IPhO 1979 · Theory problem 1

Landing a spacecraft from a lunar orbit

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A space rocket with mass \(M=12\mathrm{t}\) is moving around the Moon along the circular orbit at the height of \(h=100\ \mathrm{km}\). The engine is activated for a short time to pass at the lunar landing orbit. The velocity of the ejected gases \(u=10^4\ \mathrm{m/s}\). The Moon radius \(R_M=1{,}7\cdot10^3\ \mathrm{km}\), the acceleration of gravity near the Moon surface \(g_M=1.7\ \mathrm{m/s^2}\)

Figure 1

Fig.1

Figure 2

Fig.2

1). What amount of fuel should be spent so that when activating the braking engine at point A of the trajectory, the rocket would land on the Moon at point B (Fig.1)?

2). In the second scenario of landing, at point A the rocket is given an impulse directed towards the center of the Moon, to put the rocket to the orbit meeting the Moon surface at point C (Fig.2). What amount of fuel is needed in this case?

Official English text of the problem.

Answers

  • 1). \(\boxed{\Delta m_1\simeq29\ \mathrm{kg}}\), for a tangential braking velocity change of \(23.77\ \mathrm{m/s}\).
  • 2). \(\boxed{\Delta m_2\simeq1.2\times10^2\ \mathrm{kg}}\), for an inward radial velocity change of \(97.18\ \mathrm{m/s}\).

The ideal rocket equation gives approximately \(28.49\ \mathrm{kg}\) and \(116.05\ \mathrm{kg}\), respectively, before rounding to the precision of the supplied data.

Solution

Treat the Moon as spherical, with an inverse-square gravitational field. The engine operates briefly enough that the spacecraft’s position and the gravitational impulse during the burn can be neglected. The exhaust speed \(u\) is measured relative to the spacecraft.

Write \[ R=R_M=1.7\times10^6\ \mathrm{m},\qquad r_0=R+h=1.8\times10^6\ \mathrm{m}. \] The Moon’s gravitational parameter is \[ \mu=GM_{\text{Moon}}=g_MR^2 =4.913\times10^{12}\ \mathrm{m^3/s^2}. \] Thus the initial circular speed is \[ v_0=\sqrt{\frac{\mu}{r_0}} \simeq1652.10\ \mathrm{m/s}. \]

For a burn producing a velocity change of magnitude \(\Delta v\), conservation of momentum gives the ideal rocket equation: \[ \Delta v=u\ln\frac{M}{M-\Delta m}. \] Consequently, \[ \boxed{\Delta m=M\left(1-e^{-\Delta v/u}\right)}. \] When \(\Delta v/u\ll1\), this becomes \[ \Delta m\simeq\frac{M\Delta v}{u}. \]

Here, reaching the surface is understood as the requested orbital contact. The spacecraft retains a nonzero tangential velocity there; a soft landing would require further braking.

1). Tangential braking at A, reaching B

The braking impulse acts opposite the initial velocity. Immediately afterward, the velocity remains tangential but is smaller than the circular speed. Therefore A is the apocentre of the new elliptical orbit. Its diametrically opposite pericentre must be B, at radius \(R\).

Let \(v_A\) and \(v_B\) be the speeds on this ellipse at A and B. Conservation of angular momentum per unit mass gives \[ r_0v_A=Rv_B. \] Conservation of mechanical energy per unit mass gives \[ \frac{v_A^2}{2}-\frac{\mu}{r_0} = \frac{v_B^2}{2}-\frac{\mu}{R}. \] Eliminating \(v_B\), \[ v_A^2\left(\frac{r_0^2}{R^2}-1\right) = 2\mu\left(\frac1R-\frac1{r_0}\right), \] so \[ v_A^2=\frac{2\mu R}{r_0(r_0+R)}. \]

The required reduction in velocity is therefore \[ \boxed{ \Delta v_1 =v_0-v_A =\sqrt{\frac{\mu}{r_0}} \left(1-\sqrt{\frac{2R}{r_0+R}}\right) }. \] Numerically, \[ \Delta v_1 =1652.10\left(1-\sqrt{\frac{3.4}{3.5}}\right) \simeq23.77\ \mathrm{m/s}. \] Hence \[ \Delta m_1 =12000\left(1-e^{-23.7725/10000}\right)\ \mathrm{kg} \simeq28.49\ \mathrm{kg}, \] or, to sensible significant figures, \[ \boxed{\Delta m_1\simeq29\ \mathrm{kg}}. \]

The radius decreases from \(r_0\) to \(R\) along the transfer, so B is the first point where the orbit touches the lunar surface.

2). Radial impulse at A, reaching C

Let the inward velocity imparted by the engine be \(w\). Immediately after the burn, \[ v_r=-w,\qquad v_t=v_0. \] The velocity change has magnitude \[ \Delta v_2=w. \]

Because the impulse is radial, the specific angular momentum is unchanged: \[ \ell=r_0v_0,\qquad \ell^2=\mu r_0. \]

Let \(\theta\) be the angle swept around the Moon from A in the direction of motion. The required contact point C has \[ \theta_C=\frac{\pi}{2}. \] To determine the orbit, put \(q(\theta)=1/r(\theta)\). Angular momentum conservation gives \[ \dot r=-\ell q', \] where primes denote derivatives with respect to \(\theta\). Newton’s radial equation for inverse-square gravity then yields \[ q''+q=\frac{\mu}{\ell^2}=\frac1{r_0}. \]

At A, \[ q(0)=\frac1{r_0},\qquad q'(0)=\frac{w}{\ell}. \] The solution satisfying these conditions is \[ q(\theta)=\frac1{r_0}+\frac{w}{\ell}\sin\theta, \] or \[ \boxed{ r(\theta)=\frac{r_0}{1+(w/v_0)\sin\theta} }. \]

Imposing \(r(\pi/2)=R\) gives \[ R=\frac{r_0}{1+w/v_0}, \] and therefore \[ \boxed{ \Delta v_2=w =v_0\left(\frac{r_0}{R}-1\right) =v_0\frac{h}{R} =h\sqrt{\frac{g_M}{R+h}} }. \]

The radius decreases continuously between A and C, and its derivative vanishes at C. Thus C is the pericentre and the first contact with the lunar surface.

Numerically, \[ \Delta v_2 =1652.10\,\frac{100}{1700} \simeq97.18\ \mathrm{m/s}. \] The required fuel is \[ \Delta m_2 =12000\left(1-e^{-97.1825/10000}\right)\ \mathrm{kg} \simeq116.05\ \mathrm{kg}, \] hence \[ \boxed{\Delta m_2\simeq1.2\times10^2\ \mathrm{kg}}. \]

Using the small-fuel approximation gives \(28.53\ \mathrm{kg}\) for part 1 and \(116.62\ \mathrm{kg}\) for part 2, consistent with the rounded answers.

The solution was checked against the official solution.