IPhO 2005 · Theory problem 1

An Ill Fated Satellite

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Th 1 AN ILL FATED SATELLITE

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.

To obtain numerical values use: Earth radius \(R_T=6.37\cdot10^6\ \mathrm{m}\), 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.

Figure 1

Image: ESA

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

Question 1

1.1 (0.3) Compute the numerical value of \(r_0\).

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.

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 \(R_T\).

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.

Figure 2

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 \(\beta\).

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

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 \(\beta=1/4\).

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 \(\beta=1/4\).

\(^{1}\) Its revolution period is \(T_0\).

\(^{2}\) See the “hint”.

Question 3

3.1 (0.5) Calculate the minimum boost parameter, \(\beta_{\mathrm{esc}}\), needed for the satellite to escape Earth gravity.

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 \(r_0\).

Question 4

Suppose \(\beta>\beta_{\mathrm{esc}}\).

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

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).

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

Figure 3

F-2

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)

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

Figure 4

F-3

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:

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

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 \(0\leq\varepsilon<1\), 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.

Official English text of the problem.

Answers

  • 1.1 \[ r_0=\left(\frac{gR_T^2T_0^2}{4\pi^2}\right)^{1/3} =\boxed{4.22\times10^7\ \mathrm{m}}. \]

  • 1.2 \[ v_0=\sqrt{\frac{gR_T^2}{r_0}} =\boxed{3.07\times10^3\ \mathrm{m\,s^{-1}}}. \]

  • 1.3 \[ \boxed{L_0=\frac{mgR_T^2}{v_0}}, \qquad \boxed{E_0=-\frac12mv_0^2}. \]

  • 2.1 \[ \boxed{l=r_0,\qquad \varepsilon=\beta}. \]

  • 2.2 \[ \boxed{\alpha=\frac{\pi}{2}=90^\circ}. \] This applies for \(0<\beta<1\); at \(\beta=0\), the orbit is circular and has no distinguished major axis.

  • 2.3 \[ \boxed{r_{\min}=\frac{r_0}{1+\beta}}, \qquad \boxed{r_{\max}=\frac{r_0}{1-\beta}}. \] For \(\beta=1/4\), \[ \boxed{r_{\min}=3.38\times10^7\ \mathrm{m}}, \qquad \boxed{r_{\max}=5.63\times10^7\ \mathrm{m}}. \]

  • 2.4 \[ \boxed{T=\frac{T_0}{(1-\beta^2)^{3/2}}}. \] For \(\beta=1/4\), \(\boxed{T=26.4\ \mathrm{h}}\).

  • 3.1 \[ \boxed{\beta_{\mathrm{esc}}=1}. \]

  • 3.2 \[ \boxed{r'_{\min}=\frac{r_0}{2}}. \]

  • 4.1 \[ \boxed{v_\infty=v_0\sqrt{\beta^2-1}}. \]

  • 4.2 \[ \boxed{b=\frac{r_0}{\sqrt{\beta^2-1}}}. \]

  • 4.3 \[ \boxed{\phi=\frac{\pi}{2}+\arccos\!\left(\frac1\beta\right) =\pi-\arcsin\!\left(\frac1\beta\right)}. \] For \(\beta=\frac32\beta_{\mathrm{esc}}=\frac32\), \[ \boxed{\phi=138.2^\circ}. \]

Solution

Let \[ \mu=GM=gR_T^2, \] where \(M\) is Earth's mass. We use Newtonian gravity for a fixed, spherical Earth, with \(m\ll M\), and neglect atmospheric drag and other bodies. The engine burn is treated as instantaneous, and the satellite's mass as constant. Gravitational potential energy is zero at infinity.

1.1 — Radius of the geosynchronous orbit

For a circular orbit, gravitational acceleration provides the centripetal acceleration: \[ \frac{\mu}{r_0^2}=\frac{v_0^2}{r_0}. \] The orbital period is \(T_0\), so \[ v_0=\frac{2\pi r_0}{T_0}. \] Combining these equations gives \[ r_0^3=\frac{\mu T_0^2}{4\pi^2}. \] Thus \[ \boxed{r_0=\left(\frac{gR_T^2T_0^2}{4\pi^2}\right)^{1/3}}. \]

Using \(T_0=24.0\times3600\ \mathrm{s}\), \[ r_0= \left[ \frac{9.81(6.37\times10^6)^2(24.0\times3600)^2}{4\pi^2} \right]^{1/3} =4.222197\ldots\times10^7\ \mathrm{m}. \] Therefore, \[ \boxed{r_0=4.22\times10^7\ \mathrm{m}}. \]

1.2 — Initial orbital velocity

The circular-orbit condition gives \[ v_0^2=\frac{\mu}{r_0}=\frac{gR_T^2}{r_0}. \] Hence \[ \boxed{v_0=\sqrt{\frac{gR_T^2}{r_0}}}. \] Numerically, \[ v_0=3070.47\ldots\ \mathrm{m\,s^{-1}} =\boxed{3.07\times10^3\ \mathrm{m\,s^{-1}}}. \]

1.3 — Initial angular momentum and mechanical energy

The velocity is perpendicular to the radius, so the angular momentum magnitude is \[ L_0=mr_0v_0. \] Since \(r_0=\mu/v_0^2\), \[ \boxed{L_0=\frac{m\mu}{v_0}=\frac{mgR_T^2}{v_0}}. \]

The mechanical energy is \[ E_0=\frac12mv_0^2-\frac{\mu m}{r_0}. \] Using \(\mu/r_0=v_0^2\), \[ \boxed{E_0=-\frac12mv_0^2}. \]

2.1 — Semi-latus-rectum and eccentricity after the impulse

Let \(\hat{\mathbf r}\) point radially outward at the misfire and \(\hat{\mathbf t}\) point along the original velocity. Immediately after the impulse, \[ \mathbf v=v_0\hat{\mathbf t}-\beta v_0\hat{\mathbf r}. \] The two components are perpendicular, giving \[ v^2=v_0^2(1+\beta^2). \]

The impulse is radial and therefore has zero angular impulse about Earth's centre. Consequently, \[ L=L_0=mr_0v_0. \]

The radius remains \(r_0\) during the instantaneous burn, so the new energy is \[ E=\frac12mv_0^2(1+\beta^2)-\frac{\mu m}{r_0} =\frac12mv_0^2(\beta^2-1). \] Thus \(E<0\) for \(\beta<1\), as required for a bound orbit.

From the supplied orbital formulas, \[ l=\frac{L^2}{\mu m^2} =\frac{r_0^2v_0^2}{\mu} =\boxed{r_0}, \] and \[ \varepsilon^2 =1+\frac{2EL^2}{\mu^2m^3} =1+(\beta^2-1) =\beta^2. \] Because \(\beta\) is a nonnegative speed ratio, \[ \boxed{\varepsilon=\beta}. \]

These expressions for \(l\) and \(\varepsilon\) also hold for the parabolic and hyperbolic cases considered later.

2.2 — Orientation of the major axis

At the instant of the misfire, \(r=r_0=l\). Substitution into the polar equation gives \[ r_0=\frac{r_0}{1-\beta\cos\theta_0}. \] For \(\beta>0\), this requires \[ \cos\theta_0=0. \] The polar reference axis is the major-axis direction, so the position vector at the misfire is perpendicular to that axis: \[ \boxed{\alpha=\frac{\pi}{2}=90^\circ}. \]

To identify the branch, choose \(\theta\) to increase in the direction of orbital motion. Then \[ \frac{dr}{d\theta} =-\frac{r_0\beta\sin\theta}{(1-\beta\cos\theta)^2}. \] The initial radial velocity is inward, so \(dr/d\theta<0\). Therefore, \[ \theta_0=\frac{\pi}{2}. \] Perigee occurs subsequently at \(\theta=\pi\). In the orientation of Figure F-1, the major axis is horizontal and the perigee is to the left of Earth.

For \(\beta=0\), the orbit remains circular, and no unique major-axis direction exists.

2.3 — Perigee and apogee

For \[ r(\theta)=\frac{r_0}{1-\beta\cos\theta}, \] the smallest radius occurs when the denominator is largest, at \(\theta=\pi\). The largest radius occurs at \(\theta=0\). Hence \[ \boxed{r_{\min}=\frac{r_0}{1+\beta}}, \qquad \boxed{r_{\max}=\frac{r_0}{1-\beta}}. \]

For \(\beta=1/4\), \[ r_{\min}=\frac45r_0 =\boxed{3.38\times10^7\ \mathrm{m}}, \] \[ r_{\max}=\frac43r_0 =\boxed{5.63\times10^7\ \mathrm{m}}. \] Both are distances from Earth's centre.

2.4 — Period of the new orbit

The ellipse's semi-major axis is \[ a=\frac{r_{\min}+r_{\max}}{2} =\frac{r_0}{2} \left(\frac1{1+\beta}+\frac1{1-\beta}\right) =\frac{r_0}{1-\beta^2}. \]

Kepler's third law for orbits around the same central mass gives \[ \frac{T^2}{a^3}=\frac{T_0^2}{r_0^3}. \] Therefore, \[ T=T_0\left(\frac{a}{r_0}\right)^{3/2} =\boxed{\frac{T_0}{(1-\beta^2)^{3/2}}}. \]

For \(\beta=1/4\), \[ T=24.0\left(\frac{16}{15}\right)^{3/2}\ \mathrm{h} =26.4396\ldots\ \mathrm{h}, \] so \[ \boxed{T=26.4\ \mathrm{h}}. \]

3.1 — Minimum impulse for escape

A trajectory can reach infinity only if its mechanical energy is nonnegative. The limiting escape trajectory has \(E=0\): \[ \frac12mv_0^2(\beta_{\mathrm{esc}}^2-1)=0. \] Thus \[ \boxed{\beta_{\mathrm{esc}}=1}. \]

At this threshold, \(\varepsilon=1\), so the orbit is parabolic and the speed tends to zero at infinity.

3.2 — Closest approach at the escape threshold

For \(\beta=1\), \[ l=r_0,\qquad \varepsilon=1. \] The perigee distance of the parabola is therefore \[ r'_{\min}=\frac{l}{1+\varepsilon} =\boxed{\frac{r_0}{2}}. \]

The satellite initially moves inward, so it reaches this closest approach before escaping. Numerically, \[ r'_{\min}=2.11\times10^7\ \mathrm{m}>R_T, \] so the threshold trajectory clears Earth.

4.1 — Residual speed at infinity

For \(\beta>1\), the energy is positive. At infinity, the gravitational potential energy vanishes, and energy conservation gives \[ \frac12mv_\infty^2 =\frac12mv_0^2(\beta^2-1). \] Hence \[ \boxed{v_\infty=v_0\sqrt{\beta^2-1}}. \]

4.2 — Impact parameter

The impact parameter \(b\) is the perpendicular distance from Earth's centre to the outgoing asymptote.

Far from Earth, the motion approaches uniform motion along that asymptote. Its angular momentum magnitude is consequently \[ L=mbv_\infty. \] Conservation of angular momentum gives \[ mbv_\infty=mr_0v_0. \] Using the result of 4.1, \[ \boxed{b=\frac{r_0v_0}{v_\infty} =\frac{r_0}{\sqrt{\beta^2-1}}}. \]

4.3 — Direction of escape

Use the same polar convention as in 2.2: \[ r(\theta)=\frac{r_0}{1-\beta\cos\theta}, \qquad \theta_0=\frac{\pi}{2}, \] with \(\theta\) increasing along the motion and perigee at \(\theta=\pi\).

At infinity, the denominator tends to zero: \[ \cos\theta_\infty=\frac1\beta. \] Define \[ \delta=\arccos\!\left(\frac1\beta\right), \qquad 0<\delta<\frac{\pi}{2}. \] After passing perigee, the outgoing branch reaches \[ \theta_\infty=2\pi-\delta. \]

The tangential speed is \(L/(mr)\), which tends to zero as \(r\to\infty\). Thus the limiting velocity direction coincides with the limiting outward radial direction.

The angle \(\phi\) marked on the right of Figure F-2 is the smaller angle between the original outward radius and this outgoing direction. Therefore, \[ \phi =2\pi-(\theta_\infty-\theta_0) =2\pi-\left(2\pi-\delta-\frac{\pi}{2}\right). \] Hence \[ \boxed{\phi=\frac{\pi}{2}+\arccos\!\left(\frac1\beta\right) =\pi-\arcsin\!\left(\frac1\beta\right)}. \] The relevant branch satisfies \[ \frac{\pi}{2}<\phi<\pi. \]

For \(\beta=\frac32\beta_{\mathrm{esc}}=\frac32\), \[ \phi=90^\circ+\arccos\!\left(\frac23\right) =90^\circ+48.1897\ldots^\circ, \] giving \[ \boxed{\phi=138.2^\circ}. \]

The escape formulas describe a collision-free Kepler trajectory. For a finite Earth this additionally requires \(r_0/(1+\beta)>R_T\), which holds for the specified numerical example.

The solution was checked against the official solution.