IPhO 1999 · Theory problem 3

A space probe to Jupiter

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We consider in this problem a method frequently used to accelerate space probes in the desired direction. The space probe flies by a planet, and can significantly increase its speed and modify considerably its flight direction, by taking away a very small amount of energy from the planet's orbital motion. We analyze here this effect for a space probe passing near Jupiter.

The planet Jupiter orbits around the Sun along an elliptical trajectory, that can be approximated by a circumference of average radius \(R\); in order to proceed with the analysis of the physical situation we must first:

  1. Find the speed \(V\) of the planet along its orbit around the Sun. [1.5 points]
  2. When the probe is between the Sun and Jupiter (on the segment Sun-Jupiter), find the distance from Jupiter where the Sun's gravitational attraction balances that by Jupiter. [1 point]

A space probe of mass \(m=825\ \mathrm{kg}\) flies by Jupiter. For simplicity assume that the trajectory of the space probe is entirely in the plane of Jupiter's orbit; in this way we neglect the important case in which the space probe is expelled from Jupiter’s orbital plane.

We only consider what happens in the region where Jupiter's attraction overwhelms all other gravitational forces.

In the reference frame of the Sun's center of mass the initial speed of the space probe is \(v_0=1.00\cdot10^4\ \mathrm{m/s}\) (along the positive \(y\) direction) while Jupiter's speed is along the negative \(x\) direction (see figure 1); by "initial speed" we mean the space probe speed when it's in the interplanetary space, still far from Jupiter but already in the region where the Sun's attraction is negligible with respect to Jupiter's. We assume that the encounter occurs in a sufficiently short time to allow neglecting the change of direction of Jupiter along its orbit around the Sun. We also assume that the probe passes behind Jupiter, i.e. the \(x\) coordinate is greater for the probe than for Jupiter when the \(y\) coordinate is the same.

Figure 1

Figure 1: View in the Sun center of mass system. O denotes Jupiter’s orbit, s is the space probe.

  1. Find the space probe's direction of motion (as the angle \(\varphi\) between its direction and the \(x\) axis) and its speed \(v'\) in Jupiter's reference frame, when it's still far away from Jupiter. [2 points]
  2. Find the value of the space probe's total mechanical energy \(E\) in Jupiter's reference frame, putting – as usual – equal to zero the value of its potential energy at a very large distance, in this case when it is far enough to move with almost constant velocity owing to the smallness of all gravitational interactions. [1 point]

The space probe's trajectory in the reference frame of Jupiter is a hyperbola and its equation in polar coordinates in this reference frame is

\[ \frac{1}{r} = \frac{GM}{v'^2b^2} \left( 1+\sqrt{1+\frac{2Ev'^2b^2}{G^2M^2m}}\cos\theta \right). \tag{1} \]

where \(b\) is the distance between one of the asymptotes and Jupiter (the so called impact parameter), \(E\) is the probe’s total mechanical energy in Jupiter’s reference frame, \(G\) is the gravitational constant, \(M\) is the mass of Jupiter, \(r\) and \(\theta\) are the polar coordinates (the radial distance and the polar angle).

Figure 2 shows the two branches of a hyperbola as described by equation (1); the asymptotes and the polar co-ordinates are also shown. Note that equation (1) has its origin in the "attractive focus" of the hyperbola. The space probe's trajectory is the attractive trajectory (the emphasized branch).

Figure 2

Figure 2

  1. Using equation (1) describing the space probe's trajectory, find the total angular deviation \(\Delta\theta\) in Jupiter’s reference frame (as shown in figure 2) and express it as a function of initial speed \(v'\) and impact parameter \(b\). [2 points]
  2. Assume that the probe cannot pass Jupiter at a distance less than three Jupiter radii from the center of the planet; find the minimum possible impact parameter and the maximum possible angular deviation. [1 point]
  3. Find an equation for the final speed \(v''\) of the probe in the Sun's reference frame as a function only of Jupiter’s speed \(V\), the probe’s initial speed \(v_0\) and the deviation angle \(\Delta\theta\). [1 point]
  4. Use the previous result to find the numerical value of the final speed \(v''\) in the Sun's reference frame when the angular deviation has its maximum possible value. [0.5 points]

Hint

Depending on which path you follow in your solution, the following trigonometric formulas might be useful:

\[ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta \]

\[ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta \]

Physical constants and general data

In addition to the numerical data given within the text of the individual problems, the knowledge of some general data and physical constants may be useful, and you may find them among the following ones. These are nearly the most accurate data currently available, and they have thus a large number of digits; you are expected, however, to write your results with a number of digits that must be appropriate for each problem.

Speed of light in vacuum: \(c=299792458\ \mathrm{m\cdot s^{-1}}\)
Magnetic permeability of vacuum: \(\mu_0=4\pi\cdot10^{-7}\ \mathrm{H\cdot m^{-1}}\)
Dielectric constant of vacuum: \(\varepsilon_0=8.8541878\ \mathrm{pF\cdot m^{-1}}\)
Gravitational constant: \(G=6.67259\cdot10^{-11}\ \mathrm{m^3/(kg\cdot s^2)}\)
Gas constant: \(R=8.314510\ \mathrm{J/(mol\cdot K)}\)
Boltzmann's constant: \(k=1.380658\cdot10^{-23}\ \mathrm{J\cdot K^{-1}}\)
Stefan's constant: \(\sigma=56.703\ \mathrm{nW/(m^2\cdot K^4)}\)
Proton charge: \(e=1.60217733\cdot10^{-19}\ \mathrm{C}\)
Electron mass: \(m_e=9.1093897\cdot10^{-31}\ \mathrm{kg}\)
Planck's constant: \(h=6.6260755\cdot10^{-34}\ \mathrm{J\cdot s}\)
Base of centigrade scale: \(T_{\mathrm K}=273.15\ \mathrm{K}\)
Sun mass: \(M_{\mathrm S}=1.991\cdot10^{30}\ \mathrm{kg}\)
Earth mass: \(M_{\mathrm E}=5.979\cdot10^{24}\ \mathrm{kg}\)
Mean radius of Earth: \(r_{\mathrm E}=6.373\ \mathrm{Mm}\)
Major semiaxis of Earth orbit: \(R_{\mathrm E}=1.4957\cdot10^{11}\ \mathrm{m}\)
Sidereal day: \(d_{\mathrm S}=86.16406\ \mathrm{ks}\)
Year: \(y=31.558150\ \mathrm{Ms}\)
Standard value of the gravitational field at the Earth surface: \(g=9.80665\ \mathrm{m\cdot s^{-2}}\)
Standard value of the atmospheric pressure at sea level: \(p_0=101325\ \mathrm{Pa}\)
Refractive index of air for visibile light, at standard pressure and \(15\,^\circ\mathrm C\): \(n_{\mathrm{air}}=1.000277\)
Solar constant: \(S=1355\ \mathrm{W\cdot m^{-2}}\)
Jupiter mass: \(M=1.901\cdot10^{27}\ \mathrm{kg}\)
Equatorial Jupiter radius: \(R_{\mathrm B}=69.8\ \mathrm{Mm}\)
Average radius of Jupiter’s orbit: \(R=7.783\cdot10^{11}\ \mathrm{m}\)
Jovian day: \(d_{\mathrm J}=35.6\ \mathrm{ks}\)
Jovian year: \(y_{\mathrm J}=374.32\ \mathrm{Ms}\)
\(\pi\): \(3.14159265\)

Official English text of the problem.

Answers

  1. Jupiter’s orbital speed is \[ \boxed{V=\sqrt{\frac{GM_{\mathrm S}}{R}} \simeq1.31\times10^4\ \mathrm{m\,s^{-1}}.} \]

  2. The distance from Jupiter toward the Sun is \[ \boxed{d=\frac{R}{1+\sqrt{M_{\mathrm S}/M}} \simeq2.33\times10^{10}\ \mathrm m.} \]

  3. The initial velocity in Jupiter’s frame is \(\mathbf v'_{\mathrm i}=(V,v_0)\). Thus \[ \boxed{\varphi=\tan^{-1}\!\left(\frac{v_0}{V}\right)=37.4^\circ,\qquad v'=\sqrt{V^2+v_0^2}=1.65\times10^4\ \mathrm{m\,s^{-1}}.} \] The angle is measured counterclockwise from the positive \(x\) axis.

  4. The total mechanical energy in Jupiter’s frame is \[ \boxed{E=\frac12m(V^2+v_0^2)=1.12\times10^{11}\ \mathrm J.} \]

  5. The angular deviation is \[ \boxed{\Delta\theta =2\tan^{-1}\!\left(\frac{GM}{b\,v'^2}\right).} \]

  6. With \(r_0=3R_{\mathrm B}\), \[ \boxed{b_{\min} =\sqrt{r_0^2+\frac{2GMr_0}{v'^2}} =4.90\times10^8\ \mathrm m,} \] \[ \boxed{\Delta\theta_{\max} =2\sin^{-1}\!\left(\frac{1}{1+r_0v'^2/(GM)}\right) =87.4^\circ\ \simeq1.53\ \mathrm{rad}.} \]

  7. The final speed in the Sun’s frame is \[ \boxed{v''= \sqrt{v_0^2+2V^2(1-\cos\Delta\theta) +2Vv_0\sin\Delta\theta}.} \]

  8. At the maximum allowed deviation, \[ \boxed{v''=2.62\times10^4\ \mathrm{m\,s^{-1}} =26.2\ \mathrm{km\,s^{-1}}.} \]

Solution

We use Newtonian gravitation and the approximations specified in the problem. Jupiter’s orbit is circular, and its velocity is effectively constant during the encounter. The probe’s mass is negligible compared with Jupiter’s mass. During the flyby we neglect the Sun’s force, so Jupiter’s frame is treated as inertial and the probe moves in the potential \[ U(r)=-\frac{GMm}{r}. \] The initial and final velocities refer to the asymptotic velocities of this two-body approximation. Numerical calculations retain unrounded intermediate values.

1. Jupiter’s orbital speed

The Sun’s gravitational attraction supplies Jupiter’s centripetal force: \[ \frac{GM_{\mathrm S}M}{R^2}=\frac{MV^2}{R}. \] Therefore \[ V=\sqrt{\frac{GM_{\mathrm S}}{R}}. \] Using the supplied constants, \[ V= \sqrt{\frac{(6.67259\times10^{-11})(1.991\times10^{30})} {7.783\times10^{11}}} =1.30650\times10^4\ \mathrm{m\,s^{-1}}. \] Thus \[ \boxed{V=1.31\times10^4\ \mathrm{m\,s^{-1}}.} \]

The alternative circular-orbit calculation \(V=2\pi R/y_{\mathrm J}\) gives \(1.30642\times10^4\ \mathrm{m\,s^{-1}}\), agreeing at the reported precision.

2. Point where the gravitational attractions balance

Let \(d\) be the distance from Jupiter toward the Sun. The distance from the Sun is then \(R-d\). The two gravitational forces point in opposite directions, so their balance requires \[ \frac{GMm}{d^2}=\frac{GM_{\mathrm S}m}{(R-d)^2}. \] Taking positive square roots, \[ \frac{R-d}{d}=\sqrt{\frac{M_{\mathrm S}}{M}}. \] Solving, \[ d=\frac{R}{1+\sqrt{M_{\mathrm S}/M}}. \] Numerically, \[ d=\frac{7.783\times10^{11}} {1+\sqrt{(1.991\times10^{30})/(1.901\times10^{27})}} =2.33285\times10^{10}\ \mathrm m. \] Hence \[ \boxed{d=2.33\times10^{10}\ \mathrm m.} \]

3. Initial motion in Jupiter’s frame

In the Sun’s frame, the initial velocities are \[ \mathbf v_{\mathrm i}=v_0\hat{\mathbf y}, \qquad \mathbf V_{\mathrm J}=-V\hat{\mathbf x}. \] The Galilean velocity transformation gives \[ \mathbf v'_{\mathrm i} =\mathbf v_{\mathrm i}-\mathbf V_{\mathrm J} =V\hat{\mathbf x}+v_0\hat{\mathbf y}. \] Both components are positive, so the velocity lies in the first quadrant: \[ \tan\varphi=\frac{v_0}{V}, \qquad v'=\sqrt{V^2+v_0^2}. \] Numerically, \[ \varphi=\tan^{-1}\!\left(\frac{1.00\times10^4}{1.30650\times10^4}\right) =37.4306^\circ, \] \[ v'=\sqrt{(1.30650\times10^4)^2+(1.00\times10^4)^2} =1.64528\times10^4\ \mathrm{m\,s^{-1}}. \] Thus \[ \boxed{\varphi=37.4^\circ,\qquad v'=1.65\times10^4\ \mathrm{m\,s^{-1}}.} \]

4. Total mechanical energy in Jupiter’s frame

The potential energy vanishes at infinite separation. Consequently, the conserved total energy equals the initial asymptotic kinetic energy: \[ E=\frac12mv'^2=\frac12m(V^2+v_0^2). \] Substitution gives \[ E=\frac12(825)(1.64528\times10^4)^2 =1.11661\times10^{11}\ \mathrm J. \] Therefore \[ \boxed{E=1.12\times10^{11}\ \mathrm J.} \] Its positive value is consistent with an unbound hyperbolic orbit.

5. Angular deviation

Using \(E=\tfrac12mv'^2\), write equation (1) as \[ r=\frac{p}{1+e\cos\theta}, \] where \[ p=\frac{v'^2b^2}{GM}, \qquad e=\sqrt{1+\frac{b^2v'^4}{G^2M^2}}>1. \]

At either asymptote, \(r\to\infty\), so \[ 1+e\cos\theta=0. \] Let the asymptotic polar angles be \(\pm\theta_\infty\), where \[ \cos\theta_\infty=-\frac1e, \qquad \frac{\pi}{2}<\theta_\infty<\pi. \]

On the incoming asymptote, the velocity points opposite to the radial direction. Its angle relative to the periapsis axis is therefore \(\pi-\theta_\infty\). On the outgoing asymptote, the velocity points along the radial direction and has angle \(\theta_\infty\). Their difference is \[ \Delta\theta =\theta_\infty-(\pi-\theta_\infty) =2\theta_\infty-\pi. \] It follows that \[ \sin\frac{\Delta\theta}{2} =\sin\left(\theta_\infty-\frac{\pi}{2}\right) =-\cos\theta_\infty =\frac1e. \] Since \(0<\Delta\theta/2<\pi/2\), \[ \tan\frac{\Delta\theta}{2} =\frac{1}{\sqrt{e^2-1}} =\frac{GM}{bv'^2}. \] Thus \[ \boxed{\Delta\theta =2\tan^{-1}\!\left(\frac{GM}{bv'^2}\right).} \]

Equivalently, \[ \Delta\theta =2\sin^{-1}\!\left[ \frac{1}{\sqrt{1+b^2v'^4/(G^2M^2)}} \right]. \]

6. Minimum impact parameter and maximum deviation

Let \(r_{\mathrm p}\) be the closest distance to Jupiter’s center, and \(v_{\mathrm p}\) the speed there. Conservation of energy gives \[ \frac12mv_{\mathrm p}^2-\frac{GMm}{r_{\mathrm p}} =\frac12mv'^2, \] so \[ v_{\mathrm p}^2=v'^2+\frac{2GM}{r_{\mathrm p}}. \]

The force is central, so angular momentum about Jupiter is conserved. At closest approach the velocity is perpendicular to the radius: \[ mbv'=mr_{\mathrm p}v_{\mathrm p}. \] Squaring and substituting the energy result, \[ b^2v'^2 =r_{\mathrm p}^2\left(v'^2+\frac{2GM}{r_{\mathrm p}}\right), \] which yields \[ b^2=r_{\mathrm p}^2+\frac{2GMr_{\mathrm p}}{v'^2}. \]

This expression increases with \(r_{\mathrm p}\). Hence the smallest permitted impact parameter occurs at \[ r_{\mathrm p}=r_0=3R_{\mathrm B} =3(69.8\times10^6)\ \mathrm m. \] Therefore \[ b_{\min} =\sqrt{r_0^2+\frac{2GMr_0}{v'^2}}. \] Using \[ GM=1.268459359\times10^{17}\ \mathrm{m^3\,s^{-2}}, \] we obtain \[ b_{\min}=4.89996\times10^8\ \mathrm m, \] or \[ \boxed{b_{\min}=4.90\times10^8\ \mathrm m.} \]

The deflection decreases as \(b\) increases, so its maximum occurs at \(b_{\min}\). Thus \[ \Delta\theta_{\max} =2\tan^{-1}\!\left(\frac{GM}{b_{\min}v'^2}\right). \]

To express this directly through the closest distance, substitute the relation for \(b_{\min}^2\) into the eccentricity: \[ e_{\min}^2 =1+\frac{b_{\min}^2v'^4}{G^2M^2} =\left(1+\frac{r_0v'^2}{GM}\right)^2. \] Hence \[ e_{\min}=1+\frac{r_0v'^2}{GM}=1.44687 \] and \[ \Delta\theta_{\max} =2\sin^{-1}\!\left(\frac1{e_{\min}}\right) =1.52615\ \mathrm{rad} =87.4421^\circ. \] Therefore \[ \boxed{\Delta\theta_{\max}=87.4^\circ.} \]

7. Final speed in the Sun’s frame

Energy conservation in Jupiter’s frame ensures that the outgoing asymptotic speed is still \(v'\).

Because the probe passes behind Jupiter, the gravitational deflection rotates its velocity counterclockwise in the given \(x,y\) coordinates. The outgoing relative velocity consequently makes angle \(\varphi+\Delta\theta\) with the positive \(x\) axis: \[ \mathbf v'_{\mathrm f} =v'\cos(\varphi+\Delta\theta)\hat{\mathbf x} +v'\sin(\varphi+\Delta\theta)\hat{\mathbf y}. \]

Using \[ v'\cos\varphi=V,\qquad v'\sin\varphi=v_0, \] the addition formulas give \[ v'_{\mathrm f,x}=V\cos\Delta\theta-v_0\sin\Delta\theta, \] \[ v'_{\mathrm f,y}=V\sin\Delta\theta+v_0\cos\Delta\theta. \]

Transforming back to the Sun’s frame requires adding Jupiter’s velocity \(-V\hat{\mathbf x}\): \[ v''_x=V(\cos\Delta\theta-1)-v_0\sin\Delta\theta, \] \[ v''_y=V\sin\Delta\theta+v_0\cos\Delta\theta. \] Therefore \[ \begin{aligned} v''^2 &=(v''_x)^2+(v''_y)^2\\ &=V^2\left[(\cos\Delta\theta-1)^2+\sin^2\Delta\theta\right]\\ &\quad+v_0^2\left[\sin^2\Delta\theta+\cos^2\Delta\theta\right] +2Vv_0\sin\Delta\theta\\ &=2V^2(1-\cos\Delta\theta)+v_0^2+2Vv_0\sin\Delta\theta. \end{aligned} \] Thus the requested expression is \[ \boxed{ v''=\sqrt{v_0^2+2V^2(1-\cos\Delta\theta) +2Vv_0\sin\Delta\theta}. } \]

As a consistency check, setting \(\Delta\theta=0\) gives \(v''=v_0\).

8. Final speed at maximum angular deviation

Substituting the unrounded values \[ V=1.30650\times10^4\ \mathrm{m\,s^{-1}}, \qquad v_0=1.00\times10^4\ \mathrm{m\,s^{-1}}, \qquad \Delta\theta_{\max}=1.52615\ \mathrm{rad}, \] gives \[ \begin{aligned} v'' &=\Big[ (1.00\times10^4)^2 +2(1.30650\times10^4)^2(1-\cos1.52615)\\ &\qquad +2(1.30650\times10^4)(1.00\times10^4)\sin1.52615 \Big]^{1/2}\\ &=2.62144\times10^4\ \mathrm{m\,s^{-1}}. \end{aligned} \] Hence \[ \boxed{v''=2.62\times10^4\ \mathrm{m\,s^{-1}} =26.2\ \mathrm{km\,s^{-1}}.} \]

Jupiter’s gravity changes the direction of the probe’s asymptotic velocity in Jupiter’s frame. This change produces the increase in speed and kinetic energy in the Sun’s frame, supplied by a negligible reduction in Jupiter’s orbital energy.

The solution was checked against the official solution.