COM}}|=r_0.

Since internal forces cancel, Newton’s second law for the system gives

2m\mathbf a_{\mathrm{COM}}=\mathbf F_++\mathbf F_-=m\mathbf g_++m\mathbf g_-\quad\Longrightarrow\quad\mathbf a_{\mathrm{COM}}=\frac{\mathbf g_++\mathbf g_-}2\quad\Longrightarrow\quad a_{\mathrm{COM}}=\frac{g_++g_-}2.

We now compute the gravitational acceleration at the endpoints using the binomial expansion, noting that $\ell/r_0\ll1$ :

g(r_0\pm\ell)=\frac{G\mathfrak M_\oplus}{(r_0\pm\ell)^2}=\frac{G\mathfrak M_\oplus}{r_0^2}\left(1\pm\frac\ell{r_0}\right)^{-2}=\frac{G\mathfrak M_\oplus}{r_0^2}\left[1\mp2\frac\ell{r_0}+3\frac{\ell^2}{r_0^2}+\cdots\right].

“Averaging” the two values cancels the odd term:

a_{\mathrm{COM}}=\frac{g(r_0+\ell)+g(r_0-\ell)}2\approx\frac{G\mathfrak M_\oplus}{r_0^2}\left(1+3\frac{\ell^2}{r_0^2}\right).

Alternative way. Let us denote $x=\ell/r_0\ll1$ . We start by simplifying the expression:

\frac1{(1+x)^2}+\frac1{(1-x)^2}=\frac{(1-x)^2+(1+x)^2}{(1+x)^2(1-x)^2}=2\cdot\frac{1+x^2}{(1-x^2)^2}=2\cdot\frac{1+x^2}{1-2x^2+x^4}=

=2\cdot\frac{(1+x^2)(1+2x^2)}{(1-2x^2+x^4)(1+2x^2)}=2\cdot\frac{\boxed{1+3x^2}+2x^4}{\boxed1+(2x^2-3)x^4}\overset{\text{up to }x^2}\approx2\cdot(1+3x^2).

This result simplifies the calculation of the center-of-mass acceleration:

a_{\mathrm{COM}}=\frac12(g_++g_-)=G\mathfrak M_\oplus\cdot\frac12\left[\frac1{(r_0+\ell)^2}+\frac1{(r_0-\ell)^2}\right]=

=\frac{G\mathfrak M_\oplus}{r_0^2}\cdot\frac12\left[\frac1{(1+x)^2}+\frac1{(1-x)^2}\right]\approx\frac{G\mathfrak M_\oplus}{r_0^2}\left(1+3\frac{\ell^2}{r_0^2}\right).

For a circular COM orbit,

a_{\mathrm{COM}}=\frac{v^2}{r_0}\quad\Longrightarrow\quad v^2\approx\frac{G\mathfrak M_\oplus}{r_0}\left(1+3\frac{\ell^2}{r_0^2}\right)\quad\Longrightarrow\quad v\approx\sqrt{\frac{G\mathfrak M_\oplus}{r_0}}\left(1+\frac38\frac{L^2}{(R_\oplus+h)^2}\right).

Thus, $[\text{SOME NUMBER}]=\frac38$ , and the relative correction to the velocity is

\frac{v-v_0}{v_0}=\frac38\left(\frac L{R_\oplus+h}\right)^2=\frac38\times\left(\frac{40\ \mathrm m}{6.371\cdot10^6\ \mathrm m+0.4\cdot10^6\ \mathrm m}\right)^2=1.3\cdot10^{-11}.

c) When the rod is tilted by a small angle $\theta$ , each end feels a force that is

slightly stronger/weaker because of being nearer/farther, and
points in a direction turned by a small angle because the end is shifted sideways.

The combination of these effects produces a net torque about the center of mass, which acts as a restoring torque:

M=F_+\cdot\frac L2\sin\theta_+-F_-\cdot\frac L2\sin\theta_-,

where

F_\pm=m\cdot g(r_0\pm\ell\cos\theta),\qquad\theta_\pm\approx\theta\mp\frac{\ell\sin\theta}{r_0}.

We begin by expanding the gravitational acceleration at each end of the rod $g(r_0\pm\ell\cos\theta)$ , noting that $\ell/r_0\ll1$ :

g(r_0\pm\ell\cos\theta)=\frac{G\mathfrak M_\oplus}{(r_0\pm\ell\cos\theta)^2}=\frac{G\mathfrak M_\oplus}{r_0^2}\left(1\pm\frac{\ell\cos\theta}{r_0}\right)^{-2}=\frac{G\mathfrak M_\oplus}{r_0^2}\left[1\mp2\frac{\ell\cos\theta}{r_0}+\cdots\right].

For small angles $\theta\ll1$ , we use the approximations $\sin\theta\approx\theta$ , $\cos\theta\approx1$ . The restoring torque about the center of mass, calculated to first order, is

M\approx\frac{G\mathfrak M_\oplus m}{r_0^2}\cdot\ell\theta\cdot\left[\left(1-2\frac\ell{r_0}\right)\cdot\left(1-\frac\ell{r_0}\right)-\left(1+2\frac\ell{r_0}\right)\cdot\left(1+\frac\ell{r_0}\right)\right]\approx-\frac{6G\mathfrak M_\oplus m}{r_0^2}\frac{\ell^2}{r_0}\theta.

The moment of inertia of the two end masses about the COM in the orbital plane is

J=2m\ell^2.

Using the torque above, we obtain the equation of motion for a harmonic oscillator

J\ddot\theta=M\quad\Longrightarrow\quad\ddot\theta+3\frac{G\mathfrak M_\oplus}{r_0^3}\theta=0.

Therefore, the angular frequency and period of small librations are:

\omega_{\mathrm{osc}}=\sqrt3\cdot\sqrt{\frac{G\mathfrak M_\oplus}{r_0^3}}=\sqrt3\omega_0,

T_{\mathrm{osc}}=\frac{2\pi}{\omega_{\mathrm{osc}}}=\frac{T_0}{\sqrt3}=\frac{5.5\cdot10^3\ \mathrm s}{1.73}=3.2\cdot10^3\ \mathrm s=53^{\mathrm m}\approx0.9^{\mathrm h}.

Elegant shortcut. In the reference frame co-rotating with the rod’s center of mass, the effective potential combines Earth’s gravity and the centrifugal potential:

U(r)=-\frac{G\mathfrak M_\oplus}r-\frac12\omega_0^2r^2.

The center of mass is in mechanical equilibrium at the orbital radius $r=r_0$ , satisfying:

U'(r)\big|_{r=r_0}=0=\frac{G\mathfrak M_\oplus}{r_0^2}-\omega_0^2r_0\quad\Longrightarrow\quad\omega_0^2=\frac{G\mathfrak M_\oplus}{r_0^3},

which is the standard expression for the orbital angular velocity.

The tidal acceleration near $r_0$ is given by the second derivative of the potential. To first order:

a_{\mathrm{tidal}}(r)\approx U''(r_0)\cdot(r-r_0)=\left(-\frac{2G\mathfrak M_\oplus}{r_0^3}-\omega_0^2\right)\cdot(r-r_0)=-\frac{3G\mathfrak M_\oplus}{r_0^3}\cdot(r-r_0).

For small angular deflections, the system can be modeled as two simple pendulums of length $L/2$ , oscillating in the effective tidal field, each in a uniform field with acceleration

g_{\mathrm{eff}}=\frac{3G\mathfrak M_\oplus}{r_0^3}\cdot\frac L2.

Complete effective tidal pendulum diagram: both masses, tilted full rod, fixed center label/arrow, dashed reference line, all upper/lower blue arrows and both geff labels.

The corresponding oscillation period is

T_{\mathrm{osc}}=2\pi\sqrt{\frac{L/2}{g_{\mathrm{eff}}}}=2\pi\sqrt{\frac{r_0^3}{3G\mathfrak M_\oplus}}=\frac{T_0}{\sqrt3}.

Marking Scheme:

Question (a) Orbital period
- Correct expression for the period — 2 pt.
- Calculations and correct answer — 2 pt.
Question (b) Correction to circular speed
- Correct Newton’s 2nd law — 2 pt.
- Correct series expansion — 4 pt.
- Correct numerical result — 2 pt.
Question (c) Period of small oscillations
- Correct physical laws and relations that can be used to derive the differential equation for harmonic oscillations — 4 pt.
- Correct harmonic motion equation — 2 pt.
- Correct numerical result — 2 pt.

Теми и трудност

Оценена трудност: 4/5.

Небесна механика · Механика · Гравитационни системи · Механични трептения · Въртене на твърдо тяло

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: OWAO_2025_Theoretical_Problems.pdf · официални решения: OWAO_2025_Theoretical_Solutions.pdf