Products

Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

Holooly Ads. Manager

Advertise your business, and reach millions of students around the world.

Holooly Tables

All the data tables that you may search for.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Holooly Help Desk

Need Help? We got you covered.

Q. 6.13

A spacecraft is in a 500 km by 10 000 km altitude geocentric orbit which intersects the equatorial plane at a true anomaly of 120º (see Figure 6.34). If the inclination to the equatorial plane is 15º, what is the minimum velocity increment required to make this an equatorial orbit?

Verified Solution

The orbital parameters are

$e=\frac{r_{A}-r_{P}}{r_{A}+r_{P}}=\frac{(6378 + 10 000) − (6378 + 500)}{(6378 + 10 000) + (6378 + 500)}= 0.4085$

$r_{P}=\frac{h^{2}}{\mu }\frac{1}{1+e\cos (0)}\Rightarrow 6878 =\frac{h^{2}}{398600}\frac{1}{1+0.4085}\Rightarrow h = 62 141 km/s$

The radial coordinate and velocity components at points B and C, on the line of intersection with the equatorial plane, are

$r_{B}=\frac{h^{2}}{\mu }\frac{1}{1+e\cos \theta _{B}}=\frac{62141^{2}}{398600}\frac{1}{1+0.4085\cdot \cos120^{\circ}}=12 174 km$

$v_{\bot _{B}}=\frac{h}{r_{B}}= \frac{62141}{12174}= 5.1043 km/s$

$v_{r _{B}}=\frac{\mu }{h}e\sin \theta _{B}=\frac{398600}{62141}· 0.4085 ·\sin 120^{\circ}= 2.2692 km/s$

and

$r_{C}=\frac{h^{2}}{\mu }\frac{1}{1+e\cos \theta _{C}}=\frac{62141^{2}}{398600}\frac{1}{1+0.4085\cdot \cos300^{◦}}=8044.6 km$

$v_{\bot _{C}}=\frac{h}{r_{C}}= \frac{62141}{8044.6}= 7.7246 km/s$

$v_{r _{C}}=\frac{\mu }{h} e\sin \theta _{C}=\frac{398600}{62141}· 0.4085 ·\sin 300^{◦} = −2.2692 km/s$

All we wish to do here is rotate the plane of the orbit rigidly around the node line BC. The impulsive maneuver must occur at either B or C. Equation 6.19 applies, and since the radial and perpendicular velocity components remain fixed, it reduces to

$\Delta v=\sqrt{(v_{r_{2}}-v_{r_{1}})^{2}+v^{2}_{\bot _{1}} +v^{2}_{\bot _{2}} -2v_{\bot _{1}}v_{\bot _{2}}\cos δ}$ (6.19)

$\Delta v=v_{\bot }\sqrt{2(1-\cosδ )}$

where δ = 15°. For the minimum Δv, the maneuver must be done where $v_{⊥}$ is smallest, which is at B, the point farthest from the center of attraction F. Thus,

$\Delta v=5.1043\sqrt{2(1-\cos 15^{◦})}= 1.3325 km/s$