NASA physics

theotherguy

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I recentley got accepted as a NASA high school aerospace scholar. This is a special program hosted by NASA in which thousands of students from across the country submit an essay and a transcript to senators, who then select the top students to compete in a rigourous program designed to prime students for aerospace engineering.

I am supposed to do ten long and complicated projects over the course of the year and then attend a summer program with a select group of students and discuss with NASA scientists the probelms facing space travel.


However, I have done all of my first project except for this one question, which I simply don't know how to answer. I recognize it as a simple acceleration equation, but it has some terms that I don't quite understand, as we havent covered orbits in my physics class yet. My physics teacher is out on a holiday and none of my peers know how to do this either. Can anyone help me understand how to do this problem?

Part Two: Shuttle Math The De-orbit Burn
When it is time for the Space Shuttle to come in for a landing, it needs to reenter the Earth's atmosphere at the right time and place so that it can land successfully at the designated landing strip (Kennedy Space Center in Florida, Edwards Air Force Base in California, or the White Sands Test Facility in New Mexico).

Depending on the Shuttle's altitude, it's de-orbit burn must occur for the correct length of time for the Shuttle to begin its descent at the right speed and in the correct location. The de-orbit burn is done against the direction of travel. The Shuttle keeps going in the same direction but slows down due to the drag on the spacecraft as it enters the atmosphere.

De-orbit maneuvers are usually done to lower the perigee of the orbit to 60 miles (or less). The Orbiter is captured and re-enters as it passes into the atmosphere at this altitude.

There is a change of 1 mile for every 2 feet per second (fps) change in velocity when you are below a 500-mile altitude above the Earth.

Determine the change in velocity (delta-V) that the Shuttle will need to make if it is at an altitude of 220 miles above the Earth at apogee and 215 miles above the Earth at perigee, and needs to drop the perigee to an altitude of 60 miles.

I'm sorry if I sound lazy, but I've been looking up information on how to do this for some time and I can't seem to figure it out.
 
Don't worry, we're all physicists here.
Haha, I took one physics course because it fit my major, and we never learned about apogee or perigee. No idea.
 
are the first three sentences not irrelevant?

"There is a change of 1 mile for every 2 feet per second (fps) change in velocity when you are below a 500-mile altitude above the Earth."

so for every mile you wish to lower the perigee you need to slow down the orbiting craft by 2f/s

"Determine the change in velocity (delta-V) that the Shuttle will need to make if it is at an altitude of 220 miles above the Earth at apogee and 215 miles above the Earth at perigee, and needs to drop the perigee to an altitude of 60 miles."

i would assume that you simply wish to lower the perigee from 215 to 60 miles, a difference of 155 miles, for each mile you must slow the craft down by 2f/s

155 * 2 = 310f/s

don't shout at me if i've made a fundamental error, i avoided astronomy like the plague when i did my physics degree and did nuclear instead:p
 
I found an astronomy site and found out what a retrograde burn was, and discoverd that it simply lowers the perigee by a factor of X, at which point it all made sense to me and I came out with 310 ft/s


EDIT:

ooh, that was right!? Thank you.
 
What kind of super NASA physics showcase student can't multiply 155 by 2?
 
Can't you apply the Newton equations of motion?
 
What kind of super NASA physics showcase student can't multiply 155 by 2?

The terms confused me, and I misread the part about "the change is 1 mile for every two feet per second"...I didnt know it meant 1 mile in altitude
 
The terms confused me, and I misread the part about "the change is 1 mile for every two feet per second"...I didnt know it meant 1 mile in altitude

I hope you're not launching the next probe to mars... was that 0.002 feet per second, or 0.002 km per hour?
 
Llama,
Let's put it this way. 5 cents is not the same as 5 dollars. I understand that part. Similarly, 1 cent is not the same as 1 dollar. But those are WHOLE NUMBERS. We are talking about fractions here. Remember that 1 cent = $0.01. It's not a whole number anymore. So 0.002 cents = $0.002, because it is a fraction. Call it a percentage or whatever you want. It's the same.
 
That makes no sense. There are one hundred cents in a dollar, so x cents will be equal to the same number of dollars / 100.

Eg x = 2 cents
2 / 100 = 0.02 = X Dollars
 
Llama,
Let's put it this way. 5 cents is not the same as 5 dollars. I understand that part. Similarly, 1 cent is not the same as 1 dollar. But those are WHOLE NUMBERS. We are talking about fractions here. Remember that 1 cent = $0.01. It's not a whole number anymore. So 0.002 cents = $0.002, because it is a fraction. Call it a percentage or whatever you want. It's the same.
You're a troll.

No-body is that big an idiot.
 
Solaris, the difference between intelligence and stupidity is that intelligence has it's limits.
 
To the original poster, that?s cool that you got selected and everything. I know this is a lot of fellatio coming from me and such but I knew you were going somewhere from all those posts.
 
Solaris and Llama don't have a sense of humor. Don't you guys remember the "Verizon can't do math" thread? The Verizon customer service reps actually insisted that 0.002 cents = $0.002, so Solaris is actually wrong: somebody is that big of an idiot. Here, convince yourselves: http://www.youtube.com/watch?v=Gp0HyxQv97Q
 
I admit to joking. I expected :LOL:, :thumbs: and :E ; not :angry:, :flame: and :sniper:. I guess my sense of humor sucks. Sorry :(

I laughed. I thought Llama's reaction was hilarious. He's dumber than you :p

j/k <3 all
 
the whole 0.002 reference seems to have gone over everyone's head
 
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