My blog has moved!

You should be automatically redirected in 6 seconds. If not, visit
http://www.ledger-enquirer.com/bulldogs-blog/
and update your bookmarks.

Wednesday, April 6, 2011

For those who are wondering ...

... Right now the plan is to do a mailbag next week in advance of the spring game, and one the following week after G-Day.

I have a bunch of content to fill the blog the next few days, and will be off this weekend. (Headed back to D.C. to fulfill a personal obligation.)

In the meantime, faithful reader "Albert" sent in this mailbag question earlier in the week, and because I'm too smart to answer it, I'll post it in case anyone wants to take a swing at it:

Problem: A satellite is injected into an orbit at radius r from the center of attraction. The velocity is exactly the desired speed to go into circular orbit but the flight path angle γ≠0. The flight path angle is the angle between the velocity vector and the plane normal to the position vector.

a. Develop the relationship between the resulting eccentricity and sinγ.

b. Develop a relationship between the resulting periapsis distance and sinγ that is valid even for the rectilinear orbit case. Likewise for apoapsis distance.


As always, please show your work.

29 comments:

NCDawg said...

A Russian satellite or an American satellite?

Paul said...

I ain't even smart enough to know if that's a real question.

Anonymous said...

it's a trick questions. no country has ever sent anything into space.

BullyMack said...

err...I'll go with B...Baby Jesus. Baby Jesus, final answer.

Anonymous said...

hehe..."rectil" inear

Andy said...

It's simple physics. Calculate the velocity, v, in relation to the trajectory, t, in which g, gravity, of course remains a constant. It's not complicated.

Sandfly, GA said...

Let m be the mass of the satellite, even though we don't need it. Let a be the semi-major axis. Let e be the eccentricity.

For a circular orbit: v_0^2 = GM/r_0 (*1)

For our setup: v_0^2 = (v_0 sin γ)^2 + (v_0 cos γ)^2

For the extreme points in the orbit, there is no inward or outward motion.
So by the conservation of angular momentum, we must have

v_± = (r_0 v_0 cos γ) / r_± = (√(G M r_0) cos γ) / r_± (*2)

But from conversation of energy we have

m v_0²/ 2 - G M m/r_0 = m v_±² / 2 - G M m/r_±
or
v_±² - v_0² = 2GM (1/r_± - 1/r_0) (*3)

Combining *1, *2 and *3 we get

(G M r_0 cos² γ) / r_±² - GM/r_0 = 2GM (1/r_± - 1/r_0)
or
r_± = r_0 ± r_0 √(sin² γ) (*4)

From r_± = a(1 ± e) we see that we have:

e = | sin γ |

PS. Google is great.

bigeasydawg said...

it said "an orbit" not necessarily "earth orbit" so gravity is not a constant.

bigeasydawg said...

but i'm talkin out of my but and could be an idiot.

K said...

Gravity is a constant for a given center of orbit. It might not the same as what it would be orbiting earth, but it's there.

David said...

Is this a GT blog?

Bulldawg said...

David-

Ha!

Good one!

Bourbon Dawgwalker said...

Potato. The answer is Potato.

Tim the Enchanter said...

African or European satellite?

Next Question:
What is your favorite color?

Sir Galahad said...

Blue.

No, Yel-ahhhhhhhh

Anonymous said...

Take it to the Trade School over on North Avenue.

UGA grads don't need to know that stuff. We just need to know how to hire and fire nerds and how to keep from paying them too much.

Suomynons said...

Six

Anonymous said...

C All the above

Anonymous said...

A. Give me a bottle of good wine and a Frank Sinatra song.

B. My ex-wife got my periapsis and my siny, even now, we are too close to each other.

Anonymous said...

about 3.50

PTC DAWG said...

To answer the first question..not wondering.

dawgofdasouth said...

sandfly is that BOBO's playbook you just threw out there?

Booger Presley said...

Gimme your lunch money!

Unknown said...

hey seth,enjoy your blog.big dog fan from toccoa living in frederick md and working as a golf course supt @ va national in bluemont about 45 min west of dc.bring your clubs next week and i will hook you up.i would love to meet you.call maint and ask for bart.

Anonymous said...

220, 221, whatever it takes.

Anonymous said...

I have pondered and drank dark liquor for a while until I felt a warm sensation run down my leg. I'll have the answer tomorrow!

bizzaroneck said...

Hey Seth, question for the mailbag. Has Richt changed how much information that he is giving out about spring practice this year? It seems like no one has any good spring practice notes and in previous years you could practically find out what color powerade they were drinking.

Seth Emerson said...

Bizzanoreck, feel free to ask that again next week when I make the call for the mailbag. It's a good question worth answering, and I wouldn't want to forget it by next week.

Anonymous said...

Question for the mail bag. I'm assuming that Kwame Geathers will be starting with the first team Defense next Saturday at Nose. Who will be playing with the second team if this is the case?