I’m a part of the Texas High School Aerospace Scholars program, an online science course that the Johnson Space Center is conducting. All of the lessons consist of an essay, math problem, and occasionally a graphic. The first lesson doesn’t quite explain the math problem that well, and many of the other students are confused. I can’t simply tell you the full question because I’m supposed to be doing this myself, but I just need clarification as to what it’s asking for.
If an orbiting craft needs to reduce its perigee to a certain altitude from its current altitude, how would I find the change in velocity the craft must make? I’m provided with both the initial and final distances, along with the mile:fps change in velocity ratio for these altitudes. Two of the other students have mention the Braking Distance formula of Vf^2=Vo^2+2ad, but I don’t see how to use it.
“I’m provided with both the initial and final distances, along with the mile:fps change in velocity ratio for these altitudes.”
Sounds like to me it is a little more simple than they are making it out to be. If you have the ratio, you can figure out the initial and final velocities from that ratio table. Change in velocity is just the difference. Now if you need to calculate a rate of change of velocity (constant deceleration), then the equation your friends gave would be correct.
I am not a Rocket scientist, but I was a Physics tutor in College. Without the exact details of the problem, that is about all I can help. If you would like some further assistance, feel free to PM me and I can discuss further.
When we learned about satellite motion in my physics class last year, to find velocity at a certain altitude we used the equation
v = sqrt(G*m/r)
Where G is the gravitational constant, 6.67x 10^-11 Nm2/kg2
m is the mass (kg) of the body/planet the satellite is orbiting, and r is the radius (altitude) of orbit in meters.
You could use the given distances to find the velocity at each altitude, subtract your answers to find the change, and convert to the proper units.
This is true if you are definitely balancing centripetal Acceleration with gravitational pull. From the problem statement he gave though, it sounds like they gave him a table or ratio to calculate those values.
The page says, “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.” Sorry if I wasn’t clear. I didn’t mean they actually gave me two ratios.
I think it’s much easier than it is made out to be, and others have mentioned that using that exact equation isn’t even necessary. At the moment, I’m just stumped on the velocities.
EDIT: Delta-V, change in velocity, is defined as (Vf-Vo), correct? If so, would Delta-V or the ‘a’ in the above mentioned formula be the given ratio?
Say initial altitude is 500 miles and final altitude is 400 miles, the change in altitude would be 100 miles therefore the change in velocity would be:
(100 miles)/(1 mile/2 fps)=200 fps change in velocity.
Thanks for the help. I believe I’ve got it down now. Your information and a bit of help from JVN finally got me to understand the problem, and it is ridiculously simple. Now I just need to do my essay. :).
Again, thanks a lot for your time and information,
Parker
P.S.- I actually had the answer a few days ago when I had attempted it, but it felt too simple so I check the TAS forums to see many other students confused by it, hence the beginning of over thinking.
Well, seeing as this is a great place to do so, I want to make sure I did this next problem correctly for self-assurance.
Using the change (delta) in velocity that must be used to lower the perigee to a 60-mile altitude (This was your answer to the Shuttle math question for Lesson 1) and assuming the Orbiter’s OMS engines have a combined force (thrust) of 12,000 lbs and the Shuttle has a weight of 250,000 lbs (with a full cargo bay), use the equations below to compute the length (or time) of the burn necessary in minutes.
f = ma force equals mass times acceleration and t = v/a time equals velocity divided by acceleration
Your acceleration will be in G’s, where 1 G = 32 feet per second per second (this is how far an object travels due to the force of gravity in a vacuum).
Above is the second math problem. Below is the work I did myself.
f/m=a <- New formula
12,000 lbs / 250,000 lbs =a
.048 G =a
.048 G * (32 fps/s) =a <- Converting away from Gs
(1.536 fps/s) =a
Your calculation is correct, but I would be careful with the use of English units in general. I would convert everything to SI units (Newtons, meters, etc.) simply because working with English units can be a bit tricky, particularly when doing calculations with Newton’s 2nd law.
The problem I see with your calculations is that you are mixing pounds as force and pounds as mass. Pound generally means force, but there is a unit called the Avoirdupois pound that refers to mass. When you use Fnet = ma, the calculation only works if the units for mass and force match, kilograms with newtons, for example.
I know it sounds tedious, but when faced with a problem given in English units, I recommend converting your units to SI, use your equations to make your calculations, and then convert back for the answer. I tell my physics students that the extra brain wringing required to use English units in calculations ends up being much worse than doing the unit conversion twice.
When you do the conversion, and you divide the force by the mass, your answer is in units of acceleration, as you would expect. In the case of this problem, you divide and get an answer in G’s, which can be a bit awkward if you don’t know how to deal with it. You handled it without a problem though, so congrats!
short answer: Good work…you should think about limiting the number of decimal places you report in your final answer though
Thanks very much for looking over it, I’ll keep what you said in mind. I stared at Google Calculator and the question for about 15 minutes straight before I decided to go with my attempt at converting the Gs and such. It’s a very confusing concept for me, but I’m glad I got lucky. :rolleyes:
