Answer to Question #156138 in Quantum Mechanics for Anand

Question

 A satellite of mass 2500 kg is orbiting the Earth in an elliptical orbit. At the apagee, the 

altitude of the satellite is 3000 km, while at the perigee its altitude is 1000 km. Calculate 

the energy and angular momentum of the satellite and its speed at the apagee and perigee.

Answer 1

(a) We can find the energy of the satellite from the law of konservation of energy:

$$E_{sat}=KE_a+PE_a=KE_p+PE_p,$$ $$E_{sat}=\dfrac{1}{2}mv_a^2-\dfrac{GMm}{r_a},$$

$E_{sat}=\dfrac{1}{2}\cdot 2500\ kg\cdot(6114\ \dfrac{m}{s})^2-\dfrac{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot5.97\cdot10^{24}\ kg \cdot2500\ kg}{9.38\cdot10^6\ m}=-5.94\cdot10^{10}\ J.$

(b) We can find the angular momentum of the satellite from the formula:

$$L=mv_ar_a=mv_pr_p,$$

here, $m=2500\ kg$ is the mass of the satellite, $r_a=(R_E+h_a)$ is the apagee distance, $r_p=(R_E+h_p)$ is the perigee distance, $R_E=6.38\cdot10^6\ m$ is the radius of the Earth, $h_a=3.0\cdot10^6\ m$ is the altitude of apagee, $h_p=1.0\cdot10^6\ m$ is the altitude of perigee, $v_a$, $v_p$ is the speed of the satellite at apogee and perigee, respectivelly.

Let's first calculate $r_a$ and $r_p$:

$$r_a=(6.38\cdot10^6\ m+3.0\cdot10^6\ m)=9.38\cdot10^6\ m,$$ $$r_p=(6.38\cdot10^6\ m+1.0\cdot10^6\ m)=7.38\cdot10^6\ m.$$

Let's also find semi-major axis of the satellite:

$$a=\dfrac{r_a+r_p}{2},$$ $$a=\dfrac{9.38\cdot10^6\ m+7.38\cdot10^6\ m}{2}=8.38\cdot10^6\ m.$$

Then, we can find the speed of the satellite at apogee and perigee from the vis-viva equation:

$$v=\sqrt{GM(\dfrac{2}{r}-\dfrac{1}{a})}.$$

Then, the speed of the satellite at the apagee will be:

$v_a=\sqrt{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot5.97\cdot10^{24}\ kg\cdot(\dfrac{2}{9.38\cdot10^6\ m}-\dfrac{1}{8.38\cdot10^6\ m})}=6114\ \dfrac{m}{s}.$

$v_p=\sqrt{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot5.97\cdot10^{24}\ kg\cdot(\dfrac{2}{7.38\cdot10^6\ m}-\dfrac{1}{8.38\cdot10^6\ m})}=7771\ \dfrac{m}{s}.$

Finally, we can calculate the angular momentum of the satellite:

$$L=2500\ kg\cdot 6114\ \dfrac{m}{s}\cdot 9.38\cdot10^6\ m=1.43\cdot10^{14}\ \dfrac{kgm^2}{s}.$$

Answer:

a) $E_{sat}=-5.94\cdot10^{10}\ J.$

b) $L=1.43\cdot10^{14}\ \dfrac{kgm^2}{s}.$

c) $v_a=6114\ \dfrac{m}{s}, v_p=7771\ \dfrac{m}{s}.$