§AERE 351 Homework 3

Sep 25, 2025

§1.

Target:
v=μh1+2ecos⁡θ+e2 v = \frac{\mu}{h} \sqrt{1 + 2 e \cos \theta + e^2}
The velocity has two components, the transversal and radial velocities. Adding them up should result in the total velocity in the vector format. As for the magnitude, since the equations from Lecture 5 are scalars, the Pythagorean theorem must be used. The velocities respectively are:
vθ=μp(1+ecos⁡θ) v_\theta = \sqrt{\frac{\mu}{p}} (1 + e \cos \theta)
vr=μpesin⁡θ v_r = \sqrt{\frac{\mu}{p}} e \sin \theta
The magnitude:
v=vθ2+vr2 v = \sqrt{v_\theta^2 + v_r^2}
v=μp(1+ecos⁡θ)2+(esin⁡θ)2 v = \sqrt{\frac{\mu}{p}} \sqrt{(1 + e \cos \theta)^2 + (e \sin \theta)^2}
v=μp1+2ecos⁡θ+e2cos⁡2θ+e2sin⁡2θ v = \sqrt{\frac{\mu}{p}} \sqrt{1 + 2 e \cos \theta + e^2 \cos^2 \theta + e^2 \sin^2 \theta}
v=μp1+2ecos⁡θ+e2(cos⁡2θ+sin⁡2θ) v = \sqrt{\frac{\mu}{p}} \sqrt{1 + 2 e \cos \theta + e^2 \cancel{(\cos^2 \theta + \sin^2 \theta)}}
v=μp1+2ecos⁡θ+e2 v = \sqrt{\frac{\mu}{p}} \sqrt{1 + 2 e \cos \theta + e^2}
This is very close the target but the coefficient needs changing. Fortunately:
p=h2μ p = \frac{h^2}{\mu}
Hence:
μp=μh2μ=μ2h2=μh \sqrt{\frac{\mu}{p}} = \sqrt{\frac{\mu}{\frac{h^2}{\mu}}} = \sqrt{\frac{\mu^2}{h^2}} = \frac{\mu}{h}
Thus:
v=μh1+2ecos⁡θ+e2 \boxed{v = \frac{\mu}{h} \sqrt{1 + 2 e \cos \theta + e^2}}

§2.

Given:
r=[7000,−2000,−4000]Tkm r = [7000, -2000, -4000]^T km
v=[3,−6,5]Tkm/s v = [3, -6, 5]^T km/s
hh is the specific angular momentum and I am going to need it:
h=r×v h = r \times v
h=[7000,−2000,−4000]T×[3,−6,5]T h = [7000, -2000, -4000]^T \times [3, -6, 5]^T
h=[−34000,−47000,−36000]Tkm2/s h = [-34000, -47000, -36000]^T km^2/s
I am also going to need rr but normalized:
∣r∣=70002+20002+40002=8306.62km |r| = \sqrt{7000^2 + 2000^2 + 4000^2} = 8306.62 km
r^=r∣r∣=18306.62[7000,−2000,−4000]T \hat{r} = \frac{r}{|r|} = \frac{1}{8306.62} [7000, -2000, -4000]^T
r^=[0.8427,−0.2408,−0.4815]T \hat{r} = [0.8427, -0.2408, -0.4815]^T
Solving for ee:
e=v×hμ−r^ e = \frac{v \times h}{\mu} - \hat{r}
Evaluating v×hv \times h:
v×h=[3,−6,5]T×[−34000,−47000,−36000]T v \times h = [3, -6, 5]^T \times [-34000, -47000, -36000]^T
v×h=[451000,−62000,−345000]Tkm3/s2 v \times h = [451000, -62000, -345000]^T km^3/s^2
From Wikipedia for Earth:
μ=3.986004418∗1014m3/s2=398600.4418km3/s2 \mu = 3.986004418 * 10^{14} m^3/s^2 = 398600.4418 km^3/s^2
v×hμ=[451000,−62000,−345000]T398600.4418 \frac{v \times h}{\mu} = \frac{[451000, -62000, -345000]^T}{398600.4418}
v×hμ=[1.13146,−0.155544,−0.865528]T \frac{v \times h}{\mu} = [1.13146, -0.155544, -0.865528]^T
Finally:
e=v×hμ−r^ e = \frac{v \times h}{\mu} - \hat{r}
e=[1.13146,−0.155544,−0.865528]T−[0.8427,−0.2408,−0.4815]T e = [1.13146, -0.155544, -0.865528]^T - [0.8427, -0.2408, -0.4815]^T
e=[0.28876,0.085256,−0.384028]T \boxed{e = [0.28876, 0.085256, -0.384028]^T}
It wasn't clear to me if the question wanted the value for ee as a scalar or vector so here's the scalar too:
∣e∣=0.288762+0.0852562+0.3840282 |e| = \sqrt{0.28876^2 + 0.085256^2 + 0.384028^2}
∣e∣=0.4880 \boxed{|e| = 0.4880}
As for the true anomaly, that's the angle between ee and rr:
cos⁡θ=r^⋅e^ \cos \theta = \hat{r} \cdot \hat{e}
θ=arccos⁡r^⋅e^ \theta = \arccos \hat{r} \cdot \hat{e}
I have r^\hat{r} already but not e^\hat{e}:
e^=e∣e∣=[0.28876,0.085256,−0.384028]T0.4880 \hat{e} = \frac{e}{|e|} = \frac{[0.28876, 0.085256, -0.384028]^T}{0.4880}
e^=[0.5917,0.1747,−0.7869]T \hat{e} = [0.5917, 0.1747, -0.7869]^T
The product:
r^⋅e^=[0.8427,−0.2408,−0.4815]T⋅[0.5917,0.1747,−0.7869]T \hat{r} \cdot \hat{e} = [0.8427, -0.2408, -0.4815]^T \cdot [0.5917, 0.1747, -0.7869]^T
r^⋅e^=0.8355 \hat{r} \cdot \hat{e} = 0.8355
Finally:
θ=arccos⁡r^⋅e^=arccos⁡0.8355 \theta = \arccos \hat{r} \cdot \hat{e} = \arccos 0.8355
θ=0.5818=33.33° \boxed{\theta = 0.5818 = 33.33\degree}

§3.

Here, I will be using the cc subscript represents association with the circle orbit and ee for the elliptical. The constraint:
v(θ)e=v(θ)c v(\theta)_e = v(\theta)_c
I already have vv as a function of θ\theta:
v(θ)=v=μh1+2ecos⁡θ+e2 v(\theta) = v = \frac{\mu}{h} \sqrt{1 + 2 e \cos \theta + e^2}
v(θ)e=μhe1+2eecos⁡θ+ee2 v(\theta)_e = \frac{\mu}{h_e} \sqrt{1 + 2 e_e \cos \theta + e_e^2}
v(θ)c=μhc1+2eccos⁡θ+ec2 v(\theta)_c = \frac{\mu}{h_c} \sqrt{1 + 2 e_c \cos \theta + e_c^2}
μhe1+2eecos⁡θ+ee2=μhc1+2eccos⁡θ+ec2 \frac{\cancel{\mu}}{h_e} \sqrt{1 + 2 e_e \cos \theta + e_e^2} = \frac{\cancel{\mu}}{h_c} \sqrt{1 + 2 e_c \cos \theta + e_c^2}
1he2(1+2eecos⁡θ+ee2)=1hc2(1+2eccos⁡θ+ec2) \frac{1}{h_e^2} (1 + 2 e_e \cos \theta + e_e^2) = \frac{1}{h_c^2} (1 + 2 e_c \cos \theta + e_c^2)
Since a circular orbit has no eccentricity:
ec=0 e_c = 0
1he2(1+2eecos⁡θ+ee2)=1hc2 \frac{1}{h_e^2} (1 + 2 e_e \cos \theta + e_e^2) = \frac{1}{h_c^2}
1+2eecos⁡θ+ee2=he2hc2 1 + 2 e_e \cos \theta + e_e^2 = \frac{h_e^2}{h_c^2}
2eecos⁡θ=he2hc2−ee2−1 2 e_e \cos \theta = \frac{h_e^2}{h_c^2} - e_e^2 - 1
θ=arccos⁡(he2/hc2−1−ee22ee) \boxed{\theta = \arccos \left( \frac{h_e^2 / h_c^2 - 1 - e_e^2}{2e_e} \right)}
For part (b), the goal is to find the flight path angle which is the angle between the total velocity and the transversal velocity. I will first be solving for the flight path angle of the elliptical orbit.
cos⁡γe=veθve \cos \gamma_e = \frac{v_{e\theta}}{v_e}
γe=arccos⁡veθve \gamma_e = \arccos \frac{v_{e\theta}}{v_e}
Expanding vev_e:
ve=μhe1+2eecos⁡θ+ee2 v_e = \frac{\mu}{h_e} \sqrt{1 + 2 e_e \cos \theta + e_e^2}
θ=arccos⁡(he2/hc2−1−ee22ee) \theta = \arccos \left( \frac{h_e^2 / h_c^2 - 1 - e_e^2}{2e_e} \right)
ve=μhe1+2ee(he2/hc2−1−ee22ee)+ee2 v_e = \frac{\mu}{h_e} \sqrt{1 + \cancel{2 e_e} \left( \frac{h_e^2 / h_c^2 - 1 - e_e^2}{\cancel{2e_e}} \right) + e_e^2}
ve=μhe1+he2/hc2−1−ee2+ee2 v_e = \frac{\mu}{h_e} \sqrt{\cancel{1} + h_e^2 / h_c^2 - \cancel{1} - \cancel{e_e^2} + \cancel{e_e^2}}
ve=μhehe2/hc2 v_e = \frac{\mu}{h_e} \sqrt{h_e^2 / h_c^2}
ve=μhe(he/hc) v_e = \frac{\mu}{\cancel{h_e}} (\cancel{h_e} / h_c)
ve=μhc v_e = \frac{\mu}{h_c}
Expanding vθv_\theta:
vθ=μp(1+ecos⁡θ) v_\theta = \sqrt{\frac{\mu}{p}} (1 + e \cos \theta)
p=h2μ p = \frac{h^2}{\mu}
vθ=μh(1+ecos⁡θ) v_\theta = \frac{\mu}{h} (1 + e \cos \theta)
vθv_\theta for the elliptical orbit:
veθ=μhe(1+eecos⁡θ) v_{e\theta} = \frac{\mu}{h_e} (1 + e_e \cos \theta)
veθ=μhe(1+ee(he2/hc2−1−ee22ee)) v_{e\theta} = \frac{\mu}{h_e} \left( 1 + e_e \left( \frac{h_e^2 / h_c^2 - 1 - e_e^2}{2e_e} \right) \right)
veθ=μhe(1+12(he2/hc2−1−ee2)) v_{e\theta} = \frac{\mu}{h_e} \left( 1 + \frac{1}{2} \left( h_e^2 / h_c^2 - 1 - e_e^2 \right) \right)
The fraction within the arccosine:
veθve=μhe(1+12(he2/hc2−1−ee2))μhc \frac{v_{e\theta}}{v_e} = \frac{\cancel{\frac{\mu}{h_e}} \left( 1 + \frac{1}{2} \left( h_e^2 / h_c^2 - 1 - e_e^2 \right) \right)}{\cancel{\frac{\mu}{h_c}}}
veθve=1+12(he2/hc2−1−ee2) \frac{v_{e\theta}}{v_e} = 1 + \frac{1}{2} \left( h_e^2 / h_c^2 - 1 - e_e^2 \right)
veθve=12(he2/hc2−ee2+1) \frac{v_{e\theta}}{v_e} = \frac{1}{2} \left( h_e^2 / h_c^2 - e_e^2 + 1 \right)
γe=arccos⁡veθve \gamma_e = \arccos \frac{v_{e\theta}}{v_e}
γe=arccos⁡12(he2/hc2−ee2+1) \boxed{\gamma_e = \arccos \frac{1}{2} \left( h_e^2 / h_c^2 - e_e^2 + 1 \right)}
That was the flight path angle for just the ellipse. Now, I will deal with the circle:
vc=μhc1+2eccos⁡θ+ec2 v_c = \frac{\mu}{h_c} \sqrt{1 + \cancel{2 e_c \cos \theta} + \cancel{e_c^2}}
vc=μhc v_c = \frac{\mu}{h_c}
vθ=μhc(1+ecos⁡θ) v_\theta = \frac{\mu}{h_c} (1 + \cancel{e \cos \theta})
vθ=μhc v_\theta = \frac{\mu}{h_c}
γc=arccos⁡1=0 \boxed{\gamma_c = \arccos 1 = 0}
I feel dumb. I saw this coming when I was solving for the ellipse, forgot about it, and proved it again that there is no radial velocity, just transversal for circles 🤦Nevertheless, I am unsure where I was supposed to end up using this identity I was given:
sin⁡arccos⁡−e=1−e2 \sin \arccos -e = \sqrt{1 - e^2}
I suppose, if you were to solve for γe\gamma_e using the radial velocity, a sine function show show up which you could resolve using the equation above, something that I did not run into.

§4.

Given:
vp=kva v_p = k v_a
Find:
e=e(k) e = e(k)
From Lecture 5:
vp=μp(1+e) v_p = \sqrt{\frac{\mu}{p}} (1 + e)
va=μp(1−e) v_a = \sqrt{\frac{\mu}{p}} (1 - e)
Derivation:
vp=kva v_p = k v_a
μp(1+e)=kμp(1−e) \cancel{\sqrt{\frac{\mu}{p}}} (1 + e) = k \cancel{\sqrt{\frac{\mu}{p}}} (1 - e)
1+e=k(1−e) 1 + e = k (1 - e)
1+e=k−ke 1 + e = k - k e
e+ke=k−1 e + k e = k - 1
e(1+k)=k−1 e (1 + k) = k - 1
e=k−11+k e = \frac{k - 1}{1 + k}
e=k−1k+1 \boxed{e = \frac{k - 1}{k + 1}}

§5.

Given:
μ=398600km3/s2 \mu = 398600 km^3/s^2
R=6380km R = 6380km
ω=7.29∗10−5rad/s \omega = 7.29 * 10^{−5} rad/s
Assuming a circular orbit, converting from angular velocity to transversal velocity:
v=vθ=rω v = v_\theta = r \omega
From Lecture 5:
vθ=μh(1+ecos⁡θ) v_\theta = \frac{\mu}{h} (1 + \cancel{e \cos \theta})
vθ=v=μh=rω v_\theta = v = \frac{\mu}{h} = r \omega
And since:
h=rvθ=rv h = r v_\theta = r v
μrv=rω \frac{\mu}{r v} = r \omega
μrrω=rω \frac{\mu}{r r \omega} = r \omega
μr2ω=rω \frac{\mu}{r^2 \omega} = r \omega
μω2=r3 \frac{\mu}{\omega^2} = r^3
r=(μω2)1/3 r = \left( \frac{\mu}{\omega^2} \right)^{1/3}
r=(398600km3/s2(7.29∗10−5rad/s)2)1/3=42172.3km r = \left( \frac{398600 km^3/s^2}{(7.29 * 10^{−5} rad/s)^2} \right)^{1/3} = 42172.3km
r=altitude+R r = \text{altitude} + R
altitude=r−R=42172.3km−6380km \text{altitude} = r - R = 42172.3km - 6380km
altitude=35792.3km \boxed{\text{altitude} = 35792.3km}
v=rω v = r \omega
v=42172.3km∗7.29∗10−5rad/s v = 42172.3km * 7.29 * 10^{−5} rad/s
v=3.074km/s \boxed{v = 3.074km/s}
The last part of the question regarding the maximum latitude that a sensor can detect a satellite was a head scratcher so I drew it out:
There exist latitudes of the Earth where the ground blocks the satellite. The limit should be tangential to the surface:
A couple of equations pop out of this:
α+ϕ=π \alpha + \phi = \pi
α+β+π2=π \alpha + \beta + \frac{\pi}{2} = \pi
r2=R2+A2 r^2 = R^2 + A^2
R=rcos⁡α R = r \cos \alpha
Retrospect, I am not sure why I put down ϕ\phi as I won't be using it, but oh well. α\alpha here is the latitude.
α=arccos⁡Rr \alpha = \arccos \frac{R}{r}
α=arccos⁡6380km42172.3km \alpha = \arccos \frac{6380km}{42172.3km}
α=81.30° \boxed{\alpha = 81.30\degree}

§6.

Facts:
R=6378km R = 6378km
μ=398600.4418km3/s2 \mu = 398600.4418 km^3/s^2
Given (a):
hp=300km h_p = 300km
ha=2000km h_a = 2000km
Characteristics:
rp=R+hp=6378km+300km=6678km r_p = R + h_p = 6378km + 300km = 6678km
ra=R+ha=6378km+2000km=8378km r_a = R + h_a = 6378km + 2000km = 8378km
e=ra−rpra+rp=8378km−6678km8378km+6678km=0.1129 e = \frac{r_a - r_p}{r_a + r_p} = \frac{8378km - 6678km}{8378km + 6678km} = 0.1129
a=ra+rp2=8378km+6678km2=7528km a = \frac{r_a + r_p}{2} = \frac{8378km + 6678km}{2} = 7528km
p=a(1−e2)=7528km(1−0.11292)=7432km p = a (1 - e^2) = 7528km (1 - 0.1129^2) = 7432km
vp=μp(1+e)=398600.4418km3/s27432km(1+0.1129)=8.150km/s v_p = \sqrt{\frac{\mu}{p}} (1 + e) = \sqrt{\frac{398600.4418 km^3/s^2}{7432km}} (1 + 0.1129) = 8.150km/s
va=μp(1−e)=398600.4418km3/s27432km(1−0.1129)=6.497km/s v_a = \sqrt{\frac{\mu}{p}} (1 - e) = \sqrt{\frac{398600.4418 km^3/s^2}{7432km}} (1 - 0.1129) = 6.497km/s
T=2πa3μ=2π(7528km)3398600.4418km3/s2=6500s T = 2 \pi \sqrt{\frac{a^3}{\mu}} = 2 \pi \sqrt{\frac{(7528km)^3}{398600.4418 km^3/s^2}} = 6500s
Given (b):
a=10000km a = 10000km
e=0.1129 e = 0.1129
Characteristics:
p=a(1−e2)=10000km(1−0.11292)=9873km p = a (1 - e^2) = 10000km (1 - 0.1129^2) = 9873km
T=2πa3μ=2π(10000km)3398600.4418km3/s2=9952s T = 2 \pi \sqrt{\frac{a^3}{\mu}} = 2 \pi \sqrt{\frac{(10000km)^3}{398600.4418 km^3/s^2}} = 9952s
vp=μp(1+e)=398600.4418km3/s29873km(1+0.1129)=7.071km/s v_p = \sqrt{\frac{\mu}{p}} (1 + e) = \sqrt{\frac{398600.4418 km^3/s^2}{9873km}} (1 + 0.1129) = 7.071km/s
va=μp(1−e)=398600.4418km3/s29873km(1−0.1129)=5.637km/s v_a = \sqrt{\frac{\mu}{p}} (1 - e) = \sqrt{\frac{398600.4418 km^3/s^2}{9873km}} (1 - 0.1129) = 5.637km/s
a=ra+rp2 a = \frac{r_a + r_p}{2}
2a=ra+rp 2a = r_a + r_p
ra=2a−rp r_a = 2a - r_p
e=ra−rpra+rp=2a−rp−rp2a−rp+rp=a−rpa e = \frac{r_a - r_p}{r_a + r_p} = \frac{2a - r_p - r_p}{2a - r_p + r_p} = \frac{a - r_p}{a}
ea=a−rp e a = a - r_p
rp=a−ea=10000km−0.1129∗10000km=8871km r_p = a - e a = 10000km - 0.1129 * 10000km = 8871km
ra=2a−rp=2∗10000km−8871km=11129km r_a = 2a - r_p = 2 * 10000km - 8871km = 11129km
Given (c):
rp=6678km r_p = 6678km
a=20000km a = 20000km
Characteristics:
T=2πa3μ=2π(20000km)3398600.4418km3/s2=28149s T = 2 \pi \sqrt{\frac{a^3}{\mu}} = 2 \pi \sqrt{\frac{(20000km)^3}{398600.4418 km^3/s^2}} = 28149s
ra=2a−rp=2∗20000km−6678km=33322km r_a = 2a - r_p = 2 * 20000km - 6678km = 33322km
e=ra−rpra+rp=33322km−6678km33322km+6678km=0.6661 e = \frac{r_a - r_p}{r_a + r_p} = \frac{33322km - 6678km}{33322km + 6678km} = 0.6661
p=a(1−e2)=20000km(1−0.66612)=11126km p = a (1 - e^2) = 20000km (1 - 0.6661^2) = 11126km
vp=μp(1+e)=398600.4418km3/s211126km(1+0.6661)=9.972km/s v_p = \sqrt{\frac{\mu}{p}} (1 + e) = \sqrt{\frac{398600.4418 km^3/s^2}{11126km}} (1 + 0.6661) = 9.972km/s
va=μp(1−e)=398600.4418km3/s211126km(1−0.6661)=1.999km/s v_a = \sqrt{\frac{\mu}{p}} (1 - e) = \sqrt{\frac{398600.4418 km^3/s^2}{11126km}} (1 - 0.6661) = 1.999km/s
Given (d):
h1=H(θ=126°)=1550km h_1 = H(\theta = 126\degree) = 1550km
h2=H(θ=58°)=850km h_2 = H(\theta = 58\degree) = 850km
Characteristics:
r=p1+ecos⁡θ=H+R r = \frac{p}{1 + e \cos \theta} = H + R
p1+ecos⁡(126∗(π/180))=1550+6378 \frac{p}{1 + e \cos (126 * (\pi / 180))} = 1550 + 6378
p1+ecos⁡(58∗(π/180))=850+6378 \frac{p}{1 + e \cos (58 * (\pi / 180))} = 850 + 6378
  ⟹  e=0.08245, p=7544km \implies e = 0.08245, ~ p = 7544km
p=a(1−e2) p = a (1 - e^2)
a=p1−e2=7544km1−0.082452=7596km a = \frac{p}{1 - e^2} = \frac{7544km}{1 - 0.08245^2} = 7596km
T=2πa3μ=2π(7596km)3398600.4418km3/s2=6589s T = 2 \pi \sqrt{\frac{a^3}{\mu}} = 2 \pi \sqrt{\frac{(7596km)^3}{398600.4418 km^3/s^2}} = 6589s
rp=a−ea=7596km−0.08245∗7596km=6970km r_p = a - e a = 7596km - 0.08245 * 7596km = 6970km
ra=2a−rp=2∗7596km−6970km=8222km r_a = 2a - r_p = 2 * 7596km - 6970km = 8222km
vp=μp(1+e)=398600.4418km3/s27544km(1+0.08245)=7.868km/s v_p = \sqrt{\frac{\mu}{p}} (1 + e) = \sqrt{\frac{398600.4418 km^3/s^2}{7544km}} (1 + 0.08245) = 7.868km/s
va=μp(1−e)=398600.4418km3/s27544km(1−0.08245)=6.670km/s v_a = \sqrt{\frac{\mu}{p}} (1 - e) = \sqrt{\frac{398600.4418 km^3/s^2}{7544km}} (1 - 0.08245) = 6.670km/s