Worked exercise: satellite, aphelion and perihelion

Problem

A satellite of mass $m_s = 10^{10}\ \text{kg}$ orbits the Earth ($M_T = 6\cdot 10^{24}\ \text{kg}$). At time A it is at position $\vec{r}_A = (-3\cdot 10^8,\ 8\cdot 10^8)\ \text{m}$ with velocity $\vec{v}_A = (-3000,\ 0)\ \text{m/s}$. Determine the minimum and maximum distance of the satellite from the Earth’s centre.

Initial state A: position and velocity known; we seek the two extreme points of the orbit (perihelion and aphelion).

Solution

Strategy. At the points of maximum and minimum distance $\vec{v}\perp\vec{r}$: we use together the conservation of total energy and of angular momentum about the Earth’s centre.

Step 1 — preliminary calculations. $r_A = |\vec{r}_A| = \sqrt{(3\cdot 10^8)^2 + (8\cdot 10^8)^2} = \sqrt{73}\cdot 10^8 \approx 8{,}54\cdot 10^8\ \text{m}$ $v_A = |\vec{v}_A| = 3000\ \text{m/s}$

Step 2 — total energy. $E_\text{kin,A} = \tfrac{1}{2}m_s v_A^2 = \tfrac{1}{2}\cdot 10^{10}\cdot 9\cdot 10^6 = 4{,}5\cdot 10^{16}\ \text{J}$ $E_\text{pot,A} = -\frac{GM_T m_s}{r_A} = -\frac{6{,}7\cdot 10^{-11}\cdot 6\cdot 10^{24}\cdot 10^{10}}{8{,}54\cdot 10^8} \approx -4{,}7\cdot 10^{16}\ \text{J}$ $E_\text{tot} = 4{,}5\cdot 10^{16} - 4{,}7\cdot 10^{16} = -2{,}2\cdot 10^{15}\ \text{J}$

Since $E_\text{tot} < 0$, the orbit is elliptical (bound).

Step 3 — angular momentum. $\vec{L}_A = m_s\,(\vec{r}_A\times\vec{v}_A) = 10^{10}\cdot\left[(-3\cdot 10^8)\cdot 0 - (8\cdot 10^8)\cdot(-3000)\right]$ $L = 10^{10}\cdot 2{,}4\cdot 10^{12} = 2{,}4\cdot 10^{22}\ \text{kg}\,\text{m}^2/\text{s}$

Step 4 — equations at point B (perihelion or aphelion, where $\vec{v}\perp\vec{r}$). Comparing state A with the generic state B:

	$E_\text{kin}$	$E_\text{pot,G}$	$E_\text{tot}$
A	$4{,}5\cdot 10^{16}$	$-4{,}7\cdot 10^{16}$	$-2{,}2\cdot 10^{15}$
B	$\tfrac{1}{2}m_s v_B^2$	$-\dfrac{GM_T m_s}{d_B}$	$-2{,}2\cdot 10^{15}$

	$L$
A	$2{,}4\cdot 10^{22}$
B	$m_s v_B d_B$

Step 5 — substitution. From conservation of $L$: $m_s v_B d_B = 2{,}4\cdot 10^{22}$, hence $v_B = \frac{2{,}4\cdot 10^{12}}{d_B}$

Substituting into the energy equation: $\frac{m_s}{2}\left(\frac{2{,}4\cdot 10^{12}}{d_B}\right)^2 - \frac{GM_T m_s}{d_B} = -2{,}2\cdot 10^{15}$

Multiplying by $d_B^2$ gives a quadratic equation in $d_B$: $2{,}2\cdot 10^{15}\,d_B^2 + 4\cdot 10^{24}\,d_B - 8{,}8\cdot 10^{44} = 0$

Step 6 — the two solutions. The two positive roots of the equation are the two extreme points: $\ev{d_B^\text{peri} \approx 3{,}2\cdot 10^8\ \text{m}\qquad d_B^\text{aph} \approx 5{,}5\cdot 10^8\ \text{m}}$

Check. The two solutions correspond exactly to the two points (minimum and maximum) where $\vec{v}\perp\vec{r}$: perihelion (the nearest point) and aphelion (the farthest point). At every other point of the orbit the velocity also has a radial component.

Note on the sign of $E_\text{tot}$: if $E_\text{tot}<0$ there are two positive roots (ellipse); if $E_\text{tot}=0$ there is one (parabola); if $E_\text{tot}>0$ there is one (hyperbola).

Watch the numerical data $E_\text{pot,A} = -GM_T m_s / r_A \approx -4{,}7\cdot 10^{15}\ \text{J}$ (not $10^{16}$): the textbook example contains a slip of a factor of 10. The values of the two extreme points should therefore be taken as indicative — what matters is the method (two conservation laws → quadratic equation → two points where $\vec{v}\perp\vec{r}$). Redo the calculation carefully with your own data.

With the printed values the correct calculation of the potential energy is

Links

Topics: Gravitation Concepts: Conservation of mechanical energy · Conservation of angular momentum · Universal gravitational potential energy · Orbits Skills: Conservation of energy · Conservation of angular momentum Objects: Satellite