Question

An unmanned satellite orbits the earth with a perigee radius of 10,000 km and an apogee radius of 100,000 km. Calculate:

a. the eccentricity of the orbit
b. the semimajor axis of the orbit(km)
c. the period of the orbit(hours)
d. the specific energy of the orbit(km^2/s^2)
e. the true anomaly at which the altitude is 1000km (degrees)
f. Vr and V(perpendicular) at the points found in part (e) (km/s)
g. the speed at the perigee and apogee (km/s)

Answer 1

Solution :

Given :

radius of perigee, $r_p$ = 10,000 km

radius of apogee, $r_a$ = 100,000 km

a). Eccentricity of the orbit

  $e=\frac{|r_p-r_a|}{r_p+r_a}$

$e=\frac{|10,000-100,000|}{10,000+100,000}$

$e=\frac{9}{11}$

or e = 0.818

b). Semi major axis of the orbit

  $a=\frac{r_p+r_a}{2}$

  $a=\frac{10,000+100,000}{2}$

     = 55,000 km

c). period of orbit

  $T=\frac{2\pi}{\sqrt{\mu}}\times a^{3/2}$

Replacing μ with $398600 \ km^3/s^2$

$T=\frac{2\pi}{\sqrt{398600}}\times (55,000)^{3/2}$

$T=128304.04 \ s \left(\frac{1 \ hr}{3600 \ s}\right)$

T = 35.64 hr

d). Specific energy of the orbit

$\varepsilon = -\frac{\mu}{2a}$

$\varepsilon = -\frac{398600}{2 \times 55000}$

$\varepsilon = -3.62 \ km^2/s^2$

e). the equation of the distance to the focus

$\theta = \cos^{-1}\left(\frac{a(1-e^2)}{r}-\frac{1}{e}\right)$

$\theta = \cos^{-1}\left(\frac{55000(1-(0.818)^2)}{(1000+6378)}-\frac{11}{9}\right)$

$\theta = \cos^{-1}\left(\frac{55000(0.33)}{(7378)}-\frac{11}{9}\right)$

$\theta = \cos^{-1}\left(2.4-1.2\right)$

$\theta = \cos^{-1}\left(1.2\right)$

θ = 1.002°

f).Calculating the angular momentum

$r_p=\frac{h^2}{\mu(1+e)}$

or $h=\sqrt{r_p \mu(1+e)}$

Now calculate the radial velocity

$v_r=\frac{\mu}{h} e \sin \theta$

Substituting for h,

$v_r=\frac{\mu}{h}e \sin \theta$

$v_r=\frac{e\mu \sin \theta}{\sqrt{r_p \mu(1+e)}}$

$v_r=\frac{\frac{9}{11}\sqrt{398600} \sin 20}{\sqrt{10,000 (1+0.818)}}$

$v_r= 1.30 \ km/s$

Now calculating the azimuthal velocity

$v_{\perp}=\frac{\mu}{h}(1+e \cos \theta)$

$v_{\perp}=\frac{\mu (1+e \cos \theta)}{\sqrt{r_p \mu(1+e)}}$

$v_{\perp}=\frac{\sqrt{398600} (1+0.818 \cos 20)}{\sqrt{10000(1+0.818)}}$

$v_{\perp}=7.58 \ km/s$

g). Velocity at perigee

$v_p=\frac{h}{r_p}$

$v_p=\frac{\sqrt{r_p \mu (1+e)}}{r_p}$

$v_p=\frac{\sqrt{10000 (398600) (1+0.818)}}{10000}$

$v_p=8.52 \ km/s$

Now calculate the velocity of the apogee

$v_a=\frac{h}{r_a}$

$v_a=\frac{\sqrt{r_p \mu (1+e)}}{r_a}$

$v_p=\frac{\sqrt{10000 (398600) (1+0.818)}}{100000}$

$v_a= 0.85 \ km/s$