Съдържание

IPhO 2005, Теория — Задача 1. AN ILL FATED SATELLITE

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 2005

36th International Physics Olympiad. Salamanca (España) 2005 — Th 1 AN ILL FATED SATELLITE

Условие

The most frequent orbital manoeuvres performed by spacecraft consist of velocity variations along the direction of flight, namely accelerations to reach higher orbits or brakings done to initiate re-entering in the atmosphere. In this problem we will study the orbital variations when the engine thrust is applied in a radial direction.

Grey-toned artist's image of the Earth surrounded by several inclined elliptical satellite orbits, with small satellites on the orbits and a bright point of light at the upper right. — Image: ESA

To obtain numerical values use: Earth radius $R_T = 6.37\cdot 10^{6}\ \mathrm{m}$ , Earth surface gravity $g = 9.81\ \mathrm{m/s^2}$ , and take the length of the sidereal day to be $T_0 = 24.0\ \mathrm{h}$ .

We consider a geosynchronous $^{1}$ communications satellite of mass $m$ placed in an equatorial circular orbit of radius $r_0$ . These satellites have an “apogee engine” which provides the tangential thrusts needed to reach the final orbit.

Marks are indicated at the beginning of each subquestion, in parenthesis.

Question 1

1.1 (0.3) Compute the numerical value of $r_0$ . [0,3 т.]

1.2 (0.3+0.1) Give the analytical expression of the velocity $v_0$ of the satellite as a function of $g$ , $R_T$ , and $r_0$ , and calculate its numerical value. [0,4 т.]

1.3 (0.4+0.4) Obtain the expressions of its angular momentum $L_0$ and mechanical energy $E_0$ , as functions of $v_0$ , $m$ , $g$ and $R_T$ . [0,8 т.]

Once this geosynchronous circular orbit has been reached (see Figure F-1), the satellite has been stabilised in the desired location, and is being readied to do its work, an error by the ground controllers causes the apogee engine to be fired again. The thrust happens to be directed towards the Earth and, despite the quick reaction of the ground crew to shut the engine off, an unwanted velocity variation $\Delta v$ is imparted on the satellite. We characterize this boost by the parameter $\beta = \Delta v / v_0$ . The duration of the engine burn is always negligible with respect to any other orbital times, so that it can be considered as instantaneous.

Dashed circular orbit of radius r0 around a grey Earth; at the top point the satellite m has velocity v0 pointing left and a small downward arrow Δv pointing towards the Earth. — F-1

Question 2

Suppose $\beta < 1$ .

2.1 (0.4+0.5) Determine the parameters of the new orbit $^{2}$ , semi-latus-rectum $l$ and eccentricity $\varepsilon$ , in terms of $r_0$ and $\beta$ . [0,9 т.]

2.2 (1.0) Calculate the angle $\alpha$ between the major axis of the new orbit and the position vector at the accidental misfire. [1 т.]

2.3 (1.0+0.2) Give the analytical expressions of the perigee $r_{min}$ and apogee $r_{max}$ distances to the Earth centre, as functions of $r_0$ and $\beta$ , and calculate their numerical values for $\beta = 1/4$ . [1,2 т.]

2.4 (0.5+0.2) Determine the period of the new orbit, $T$ , as a function of $T_0$ and $\beta$ , and calculate its numerical value for $\beta = 1/4$ . [0,7 т.]

$^{1}$ Its revolution period is $T_0$ .

$^{2}$ See the “hint”.

Question 3

3.1 (0.5) Calculate the minimum boost parameter, $\beta_{esc}$ , needed for the satellite to escape Earth gravity. [0,5 т.]

3.2 (1.0) Determine in this case the closest approach of the satellite to the Earth centre in the new trajectory, $r'_{min}$ , as a function of $r_0$ . [1 т.]

Dotted circle of radius r0 around the Earth; from the top point the satellite with velocity v0 to the left and Δv downwards follows a curved escape trajectory towards the lower right, ending in the asymptotic direction v∞; the angle φ is marked at the Earth between the vertical and the asymptote direction, and the impact parameter b is the distance between the asymptote and a parallel dashed line through the Earth. — F-2

Question 4

Suppose $\beta > \beta_{esc}$ .

4.1 (1.0) Determine the residual velocity at the infinity, $v_\infty$ , as a function of $v_0$ and $\beta$ . [1 т.]

4.2 (1.0) Obtain the “impact parameter” $b$ of the asymptotic escape direction in terms of $r_0$ and $\beta$ . (See Figure F-2). [1 т.]

4.3 (1.0+0.2) Determine the angle $\phi$ of the asymptotic escape direction in terms of $\beta$ . Calculate its numerical value for $\beta = \dfrac{3}{2}\beta_{esc}$ . [1,2 т.]

HINT

Under the action of central forces obeying the inverse-square law, bodies follow trajectories described by ellipses, parabolas or hyperbolas. In the approximation $m \ll M$ the gravitating mass $M$ is at one of the focuses. Taking the origin at this focus, the general polar equation of these curves can be written as (see Figure F-3)

Conic curve (part of an ellipse opening to the right) with the mass M at a focus; the radius vector r goes from M to the mass m on the curve at polar angle θ measured from a dashed horizontal axis. — F-3

r(\theta) = \frac{l}{1 - \varepsilon\,\cos\theta}

where $l$ is a positive constant named the semi-latus-rectum and $\varepsilon$ is the eccentricity of the curve. In terms of constants of motion:

l = \frac{L^2}{G M m^2} \qquad \text{and} \qquad \varepsilon = \left(1 + \frac{2 E L^2}{G^2 M^2 m^3}\right)^{1/2}

where $G$ is the Newton constant, $L$ is the modulus of the angular momentum of the orbiting mass, with respect to the origin, and $E$ is its mechanical energy, with zero potential energy at infinity.

We may have the following cases:

i) If $0 \le \varepsilon < 1$ , the curve is an ellipse (circumference for $\varepsilon = 0$ ).

ii) If $\varepsilon = 1$ , the curve is a parabola.

iii) If $\varepsilon > 1$ , the curve is a hyperbola.

COUNTRY CODE	STUDENT CODE	PAGE NUMBER	TOTAL No OF PAGES

Th 1 ANSWER SHEET

Question	Analytical results	Numerical results	Marking guideline
1.1		$r_0 =$	0.3
1.2	$v_0 =$	$v_0 =$	0.4
1.3	$L_0 =$ ; $E_0 =$		0.4; 0.4
2.1	$l =$ ; $\varepsilon =$		0.4; 0.5
2.2		$\alpha =$	1.0
2.3	$r_{max} =$ ; $r_{min} =$	$r_{max} =$ ; $r_{min} =$	1.2
2.4	$T =$	$T =$	0.7
3.1		$\beta_{esc} =$	0.5
3.2	$r'_{min} =$		1.0
4.1	$v_\infty =$		1.0
4.2	$b =$		1.0
4.3	$\phi =$	$\phi =$	1.2

Отговори

Решение

1.1 and 1.2

\left.\begin{aligned} G\frac{M_T m}{r_0^2} &= m\frac{v_0^2}{r_0}\\ v_0 &= \frac{2\pi r_0}{T_0}\\ g &= \frac{G M_T}{R_T^2}\end{aligned}\right\}\Rightarrow\left\{\begin{aligned} & r_0 = \left(\frac{g R_T^2 T_0^2}{4\pi^2}\right)^{1/3} \quad\Rightarrow\quad \boxed{r_0 = 4.22\cdot 10^{7}\ \mathrm{m}}\\ & \boxed{v_0 = R_T\sqrt{\frac{g}{r_0}}} \quad\Rightarrow\quad \boxed{v_0 = 3.07\cdot 10^{3}\ \mathrm{m/s}}\end{aligned}\right.

1.3

L_0 = r_0\, m v_0 = \frac{g R_T^2}{v_0^2} m v_0 \quad\Rightarrow\quad \boxed{L_0 = \frac{m g R_T^2}{v_0}}

E_0 = \frac{1}{2} m v_0^2 - G\frac{M_T m}{r_0} = \frac{1}{2} m v_0^2 - \frac{g R_T^2\, m}{r_0} = \frac{1}{2} m v_0^2 - m v_0^2 \quad\Rightarrow\quad \boxed{E_0 = -\frac{1}{2} m v_0^2}

2.1

The value of the semi-latus-rectum $l$ is obtained taking into account that the orbital angular momentum is the same in both orbits. That is

l = \frac{L_0^2}{G M_T\, m^2} = \frac{m^2 g^2 R_T^4}{v_0^2}\,\frac{1}{g R_T^2\, m^2} = \frac{g R_T^2}{v_0^2} = r_0 \qquad\Rightarrow\qquad \boxed{l = r_0}

The eccentricity value is

\varepsilon^2 = 1 + \frac{2 E L_0^2}{G^2 M_T^2 m^3}

where $E$ is the new satellite mechanical energy

E = \frac{1}{2} m\left(v_0^2 + \Delta v^2\right) - G\frac{M_T m}{r_0} = \frac{1}{2} m\Delta v^2 + E_0 = \frac{1}{2} m\Delta v^2 - \frac{1}{2} m v_0^2

that is

E = \frac{1}{2} m v_0^2\left(\frac{\Delta v^2}{v_0^2} - 1\right) = \frac{1}{2} m v_0^2\left(\beta^2 - 1\right)

Combining both, one gets $\boxed{\varepsilon = \beta}$

This is an elliptical trajectory because $\varepsilon = \beta < 1$ .

2.2

The initial and final orbits cross at P, where the satellite engine fired instantaneously (see Figure 4). At this point

Dashed circular orbit of radius r0 around the Earth and the new solid elliptical orbit with the Earth at its left focus; at the crossing point P at the top the velocity vector v0 (left) and Δv (down) combine; the angle α = π/2 is marked between the vertical radius and the major axis; the horizontal major axis is labelled r_min to the left of the Earth and r_max to the right. — Figure 4

r\left(\theta = \alpha\right) = r_0 = \frac{r_0}{1 - \beta\cos\alpha} \quad\Rightarrow\quad \boxed{\alpha = \frac{\pi}{2}}

2.3

From the trajectory expression one immediately obtains that the maximum and minimum values of $r$ correspond to $\theta = 0$ and $\theta = \pi$ respectively (see Figure 4). Hence, they are given by

r_{max} = \frac{l}{1-\varepsilon} \qquad r_{min} = \frac{l}{1+\varepsilon}

that is

\boxed{r_{\max} = \frac{r_0}{1-\beta}} \quad \text{and} \quad \boxed{r_{\min} = \frac{r_0}{1+\beta}}

For $\beta = 1/4$ , one gets

\boxed{r_{max} = 5.63\cdot 10^{7}\ \mathrm{m};\quad r_{min} = 3.38\cdot 10^{7}\ \mathrm{m}}

The distances $r_{max}$ and $r_{min}$ can also be obtained from mechanical energy and angular momentum conservation, taking into account that $\vec r$ and $\vec v$ are orthogonal at apogee and at perigee

E = \frac{1}{2} m v_0^2\left(\beta^2 - 1\right) = \frac{1}{2} m v^2 - \frac{g R_T^2 m}{r}

L_0 = \frac{m g R_T^2}{v_0} = m v r

What remains of them, after eliminating $v$ , is a second-degree equation whose solutions are $r_{max}$ and $r_{min}$ .

2.4

By the Third Kepler Law, the period $T$ in the new orbit satisfies that

\frac{T^2}{a^3} = \frac{T_0^2}{r_0^3}

where $a$ , the semi-major axis of the ellipse, is given by

a = \frac{r_{max} + r_{min}}{2} = \frac{r_0}{1-\beta^2}

Therefore

\boxed{T = T_0\left(1-\beta^2\right)^{-3/2}}

For $\beta = 1/4$ $\qquad \boxed{T = T_0\left(\frac{15}{16}\right)^{-3/2} = 26.4\ \mathrm{h}}$

3.1

Only if the satellite follows an open trajectory it can escape from the Earth gravity attraction. Then, the orbit eccentricity has to be equal or larger than one. The minimum boost corresponds to a parabolic trajectory, with $\varepsilon = 1$

\varepsilon = \beta \qquad\Rightarrow\qquad \boxed{\beta_{esc} = 1}

This can also be obtained by using that the total satellite energy has to be zero to reach infinity ( $E_p = 0$ ) without residual velocity ( $E_k = 0$ )

E = \frac{1}{2} m v_0^2\left(\beta_{esc}^2 - 1\right) = 0 \quad\Rightarrow\quad \beta_{esc} = 1

This also arises from $T = \infty$ or from $r_{max} = \infty$ .

3.2

Due to $\varepsilon = \beta_{esc} = 1$ , the polar parabola equation is

r = \frac{l}{1 - \cos\theta}

where the semi-latus-rectum continues to be $l = r_0$ . The minimum Earth - satellite distance corresponds to $\theta = \pi$ , where

\boxed{r'_{min} = \frac{r_0}{2}}

This also arises from energy conservation (for $E = 0$ ) and from the equality between the angular momenta ( $L_0$ ) at the initial point P and at maximum approximation, where $\vec r$ and $\vec v$ are orthogonal.

4.1

If the satellite escapes to infinity with residual velocity $v_\infty$ , by energy conservation

Hyperbolic escape trajectory starting at P above the Earth (distance r0) with velocity v0 to the left and Δv downwards; two dashed asymptotes labelled Asymptote cross on the horizontal axis at the left, each making the angle θ_asym with the axis; the angle φ is marked at the Earth, θ_asym also at the Earth between the axis and the dashed line parallel to the lower asymptote, and b is the distance between that line and the lower asymptote along which v∞ points. — Figure 5

E = \frac{1}{2} m v_0^2\left(\beta^2 - 1\right) = \frac{1}{2} m v_\infty^2 \quad\Rightarrow

\boxed{v_\infty = v_0\left(\beta^2 - 1\right)^{1/2}}

4.2

As $\varepsilon = \beta > \beta_{esc} = 1$ the satellite trajectory will be a hyperbola.

The satellite angular momentum is the same at P than at the point where its residual velocity is $v_\infty$ (Figure 5), thus

m v_0 r_0 = m v_\infty b

b = r_0\frac{v_0}{v_\infty} \quad\Rightarrow\quad \boxed{b = r_0\left(\beta^2 - 1\right)^{-1/2}}

4.3

The angle between each asymptote and the hyperbola axis is that appearing in its polar equation in the limit $r \to \infty$ . This is the angle for which the equation denominator vanishes

1 - \beta\cos\theta_{asym} = 0 \quad\Rightarrow\quad \theta_{asym} = \cos^{-1}\left(\frac{1}{\beta}\right)

According to Figure 5

\phi = \frac{\pi}{2} + \theta_{asym} \qquad\Rightarrow\qquad \boxed{\phi = \frac{\pi}{2} + \cos^{-1}\left(\frac{1}{\beta}\right)}

For $\beta = \dfrac{3}{2}\beta_{esc} = \dfrac{3}{2}$ , one gets $\boxed{\phi = 138^{o} = 2.41\ \mathrm{rad}}$

Th 1 ANSWER SHEET

Question	Basic formulas and ideas used	Analytical results	Numerical results	Marking guideline
1.1	$G\dfrac{M_T m}{r_0^2} = m\dfrac{v_0^2}{r_0}$ ; $v_0 = \dfrac{2\pi r_0}{T_0}$ ; $g = \dfrac{G M_T}{R_T^2}$		$r_0 = 4.22\cdot 10^{7}\ \mathrm{m}$	0.3
1.2		$v_0 = R_T\sqrt{\dfrac{g}{r_0}}$	$v_0 = 3.07\cdot 10^{3}\ \mathrm{m/s}$	0.3 + 0.1
1.3	$\vec L = m\,\vec r\times\vec v$ ; $E = \dfrac{1}{2} m v^2 - G\dfrac{Mm}{r}$	$L_0 = \dfrac{m g R_T^2}{v_0}$ ; $E_0 = -\dfrac{1}{2} m v_0^2$		0.4; 0.4
2.1	Hint on the conical curves	$l = r_0$ ; $\varepsilon = \beta$		0.4; 0.5
2.2			$\alpha = \dfrac{\pi}{2}$	1.0
2.3	Results of 2.1, or conservation of $E$ and $L$	$r_{max} = \dfrac{r_0}{1-\beta}$ ; $r_{min} = \dfrac{r_0}{1+\beta}$	$r_{\max} = 5.63\cdot 10^{7}\ \mathrm{m}$ ; $r_{\min} = 3.38\cdot 10^{7}\ \mathrm{m}$	1.0 + 0.2
2.4	Third Kepler's Law	$T = T_0\left(1-\beta^2\right)^{-3/2}$	$T = 26.4\ \mathrm{h}$	0.5 + 0.2
3.1	$\varepsilon = 1$ , $E = 0$ , $T = \infty$ or $r_{max} = \infty$		$\beta_{esc} = 1$	0.5
3.2	$\varepsilon = 1$ and results of 2.1	$r'_{min} = \dfrac{r_0}{2}$		1.0
4.1	Conservation of $E$	$v_\infty = v_0\left(\beta^2 - 1\right)^{1/2}$		1.0
4.2	Conservation of $L$	$b = r_0\left(\beta^2 - 1\right)^{-1/2}$		1.0
4.3	Hint on the conical curves	$\phi = \dfrac{\pi}{2} + \cos^{-1}\left(\dfrac{1}{\beta}\right)$	$\phi = 138^{\circ} = 2.41\ \mathrm{rad}$	1.0 + 0.2

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf