IPhO 2013 · Theory problem 1

The Maribo Meteorite

Introduction

A meteoroid is a small particle (typically smaller than 1 m) from a comet or an asteroid. A meteoroid that impacts the ground is called a meteorite.

On the night of 17 January 2009 many people near the Baltic Sea saw the glowing trail or fireball of a meteoroid falling through the atmosphere of the Earth. In Sweden a surveillance camera recorded a video of the event, see Fig. 1.1(a). From these pictures and eyewitness accounts it was possible to narrow down the impact area, and six weeks later a meteorite with the mass 0.025 kg was found in the vicinity of the town Maribo in southern Denmark. Measurements on the meteorite, now named Maribo, and its orbit in the sky showed interesting results. Its speed when entering the atmosphere had been exceptionally high. Its age, \(4.567\times10^9\) year, shows that it had been formed shortly after the birth of the solar system. The Maribo meteorite is possibly a part of Comet Encke.

The speed of Maribo

The fireball was moving in westerly direction, heading \(285^\circ\) relative to north, toward the location where the meteorite was subsequently found, as sketched in Fig. 1.1. The meteorite was found at a distance 195 km from the surveillance camera in the direction \(230^\circ\) relative to north.

Part Question Points
1.1 Use this and the data in Fig. 1.1 to calculate the average speed of Maribo in the time interval between frames 155 and 161. The curvature of the Earth and the gravitational force on the meteoroid can both be neglected. 1.3

Through the atmosphere and melting?

The friction from the air on a meteoroid moving in the higher atmosphere depends in a complicated way on the shape and velocity of the meteoroid, and on the temperature and density of the atmosphere. As a reasonable approximation the friction force \(F\) in the upper atmosphere is given by the expression \(F=k\rho_{\mathrm{atm}}Av^2\), where \(k\) is a constant, \(\rho_{\mathrm{atm}}\) the density of the atmosphere, \(A\) the projected cross-section area of the meteorite, and \(v\) its speed.

The following simplifying assumptions are made to analyze the meteoroid: The object entering the atmosphere was a sphere of mass \(m_{\mathrm M}=30\ \mathrm{kg}\), radius \(R_{\mathrm M}=0.13\ \mathrm m\), temperature \(T_0=200\ \mathrm K\), and speed \(v_{\mathrm M}=2.91\times10^4\ \mathrm{m/s}\). The density of the atmosphere is constant (its value 40 km above the surface of the Earth), \(\rho_{\mathrm{atm}}=4.1\times10^{-3}\ \mathrm{kg/m^3}\), and the friction coefficient is \(k=0.60\).

Part Question Points
1.2a Estimate how long time after entering the atmosphere it takes the meteoroid to have its speed reduced by 10 % from \(v_{\mathrm M}\) to \(0.90v_{\mathrm M}\). You can neglect the gravitational force on the meteoroid and assume, that it maintains its mass and shape. 0.7
1.2b Calculate how many times larger the kinetic energy \(E_{\mathrm{kin}}\) of the meteoroid entering the atmosphere is than the energy \(E_{\mathrm{melt}}\) necessary for melting it completely (see data sheet). 0.3
Figure 1

Figure 1.1 (a) Azimuth is the clockwise angular position from north in the horizontal plane, and altitude is the angular position above the horizon. A series of frames recorded by the surveillance camera in Sweden, showing the motion of Maribo as a fireball on its way down through the atmosphere. (b) The data from two frames indicating the time, the direction (azimuth) in degrees, as seen by the camera (C), and the height above the horizon (altitude) in degrees. (c) Sketch of the directions of the path (magenta arrow) of Maribo relative to north (N) and of the landing site (M) in Denmark as seen by the camera (C).

Heating of Maribo during its fall in the atmosphere

When the stony meteoroid Maribo entered the atmosphere at supersonic speed it appeared as a fireball because the surrounding air was glowing. Nevertheless, only the outermost layer of Maribo was heated. Assume that Maribo is a homogenous sphere with density \(\rho_{\mathrm{sm}}\), specific heat capacity \(c_{\mathrm{sm}}\), and thermal conductivity \(k_{\mathrm{sm}}\) (for values see the data sheet). Furthermore, when entering the atmosphere, it had the temperature \(T_0=200\ \mathrm K\). While falling through the atmosphere its surface temperature was constant \(T_{\mathrm s}=1000\ \mathrm K\) due to the air friction, thus gradually heating the interior.

After falling a time \(t\) in the atmosphere, an outer shell of Maribo of thickness \(x\) will have been heated to a temperature significantly larger than \(T_0\). This thickness can be estimated by dimensional analysis as the simple product of powers of the thermodynamic parameters: \(x\approx t^\alpha\rho_{\mathrm{sm}}^\beta c_{\mathrm{sm}}^\gamma k_{\mathrm{sm}}^\delta\).

Part Question Points
1.3a Determine by dimensional (unit) analysis the value of the four powers \(\alpha,\beta,\gamma,\) and \(\delta\). 0.6
1.3b Calculate the thickness \(x\) after a fall time \(t=5\ \mathrm s\), and determine the ratio \(x/R_{\mathrm M}\). 0.4

The age of a meteorite

The chemical properties of radioactive elements may be different, so during the crystallization of the minerals in a given meteorite, some minerals will have a high content of a specific radioactive element and others a low content. This difference can be used to determine the age of a meteorite by radiometric dating of its radioactive minerals.

As a specific example, we study the isotope \({}^{87}\mathrm{Rb}\) (element no. 37), which decays into the stable isotope \({}^{87}\mathrm{Sr}\) (element no. 38) with a half-life of \(T_{\frac12}=4.9\times10^{10}\) year, relative to the stable isotope \({}^{86}\mathrm{Sr}\). At the time of crystallization the ratio \({}^{87}\mathrm{Sr}/{}^{86}\mathrm{Sr}\) was identical for all minerals, while the ratio \({}^{87}\mathrm{Rb}/{}^{86}\mathrm{Sr}\) was different. As time passes on, the amount of \({}^{87}\mathrm{Rb}\) decreases by decay, while consequently the amount of \({}^{87}\mathrm{Sr}\) increases. As a result, the ratio \({}^{87}\mathrm{Sr}/{}^{86}\mathrm{Sr}\) will be different today. In Fig. 1.2(a), the points on the horizontal line refer to the ratio \({}^{87}\mathrm{Rb}/{}^{86}\mathrm{Sr}\) in different minerals at the time, when they are crystallized.

Figure 2

Figure 1.2 (a) The ratio \({}^{87}\mathrm{Sr}/{}^{86}\mathrm{Sr}\) in different minerals at the time \(t=0\) of crystallization (open circles) and at present time (filled circles). (b) The isochron-line for three different mineral samples taken from a meteorite at present time.

Part Question Points
1.4a Write down the decay scheme for the transformation of \({}^{87}_{37}\mathrm{Rb}\) to \({}^{87}_{38}\mathrm{Sr}\). 0.3
1.4b Show that the present-time ratio \({}^{87}\mathrm{Sr}/{}^{86}\mathrm{Sr}\) plotted versus the present-time ratio \({}^{87}\mathrm{Rb}/{}^{86}\mathrm{Sr}\) in different mineral samples from the same meteorite forms a straight line, the so-called isochron-line, with slope \(a(t)=(e^{\lambda t}-1)\). Here \(t\) is the time since the formation of the minerals, while \(\lambda\) is the decay constant inversely proportional to half-life \(T_{\frac12}\). 0.7
1.4c Determine the age \(\tau_{\mathrm M}\) of the meteorite using the isochron-line of Fig. 1.2(b). 0.4

Comet Encke, from which Maribo may originate

In its orbit around the Sun, the minimum and maximum distances between comet Encke and the Sun are \(a_{\min}=4.95\times10^{10}\ \mathrm m\) and \(a_{\max}=6.16\times10^{11}\ \mathrm m\), respectively.

Part Question Points
1.5 Calculate the orbital period \(t_{\mathrm{Encke}}\) of comet Encke. 0.6

Consequences of an asteroid impact on Earth

65 million years ago Earth was hit by a huge asteroid with density \(\rho_{\mathrm{ast}}=3.0\times10^3\ \mathrm{kg\,m^{-3}}\), radius \(R_{\mathrm{ast}}=5.0\ \mathrm{km}\), and final speed of \(v_{\mathrm{ast}}=2.5\times10^4\ \mathrm{m/s}\). This impact resulted in the extermination of most of the life on Earth and the formation of the enormous Chicxulub Crater. Assume that an identical asteroid would hit Earth today in a completely inelastic collision, and use the fact that the moment of inertia of Earth is 0.83 times that for a homogeneous sphere of the same mass and radius. The moment of inertia of a homogeneous sphere with mass \(M\) and radius \(R\) is \(\frac25MR^2\). Neglect changes in the orbit of the Earth.

Part Question Points
1.6a Let the asteroid hit the North Pole. Find the maximum change in angular orientation of the axis of Earth after the impact. 0.7
1.6b Let the asteroid hit the Equator in a radial impact. Find the change \(\Delta\tau_{\mathrm{vrt}}\) in the duration of one revolution of Earth after the impact. 0.7
1.6c Let the asteroid hit the Equator in a tangential impact in the equatorial plane. Find the change \(\Delta\tau_{\mathrm{tan}}\) in the duration of one revolution of Earth after the impact. 0.7

Maximum impact speed

Consider a celestial body, gravitationally bound in the solar system, which impacts the surface of Earth with a speed \(v_{\mathrm{imp}}\). Initially the effect of the gravitational field of the Earth on the body can be neglected. Disregard the friction in the atmosphere, the effect of other celestial bodies, and the rotation of the Earth.

Part Question Points
1.7 Calculate \(v_{\mathrm{imp}}^{\max}\), the largest possible value of \(v_{\mathrm{imp}}\). 1.6

Data sheet: Table of physical parameters

Physical parameter Value
Speed of light in vacuum \(c=2.998\times10^8\ \mathrm{m\,s^{-1}}\)
Planck's constant over \(2\pi\) \(\hbar=1.055\times10^{-34}\ \mathrm{J\,s}\)
Gravitational constant \(G=6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\)
Gravitational acceleration \(g=9.82\ \mathrm{m\,s^{-2}}\)
Elementary charge \(e=1.602\times10^{-19}\ \mathrm C\)
Electric permittivity of vacuum \(\epsilon_0=8.854\times10^{-12}\ \mathrm{C^2\,J^{-1}\,m^{-1}}\)
Electron mass \(m_{\mathrm e}=9.109\times10^{-31}\ \mathrm{kg}\)
Avogadro constant \(N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}\)
Boltzmann constant \(k_{\mathrm B}=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}\)
Stony meteorite, specific heat \(c_{\mathrm{sm}}=1.2\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Stony meteorite, thermal conductivity \(k_{\mathrm{sm}}=2.0\ \mathrm{W\,m^{-1}\,K^{-1}}\)
Stony meteorite, density \(\rho_{\mathrm{sm}}=3.3\times10^3\ \mathrm{kg\,m^{-3}}\)
Stony meteorite, melting point \(T_{\mathrm{sm}}=1.7\times10^3\ \mathrm K\)
Stony meteorite, specific melting heat \(L_{\mathrm{sm}}=2.6\times10^5\ \mathrm{J\,kg^{-1}}\)
Silver, molar mass \(M_{\mathrm{Ag}}=1.079\times10^{-1}\ \mathrm{kg\,mol^{-1}}\)
Silver, density \(\rho_{\mathrm{Ag}}=1.049\times10^4\ \mathrm{kg\,m^{-3}}\)
Silver, specific heat capacity \(c_{\mathrm{Ag}}=2.40\times10^2\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Water, molar mass \(M_{\mathrm{wa}}=1.801\times10^{-2}\ \mathrm{kg\,mol^{-1}}\)
Water, density \(\rho_{\mathrm{wa}}=0.998\times10^3\ \mathrm{kg\,m^{-3}}\)
Water, specific heat capacity \(c_{\mathrm{wa}}=4.181\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Water, heat of vaporization \(L_{\mathrm{wa}}=2.260\times10^6\ \mathrm{J\,kg^{-1}}\)
Water, boiling temperature \(T_{100}=100\,^\circ\mathrm C=373.15\ \mathrm K\)
Ice, density of glacier \(\rho_{\mathrm{ice}}=0.917\times10^3\ \mathrm{kg\,m^{-3}}\)
Steam, specific heat capacity \(c_{\mathrm{st}}=2.080\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Earth, mass of the \(m_{\mathrm E}=5.97\times10^{24}\ \mathrm{kg}\)
Earth, radius of the \(R_{\mathrm E}=6.38\times10^6\ \mathrm m\)
Sun, mass of the \(m_{\mathrm S}=1.99\times10^{30}\ \mathrm{kg}\)
Sun, radius of the \(R_{\mathrm S}=6.96\times10^8\ \mathrm m\)
Average Sun-Earth distance \(a_{\mathrm E}=1.50\times10^{11}\ \mathrm m\)

Official English text of the problem.

Answers

  • 1.1: \(\boxed{\bar v=2.86\times10^4\ \mathrm{m\,s^{-1}}=28.6\ \mathrm{km\,s^{-1}}}\).
  • 1.2a: \(\boxed{t=0.88\ \mathrm s}\).
  • 1.2b: \(\boxed{E_{\mathrm{kin}}/E_{\mathrm{melt}}\approx2.1\times10^2}\).
  • 1.3a: \(\boxed{\alpha=\delta=\tfrac12,\quad\beta=\gamma=-\tfrac12}\), so \(x\approx\sqrt{k_{\mathrm{sm}}t/(\rho_{\mathrm{sm}}c_{\mathrm{sm}})}\).
  • 1.3b: \(\boxed{x\approx1.6\ \mathrm{mm},\quad x/R_{\mathrm M}\approx0.012=1.2\%}\).
  • 1.4a: \(\boxed{{}^{87}_{37}\mathrm{Rb}\rightarrow{}^{87}_{38}\mathrm{Sr}+{}^0_{-1}e+\bar\nu_{\mathrm e}}\).
  • 1.4b: \(\boxed{Y(t)=Y_0+(e^{\lambda t}-1)X(t)}\), where \(Y={}^{87}\mathrm{Sr}/{}^{86}\mathrm{Sr}\), \(X={}^{87}\mathrm{Rb}/{}^{86}\mathrm{Sr}\), and \(\lambda=\ln2/T_{\frac12}\).
  • 1.4c: \(\boxed{\tau_{\mathrm M}\approx3.5\times10^9\ \mathrm{yr}}\), from a graph slope of approximately \(0.050\).
  • 1.5: \(\boxed{t_{\mathrm{Encke}}=1.05\times10^8\ \mathrm s\approx3.32\ \mathrm{yr}}\).
  • 1.6a: \(\boxed{\theta_{\max}\approx4.3\times10^{-8}\ \mathrm{rad}\approx2.4\times10^{-6}\ \mathrm{degree}}\).
  • 1.6b: \(\boxed{\Delta\tau_{\mathrm{vrt}}\approx+6.8\times10^{-5}\ \mathrm s}\).
  • 1.6c: \(\boxed{\Delta\tau_{\mathrm{tan}}\approx-3.6\times10^{-3}\ \mathrm s}\) for a prograde impact; \(\boxed{\Delta\tau_{\mathrm{tan}}\approx+3.8\times10^{-3}\ \mathrm s}\) for a retrograde impact.
  • 1.7: \(\boxed{v_{\mathrm{imp}}^{\max}\approx7.27\times10^4\ \mathrm{m\,s^{-1}}=72.7\ \mathrm{km\,s^{-1}}}\), understood as a limiting upper bound for a gravitationally bound body.

Solution

1.1

Let \(C\) denote the camera, \(M\) the landing site, and \(P_i\) the horizontal projection of the meteoroid’s position in frame \(i\). Write

\[ CM=D=195\ \mathrm{km},\qquad CP_i=r_i,\qquad MP_i=s_i. \]

The observed quantities are:

Frame Time \(t_i\) Azimuth \(\phi_i\) Altitude angle \(h_i\)
155 \(1.46\ \mathrm s\) \(215^\circ\) \(19.2^\circ\)
161 \(2.28\ \mathrm s\) \(221^\circ\) \(14.7^\circ\)

The horizontal trajectory has heading \(285^\circ\). Viewed from \(M\), the camera has bearing \(50^\circ\), while the earlier meteoroid positions have bearing \(105^\circ\). Hence the angle at \(M\) in triangle \(CMP_i\) is \(55^\circ\). The angle at \(C\) is \(230^\circ-\phi_i\).

The sine rule therefore gives

\[ r_i =D\frac{\sin55^\circ}{\sin(285^\circ-\phi_i)}, \qquad s_i =D\frac{\sin(230^\circ-\phi_i)}{\sin(285^\circ-\phi_i)}. \]

Since Earth’s curvature is neglected, the meteoroid’s height is

\[ z_i=r_i\tan h_i. \]

Numerically,

Frame \(r_i\) (km) \(s_i\) (km) \(z_i\) (km)
155 169.986 53.709 59.195
161 177.721 33.940 46.624

Thus the horizontal and vertical distances travelled are

\[ \Delta s=19.769\ \mathrm{km}, \qquad |\Delta z|=12.571\ \mathrm{km}. \]

Taking the trajectory over this short interval to be straight,

\[ \Delta\ell =\sqrt{(\Delta s)^2+(\Delta z)^2} =23.428\ \mathrm{km}. \]

The elapsed time is \(2.28-1.46=0.82\ \mathrm s\), so

\[ \boxed{ \bar v=\frac{23.428\ \mathrm{km}}{0.82\ \mathrm s} =28.6\ \mathrm{km\,s^{-1}}. } \]

For the subsequent parts, we use the separately specified initial speed \(v_{\mathrm M}=2.91\times10^4\ \mathrm{m\,s^{-1}}\).

1.2a

The projected area of the sphere is \(A=\pi R_{\mathrm M}^2\). Newton’s second law, with gravity neglected, gives

\[ m_{\mathrm M}\frac{dv}{dt} =-k\rho_{\mathrm{atm}}\pi R_{\mathrm M}^2v^2. \]

Separating variables and integrating from \(v_{\mathrm M}\) to \(v\),

\[ \frac1v-\frac1{v_{\mathrm M}} =\frac{k\rho_{\mathrm{atm}}\pi R_{\mathrm M}^2}{m_{\mathrm M}}t. \]

For \(v=0.90v_{\mathrm M}\),

\[ t =\frac{m_{\mathrm M}} {k\rho_{\mathrm{atm}}\pi R_{\mathrm M}^2v_{\mathrm M}} \left(\frac1{0.90}-1\right). \]

Substitution yields

\[ t= \frac{30} {(0.60)(4.1\times10^{-3})\pi(0.13)^2(2.91\times10^4)} \left(\frac1{0.90}-1\right) =0.877\ \mathrm s. \]

Therefore,

\[ \boxed{t\approx0.88\ \mathrm s.} \]

1.2b

Complete melting requires both heating the material from \(T_0\) to its melting point and supplying its latent heat of fusion. Using the constant specific heat given in the data sheet,

\[ E_{\mathrm{melt}} =m_{\mathrm M}\left[c_{\mathrm{sm}}(T_{\mathrm{sm}}-T_0)+L_{\mathrm{sm}}\right]. \]

Thus

\[ E_{\mathrm{melt}} =30\left[(1.2\times10^3)(1700-200)+2.6\times10^5\right] =6.18\times10^7\ \mathrm J. \]

The initial kinetic energy is

\[ E_{\mathrm{kin}} =\frac12m_{\mathrm M}v_{\mathrm M}^2 =\frac12(30)(2.91\times10^4)^2 =1.2702\times10^{10}\ \mathrm J. \]

Consequently,

\[ \boxed{ \frac{E_{\mathrm{kin}}}{E_{\mathrm{melt}}} =\frac{v_{\mathrm M}^2} {2[c_{\mathrm{sm}}(T_{\mathrm{sm}}-T_0)+L_{\mathrm{sm}}]} =205.5\approx2.1\times10^2. } \]

This energy comparison does not imply complete melting: the kinetic energy need not be deposited as heat throughout the meteoroid.

1.3a

Use \(M,L,\mathsf T,\Theta\) for the dimensions of mass, length, time, and temperature. The relevant dimensions are

\[ [\rho_{\mathrm{sm}}]=ML^{-3}, \qquad [c_{\mathrm{sm}}]=L^2\mathsf T^{-2}\Theta^{-1}, \qquad [k_{\mathrm{sm}}]=ML\mathsf T^{-3}\Theta^{-1}. \]

Therefore,

\[ [t^\alpha\rho_{\mathrm{sm}}^\beta c_{\mathrm{sm}}^\gamma k_{\mathrm{sm}}^\delta] = M^{\beta+\delta} L^{-3\beta+2\gamma+\delta} \mathsf T^{\alpha-2\gamma-3\delta} \Theta^{-\gamma-\delta}. \]

Equating this to the dimension \(L\) of \(x\) gives

\[ \begin{aligned} \beta+\delta&=0,\\ -3\beta+2\gamma+\delta&=1,\\ \alpha-2\gamma-3\delta&=0,\\ -\gamma-\delta&=0. \end{aligned} \]

The first and fourth equations give \(\beta=\gamma=-\delta\). The length equation then gives \(2\delta=1\), and the time equation gives \(\alpha=\delta\). Hence

\[ \boxed{\alpha=\delta=\frac12,\qquad\beta=\gamma=-\frac12.} \]

The resulting estimate is

\[ \boxed{x\approx\sqrt{\frac{k_{\mathrm{sm}}t}{\rho_{\mathrm{sm}}c_{\mathrm{sm}}}}.} \]

Dimensional analysis determines the exponents but leaves an undetermined dimensionless coefficient. The requested estimate takes this coefficient to be unity.

1.3b

The thermal diffusivity is

\[ \chi=\frac{k_{\mathrm{sm}}}{\rho_{\mathrm{sm}}c_{\mathrm{sm}}} =\frac{2.0}{(3.3\times10^3)(1.2\times10^3)} =5.05\times10^{-7}\ \mathrm{m^2\,s^{-1}}. \]

At \(t=5\ \mathrm s\),

\[ x\approx\sqrt{\chi t} =\sqrt{(5.05\times10^{-7})(5)} =1.589\times10^{-3}\ \mathrm m. \]

Thus

\[ \boxed{x\approx1.6\ \mathrm{mm}}, \qquad \boxed{ \frac{x}{R_{\mathrm M}} =\frac{1.589\times10^{-3}}{0.13} \approx0.012=1.2\%. } \]

The heated depth is small compared with the radius, consistent with only an outer layer warming appreciably during the fall.

1.4a

The atomic number increases by one while the mass number remains unchanged. This is beta-minus decay:

\[ \boxed{ {}^{87}_{37}\mathrm{Rb} \longrightarrow {}^{87}_{38}\mathrm{Sr} +{}^0_{-1}e+\bar\nu_{\mathrm e}. } \]

An electron and an electron antineutrino are emitted.

1.4b

Consider one mineral sample as a closed system after crystallization. Let \(N_{\mathrm{Rb}}(t)\), \(N_{87}(t)\), and \(N_{86}\) be the numbers of \({}^{87}\mathrm{Rb}\), \({}^{87}\mathrm{Sr}\), and \({}^{86}\mathrm{Sr}\) nuclei. The stable reference population \(N_{86}\) is constant.

Radioactive decay gives

\[ N_{\mathrm{Rb}}(t)=N_{\mathrm{Rb}}(0)e^{-\lambda t}, \qquad \lambda=\frac{\ln2}{T_{\frac12}}. \]

Each decayed rubidium nucleus produces one strontium-87 nucleus, so

\[ \begin{aligned} N_{87}(t) &=N_{87}(0)+N_{\mathrm{Rb}}(0)-N_{\mathrm{Rb}}(t)\\ &=N_{87}(0)+N_{\mathrm{Rb}}(t)(e^{\lambda t}-1). \end{aligned} \]

Dividing by \(N_{86}\), define

\[ Y(t)=\frac{N_{87}(t)}{N_{86}}, \qquad X(t)=\frac{N_{\mathrm{Rb}}(t)}{N_{86}}, \qquad Y_0=\frac{N_{87}(0)}{N_{86}}. \]

Then

\[ \boxed{Y(t)=Y_0+(e^{\lambda t}-1)X(t).} \]

All samples have the same initial strontium isotope ratio \(Y_0\) and the same age \(t\). Their present values therefore lie on a straight line with vertical intercept \(Y_0\) and slope

\[ \boxed{a(t)=e^{\lambda t}-1.} \]

At crystallization the slope is zero; it subsequently increases with age.

1.4c

The line in Fig. 1.2(b) passes approximately through

\[ (X,Y)\approx(0.05,0.702) \quad\text{and}\quad (0.25,0.712). \]

Its slope is therefore approximately

\[ a\approx\frac{0.712-0.702}{0.25-0.05}=0.050. \]

Inverting the result of part 1.4b,

\[ \tau_{\mathrm M} =\frac{\ln(1+a)}{\lambda} =\frac{T_{\frac12}}{\ln2}\ln(1+a). \]

Hence

\[ \tau_{\mathrm M} \approx \frac{4.9\times10^{10}\ \mathrm{yr}}{\ln2}\ln(1.050) =3.45\times10^9\ \mathrm{yr}. \]

Allowing for the graphical reading,

\[ \boxed{\tau_{\mathrm M}\approx3.5\times10^9\ \mathrm{yr}.} \]

Figure 1.2(b) refers to samples from “a meteorite”; its inferred age need not equal Maribo’s age quoted in the introduction.

1.5

For an elliptical orbit, the perihelion and aphelion distances sum to twice the semimajor axis:

\[ a=\frac{a_{\min}+a_{\max}}2 =\frac{4.95\times10^{10}+6.16\times10^{11}}2 =3.3275\times10^{11}\ \mathrm m. \]

Neglecting the comet’s mass compared with the Sun’s mass, Kepler’s third law gives

\[ t_{\mathrm{Encke}}^2=\frac{4\pi^2a^3}{Gm_{\mathrm S}}. \]

Thus

\[ t_{\mathrm{Encke}} =2\pi\sqrt{ \frac{(3.3275\times10^{11})^3} {(6.67\times10^{-11})(1.99\times10^{30})} } =1.0468\times10^8\ \mathrm s. \]

Using \(1\ \mathrm{yr}\approx3.156\times10^7\ \mathrm s\),

\[ \boxed{t_{\mathrm{Encke}}\approx1.05\times10^8\ \mathrm s\approx3.32\ \mathrm{yr}.} \]

1.6a

First calculate the quantities used in all three impact questions. The asteroid’s mass is

\[ m_{\mathrm{ast}} =\frac43\pi R_{\mathrm{ast}}^3\rho_{\mathrm{ast}} =\frac43\pi(5.0\times10^3)^3(3.0\times10^3) =1.571\times10^{15}\ \mathrm{kg}. \]

Earth’s moment of inertia is

\[ I_{\mathrm E} =0.83\frac25m_{\mathrm E}R_{\mathrm E}^2 =8.068\times10^{37}\ \mathrm{kg\,m^2}. \]

Taking its initial rotation period to be \(\tau_{\mathrm E}=24\ \mathrm h=86400\ \mathrm s\),

\[ \omega_{\mathrm E}=\frac{2\pi}{\tau_{\mathrm E}} =7.272\times10^{-5}\ \mathrm{s^{-1}}, \]

and its spin angular momentum is

\[ L_{\mathrm E}=I_{\mathrm E}\omega_{\mathrm E} =5.867\times10^{33}\ \mathrm{kg\,m^2\,s^{-1}}. \]

We treat the accreted asteroid as a point mass at Earth’s surface. Its size and the displacement of the common center of mass give negligible corrections.

At the North Pole, the asteroid’s position vector is parallel to Earth’s initial rotation axis. Consequently, its angular momentum

\[ \mathbf L_{\mathrm{ast}} =\mathbf r\times m_{\mathrm{ast}}\mathbf v_{\mathrm{ast}} \]

is perpendicular to Earth’s initial spin angular momentum. Its magnitude is largest for a tangential impact:

\[ L_{\mathrm{ast}}^{\max} =m_{\mathrm{ast}}R_{\mathrm E}v_{\mathrm{ast}} =2.505\times10^{26}\ \mathrm{kg\,m^2\,s^{-1}}. \]

Conservation of angular momentum then gives, to leading order in the small asteroid-to-Earth mass ratio,

\[ \tan\theta_{\max} \simeq\frac{L_{\mathrm{ast}}^{\max}}{L_{\mathrm E}}. \]

The angle is extremely small, so

\[ \boxed{ \theta_{\max} \simeq \frac{m_{\mathrm{ast}}R_{\mathrm E}v_{\mathrm{ast}}} {I_{\mathrm E}\omega_{\mathrm E}} =4.27\times10^{-8}\ \mathrm{rad} \approx2.4\times10^{-6}\ \mathrm{degree}. } \]

The change in the inertia tensor contributes only a higher-order correction to this small tilt.

1.6b

For a radial impact, the incoming asteroid has zero angular momentum about Earth’s center. After it sticks at the Equator, the moment of inertia increases to

\[ I'=I_{\mathrm E}+m_{\mathrm{ast}}R_{\mathrm E}^2. \]

Angular momentum conservation requires

\[ I_{\mathrm E}\omega_{\mathrm E} =(I_{\mathrm E}+m_{\mathrm{ast}}R_{\mathrm E}^2)\omega'. \]

Since the period is inversely proportional to angular speed,

\[ \tau' =\tau_{\mathrm E} \left(1+\frac{m_{\mathrm{ast}}R_{\mathrm E}^2}{I_{\mathrm E}}\right). \]

Therefore,

\[ \Delta\tau_{\mathrm{vrt}} =\tau_{\mathrm E}\frac{m_{\mathrm{ast}}R_{\mathrm E}^2}{I_{\mathrm E}} =\tau_{\mathrm E}\frac{m_{\mathrm{ast}}}{0.332m_{\mathrm E}}. \]

Numerically,

\[ \boxed{ \Delta\tau_{\mathrm{vrt}} =6.85\times10^{-5}\ \mathrm s \approx+6.8\times10^{-5}\ \mathrm s. } \]

The day becomes approximately \(68\ \mu\mathrm s\) longer.

1.6c

The direction of the tangential impact matters. Let \(\sigma=+1\) for a prograde impact, which adds spin angular momentum, and \(\sigma=-1\) for a retrograde impact.

Conservation of angular momentum gives

\[ I_{\mathrm E}\omega_{\mathrm E} +\sigma m_{\mathrm{ast}}R_{\mathrm E}v_{\mathrm{ast}} =(I_{\mathrm E}+m_{\mathrm{ast}}R_{\mathrm E}^2)\omega'. \]

Define the small dimensionless quantities

\[ \epsilon =\frac{m_{\mathrm{ast}}R_{\mathrm E}^2}{I_{\mathrm E}} =7.925\times10^{-10}, \qquad \eta =\frac{m_{\mathrm{ast}}R_{\mathrm E}v_{\mathrm{ast}}} {I_{\mathrm E}\omega_{\mathrm E}} =4.270\times10^{-8}. \]

The new period and its change are

\[ \tau'=\tau_{\mathrm E}\frac{1+\epsilon}{1+\sigma\eta}, \]

\[ \boxed{ \Delta\tau_{\mathrm{tan}} =\tau_{\mathrm E}\frac{\epsilon-\sigma\eta}{1+\sigma\eta}. } \]

Since both \(\epsilon\) and \(\eta\) are very small, this is also accurately

\[ \Delta\tau_{\mathrm{tan}}\approx\tau_{\mathrm E}(\epsilon-\sigma\eta). \]

Thus:

  • Prograde impact:
    \[ \boxed{\Delta\tau_{\mathrm{tan}}=-3.621\times10^{-3}\ \mathrm s\approx-3.6\ \mathrm{ms}.} \]

  • Retrograde impact:
    \[ \boxed{\Delta\tau_{\mathrm{tan}}=+3.758\times10^{-3}\ \mathrm s\approx+3.8\ \mathrm{ms}.} \]

The added moment of inertia lengthens the day in both cases. The incoming angular momentum produces the larger effect and determines the sign.

1.7

First consider the body’s motion relative to the Sun, before Earth’s gravity becomes significant. At Earth’s orbital distance, a gravitationally bound body has negative specific orbital energy:

\[ \mathcal E =\frac{u^2}{2}-\frac{Gm_{\mathrm S}}{a_{\mathrm E}}<0. \]

Its heliocentric speed is therefore bounded by the local solar escape speed:

\[ u<u_{\max} =\sqrt{\frac{2Gm_{\mathrm S}}{a_{\mathrm E}}}. \]

The limiting case is an orbit whose binding energy tends to zero.

Approximating Earth’s orbit as circular at the given mean distance, its orbital speed is

\[ v_{\mathrm E}=\sqrt{\frac{Gm_{\mathrm S}}{a_{\mathrm E}}}. \]

The largest relative approach speed occurs when the body moves directly opposite to Earth’s orbital motion. Hence

\[ v_\infty^{\max} =u_{\max}+v_{\mathrm E} =(1+\sqrt2)\sqrt{\frac{Gm_{\mathrm S}}{a_{\mathrm E}}}. \]

Numerically,

\[ v_{\mathrm E}=29.75\ \mathrm{km\,s^{-1}}, \qquad u_{\max}=42.07\ \mathrm{km\,s^{-1}}, \]

so

\[ v_\infty^{\max}=71.82\ \mathrm{km\,s^{-1}}. \]

Earth’s gravity further accelerates the body. In the terrestrial encounter, conservation of mechanical energy gives

\[ \frac12v_{\mathrm{imp}}^2-\frac{Gm_{\mathrm E}}{R_{\mathrm E}} =\frac12v_\infty^2. \]

Thus

\[ \boxed{ v_{\mathrm{imp}}^{\max} = \sqrt{ (1+\sqrt2)^2\frac{Gm_{\mathrm S}}{a_{\mathrm E}} +\frac{2Gm_{\mathrm E}}{R_{\mathrm E}} }. } \]

Earth’s surface escape speed is

\[ \sqrt{\frac{2Gm_{\mathrm E}}{R_{\mathrm E}}} =11.17\ \mathrm{km\,s^{-1}}. \]

Therefore,

\[ \boxed{ v_{\mathrm{imp}}^{\max} =\sqrt{(71.82)^2+(11.17)^2}\ \mathrm{km\,s^{-1}} \approx72.7\ \mathrm{km\,s^{-1}}. } \]

For a strictly bound orbit, the speed is strictly below this value. It can approach the value arbitrarily closely as the initial solar binding energy tends to zero.

The solution was checked against the official solution.