Question

6) (12 pts.) A meteor of mass m is approaching earth as shown on the sketch. The distance h on the sketch below is called the impact parameter. The radius of the earth is RE , and the mass of the earth is mE. Suppose the meteor has an initial speed of v0 and that the meteor started very far away (infinitely far) from the earth. Suppose the meteor just grazes the earth. You may ignore all other gravitational forces except the earth. (a) Taking the center of the Earth to be point S, calculate the torque ~τS about the center of the Earth. (b) Calculate the initial angular momentum vector ~ LS,0 about the center of the Earth
when the meteor is very far from the Earth. Express your answer in terms of h, m, and v0. (c) Calculate the final angular momentum vector ~ LS,f about the center of the Earth when the meteor is at its nearest approach to Earth (x = RE). Express your answer in terms of RE, vf, and m. Hint: The velocity of the meteor is no longer equal to v0. (d) Use energy conservation to determine the impact parameter h. Express your answer in terms of RE, G, mE, and v0.

Answer 1

(a) Taking the center of an Earth to be point S, then the torque about the center of an Earth will be given as :

Force on the meteor, $\vec{F}$ ^G_e,m = - (G m m_e / r²) $\hat{r}$

Vector from the center of an Earth to the meteor, $\vec{r}$ _s = r $\hat{r}$

The torque about S is zero, because they two vectors are anti-parallel.

Then, we get

$\vec{\tau }$ _S = $\vec{r}$ _s x $\vec{F}$ ^G_e,m

$\vec{\tau }$ _S = r $\hat{r}$ x [- (G m m_e / r²) $\hat{r}$ ]

$\vec{\tau }$ _S = 0

(b) The initial angular momentum vector about the center of an Earth which will be given as :

$\vec{L}$ _S,0 = $\vec{r}$ _s,0 x m $\vec{v}$ ₀

$\vec{L}$ _S,0 = (x₀ $\hat{i}$ + h _$\hat{j}$ ) x m v₀ $\hat{i}$

$\vec{L}$ _S,0 = - h m v₀ $\hat{k}$

(c) The final angular momentum vector about the center of an Earth which will be given as :

$\vec{L}$ _S,f = $\vec{r}$ _s,f x m $\vec{v}$ _f

$\vec{L}$ _S,f = R_e $\hat{i}$ x m v_f (- _$\hat{j}$ )

$\vec{L}$ _S,f = - R_e m v_f

(d) Using conservation of angular momentum, we have

L_initial = L_final $\Leftrightarrow$ m v₀ h = m v_f R_e

v_f = (v₀ h) / R_e

Since we are neglecting all forces on the meteor other than earth’s gravity & mechanical energy is conserved.

using conservation of energy, we have

E_initial = E_final $\Leftrightarrow$ K.E_i + P.E_i = K.E_f + P.E_f

(1/2) m v₀² = (1/2) m v_f² - (G m m_e / R_e)

(1/2) v₀² = (1/2) v_f² - (G m_e / R_e)

v₀² = v₀² (h / R_e)² - (2 G m_e / R_e)

Solving for 'h' & we get -

h = _$\sqrt{}$ R_e² + [(2 G m_e R_e) / v₀²]